Introductory Physics Volume Two
Introductory Physics Volume Two
Introductory Physics Volume Two
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1<br />
<strong>Introductory</strong> <strong>Physics</strong><br />
<strong>Volume</strong> <strong>Two</strong><br />
Saint Mary’s College<br />
Chris Ray<br />
Spring 2012<br />
cover art by Susana del Baño Ranner
2 Electric Field 1.0
1.1 Electric Charge 3<br />
Demo<br />
1 Electric Field<br />
§ 1.1 Electric Charge<br />
Sometimes when you pet a cat, the cat’s fur begins to stand up<br />
on end and if you then touch the cats nose, you and the cat are both<br />
shocked. Through petting, the cat has been charged with static electricity.<br />
The shock occurs when the cat discharges the built up static<br />
charge to your finger. She has not done this on purpose, it is simply<br />
the inevitable force of nature taking its course, so don’t get mad at the<br />
cat. Of course if you had not touched the cat’s nose she would have lost<br />
her charge more gradually through the air, so she might have reason to<br />
be mad at you for sucking all the charge from her nose.<br />
Let’s investigate static electricity further.<br />
Static Electricity: Rub a rubber rod with fur and hang it by a string.<br />
Rub a second rubber rod with fur and bring it near the first:<br />
Rubber Rods rubbed with fur<br />
Observe that the charge rods repel one another. Thus, for two rods<br />
of the same material rubbed in the same way we find there will be<br />
a repulsive force between them. Remove the second rubber rod, and<br />
replace it with a plexiglass rod that has been rubbed with the fur.<br />
Observe that the hanging rubber rod is attracted to the plexiglass rod.<br />
So it is possible by rubbing different objects with fur to cause either<br />
attractive or repulsive forces between the objects.<br />
The observations above can be explained by the existence of two<br />
types of “electric charge.” Objects with the same charge repel each<br />
other, while objects with different charges attract one another. The<br />
two types of charge, following the convention of Benjamin Franklin,<br />
are called positive and negative. Rubbing rubber with fur leaves the<br />
rubber with a negative charge, while rubbing plexiglass with fur leaves<br />
the plexiglass with a positive charge.
4 Electric Field 1.1<br />
Fact: <strong>Two</strong> Types of Charge<br />
From observations, such as the demonstration with the rubber rods<br />
and fur, we have the following rules:<br />
• There are two types of electric charge: + and -.<br />
• Objects with the same sign charge repel.<br />
• Object with oppositely signed charges attract<br />
⊙ Do This Now 1.1<br />
Your friend Albert claims that he has isolated a third type of electric charge<br />
in nature. He calls the three types of charge A, B and C. Explain how he<br />
would prove this to you. Hint: you might think about the rock-paper-scissors<br />
game, and make a table showing which pairs are attractive and which are<br />
repulsive.<br />
Does the fur rubbing against the rubber (plexiglass) create the<br />
negative (positive) electric charge? It turns out that the answer is<br />
no. Careful experiments show that whenever a negative electric charge<br />
shows up an equal amount of positive charge will be found somewhere<br />
else, and vice versa. The net charge in the universe remains zero. To<br />
see this for the rod-fur case, consider the following. Charge a rubber<br />
rod with fur, hang the rod and then bring the fur near the rod. The<br />
negatively charged rod will be attracted to the fur, showing that the<br />
fur has gained a positive charge. We can understand what is going<br />
on by considering the fundamental building blocks of normal matter:<br />
protons, neutrons and electrons. Protons have a positive charge +e,<br />
neutrons have no charge, and electrons have a negative charge −e.<br />
Uncharged objects contains an equal number of electrons and protons.<br />
A net electric charge is usually established on an object by adding<br />
or removing electrons. In the case of the fur and rod, electrons are<br />
removed form the fur and collected by the rod. This leaves the fur<br />
with a net positive charge (more protons than electrons) and the rod<br />
with a net negative charge (more electrons than protons).<br />
⊙ Do This Now 1.2<br />
Explain why the cat’s fur stands on end when it is pet.<br />
Fact: Electric Charge<br />
Electric Charge is neither created nor destroyed.<br />
Demo<br />
Charging Without Rubbing: Hang two small metal balls as indicated<br />
below:
1.1 Electric Charge 5<br />
Then charge a rubber rod with fur, place it in contact with the metal<br />
balls, and then remove it. The two balls will be observed to repel each<br />
other.<br />
Here is how we can explain the demonstration. When the rubber rod<br />
comes in contact with the metal, some of the charge (which is negative<br />
since it is from a rubber rod rubbed with fur) moves over to the metal.<br />
This is because metals easily accept or give up electric charge; electric<br />
charge moves easily through metals. If we indicate the charges using<br />
negative signs, then we can represent what happens with the following<br />
picture:<br />
- - -<br />
-<br />
-<br />
-<br />
- -<br />
- - - -<br />
-<br />
-<br />
-<br />
-<br />
-<br />
- -<br />
-<br />
-<br />
-<br />
-<br />
- -<br />
-<br />
-<br />
-<br />
The charges move from the rubber rod onto the metal balls because<br />
of the repulsive forces between them. Once isolated, the two metal<br />
spheres repel each other because they both have a net charge of the<br />
same sign.<br />
Example<br />
<strong>Two</strong> equal masses, each with mass m = 4g, are electrically charged<br />
and hung using string as shown. If the masses hang in equilibrium at<br />
an angle θ = 10 ◦ , what is the magnitude of the electric force that each<br />
mass exerts on the other?
6 Electric Field 1.1<br />
θ<br />
First draw a free body diagram for one of the masses:<br />
θ<br />
2<br />
F T<br />
F E<br />
mg<br />
Sum forces:<br />
∑ ⃗F = î(FT sin θ 2 − F E) + ĵ(F T cos θ 2 − mg) = 0<br />
This gives two equations:<br />
F E = F T sin θ 2<br />
F T cos θ 2 = mg<br />
Solving for the electric force F E gives<br />
F E = mg tan θ 2 = (.004kg)(9.8 m s<br />
) tan(5 ◦ ) = 3.4mN<br />
2<br />
Demo<br />
Moving Charges: Place an aluminum can on a table. Perviously we<br />
showed that we had two types or charge. If we produce either of these<br />
types of charge the can is attracted toward the charge.<br />
- - -<br />
-<br />
- - -
-<br />
-<br />
-<br />
-<br />
-<br />
1.2 Coulomb’s Law 7<br />
How can we understand this apparently new type of charge, a<br />
charge that is attracted to both positive and negative charges? The<br />
place to start is consider the aluminum of which the can is composed.<br />
Each aluminum atom is composed of an equal number of protons and<br />
electrons, so the net charge of the can is zero. But there is a difference<br />
between the protons and electrons. The aluminum atoms are locked<br />
together in a crystal lattice, with the protons locked together with the<br />
neutrons in the nucleus of the atoms. The protons cannot move. On the<br />
other hand some of the electrons are shared between all of the atoms,<br />
and move around freely. This is what makes aluminum a conductor, it<br />
has electrons that move freely through the bulk of the material.<br />
So when a positively charged rod is brought near the can the electrons<br />
are attracted toward the oppositely charged rod and the move<br />
within the metal a little bit toward the rod. This leaves the can with a<br />
net negative charge on the side toward the rod and a net positive charge<br />
on the side away from the rod. The net positive charge is because the<br />
protons did not move, so that when the electrons moved toward the<br />
other side of the can, the protons were left alone, unpaired.<br />
+<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
can alone<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
+<br />
-<br />
+<br />
-<br />
+<br />
+<br />
+<br />
-<br />
+<br />
-<br />
-<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
+<br />
-<br />
+<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
-<br />
can next to<br />
+ rod<br />
-<br />
-<br />
-<br />
-<br />
-<br />
+<br />
+<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
-<br />
+<br />
+<br />
+<br />
+<br />
+<br />
-<br />
-<br />
+ + + +<br />
-<br />
-<br />
-<br />
-<br />
+<br />
showing<br />
net charge<br />
+ + + +<br />
One might think that the net force would still be zero once the electrons<br />
have redistributed themselves, since we end up with a net positive<br />
charge on one side that is repelled and a net negative charge on the<br />
other side that is attracted. But it ends up that the force between two<br />
charges decreases with distance, so that the repulsive force is less than<br />
the attractive force.<br />
⊙ Do This Now 1.3<br />
Explain how a negatively charged rod also attracts the can.<br />
§ 1.2 Coulomb’s Law<br />
As we have seen there is a force between charged objects. After<br />
careful observation it was determined that the electrical force is
8 Electric Field 1.2<br />
similar to the gravitational force. Recall that the gravitational force<br />
between two massive objects is inversely proportional to the square of<br />
the distance between the objects and proportional to the weight of each<br />
object. The electric force between two charged objects is also inversely<br />
proportional to the square of the distance between the objects and is<br />
proportional to the charge on each object. This observational fact is<br />
referred to as Coulomb’s Law.<br />
Fact: Coulomb’s Law<br />
The magnitude of the force between charges q a and q b that are a<br />
distance r apart is given by<br />
F = q aq b 1<br />
4πɛ o r 2<br />
1<br />
where<br />
4πɛ o<br />
= 8.987552 × 10 9 N · m 2 /C 2 ≈ 9.0 × 10 9 N · m 2 /C 2 .<br />
It has already been noted that the direction of the<br />
electric force is determined by the sign of the charges:<br />
opposites charges attract, like charges repel.<br />
Fact: Coulomb’s Law in Vector Form<br />
The magnitude and direction of the electric force can be combined<br />
into one vector expression of Coulomb’s Law as follows:<br />
⃗F ab = q aq b ⃗r a − ⃗r b<br />
4πɛ 0 |⃗r a − ⃗r b | 3<br />
where F ⃗ ab is the force on charge a due to charge b and ⃗r a and ⃗r b<br />
are the position vectors of the two charges. Note that if we let<br />
⃗r = ⃗r a − ⃗r b that we can write<br />
⃗F ab = q aq b 1<br />
4πɛ 0 r 2 ˆr<br />
The vectors in the above definition are pictured below.<br />
a<br />
r a<br />
r ab<br />
r b<br />
b<br />
Both vectors are from the origin of the coordinate system, which can<br />
be chosen for computational convenience.
1.2 Coulomb’s Law 9<br />
Example<br />
Consider three charges as shown. We want to know the net force on<br />
charge a due to charges b and c.<br />
0.4<br />
y (m)<br />
4.0 mC<br />
r b<br />
2.0 nC<br />
r a<br />
0.2<br />
r c<br />
-3.0 nC<br />
0.0<br />
0.0<br />
We find from the diagram that<br />
0.2 0.4 0.6 0.8 1.0<br />
x (m)<br />
⃗r a = (0.2î + 0.4ĵ)m and q a = 4.0 × 10 −3 C<br />
⃗r b = (0.8î + 0.5ĵ)m and q b = 2.0 × 10 −9 C<br />
⃗r c = (0.7î + 0.2ĵ)m and q c = −3.0 × 10 −9 C<br />
so that we can compute<br />
⃗r a − ⃗r b = (−0.6î − 0.1ĵ)m and q a q b = 8.0 × 10 −12 C 2<br />
⃗r a − ⃗r c = (−0.5î + 0.2ĵ)m and q a q c = −12.0 × 10 −12 C 2<br />
so that<br />
⃗F a = F ⃗ ab + F ⃗ ac<br />
= q aq b<br />
4πɛ 0<br />
⃗r a − ⃗r b<br />
|⃗r a − ⃗r b | 3 + q aq c<br />
4πɛ 0<br />
−0.6î − 0.1ĵ<br />
⃗r a − ⃗r c<br />
|⃗r a − ⃗r c | 3<br />
−0.5î + 0.2ĵ<br />
= 0.072N − 0.108N<br />
| − 0.6î − 0.1ĵ|<br />
3<br />
| − 0.5î + 0.2ĵ| 3<br />
−0.6î − 0.1ĵ<br />
−0.5î + 0.2ĵ<br />
= 0.072N − 0.108N<br />
(0.6 2 + 0.1 2 )<br />
3/2<br />
(0.5 2 + 0.2 2 ) 3/2<br />
= 0.32N(−0.6î − 0.1ĵ) − 0.69N(−0.5î + 0.2ĵ)<br />
= (0.15î − 0.17ĵ)N<br />
The net force is to the right and down at about a 45 ◦ .<br />
If we reworked the previous example but with the charge q a doubled,<br />
then we would find that the force on the charge q a would also be<br />
doubled, ⃗ Fa = 2(0.15î − 0.17ĵ)N. In general the force on a charge is<br />
proportional to the charge. For example, in the previous example<br />
⃗F a = q a (37.5 î − 42.5 ĵ) N C .<br />
So we find that the ratio of the force and charge is a constant.<br />
⃗F a<br />
q a<br />
= (37.5 î − 42.5 ĵ) N C .
10 Electric Field 1.2<br />
This observation leads to the definition of the electric field, which occurs<br />
in the next section.<br />
When computing the net electric force it is sometimes easier to<br />
deal with the directions “by hand”. If you can tell the direction of the<br />
force simply by looking at the configuration, then there is no reason to<br />
use the vector form of Coulmb’s law. In such a situation just use the<br />
magnitude form F = qaq b 1<br />
4πɛ o r<br />
. The following example will demonstrate<br />
2<br />
the “by hand” method.<br />
Example<br />
There are three equal charges q at each vertex of an equilateral triangle<br />
with sides of length L. What is the force on each charge?<br />
Let us find the force on the charge at the top. We know that the forces<br />
are repulsive so that we can draw the free body diagram for the charge<br />
on the top.<br />
F 1<br />
F 1<br />
F 2<br />
F net<br />
F 2<br />
By inspection we see that the two forces are both at a 30 ◦ angle from<br />
the vertical, and that the horizontal components of the two forces are<br />
equal and opposite. So that when we add the two forces together, to<br />
get the net force, the horizontal components will cancel. The net force<br />
will be just the sum of the vertical components.<br />
⃗F net = F 1y ĵ + F 2y ĵ = F 1 cos 30 ◦ ĵ + F 2 cos 30 ◦ ĵ<br />
= q2 1<br />
4πɛ 0 L 2 cos 30◦ ĵ +<br />
q2 1<br />
4πɛ 0 L 2 cos 30◦ ĵ<br />
= 2 q2 1<br />
4πɛ 0 L 2 cos 30◦ ĵ<br />
By the symmetry of the configuration the force on the other two charges<br />
will be the same magnitude and also pointing directly away from the<br />
center of the group.<br />
⊲ Problem 1.1<br />
There are two +q charges and two −q charges on the corners of a square<br />
of size a as shown.
1.3 Electric Field 11<br />
+ -<br />
Compute the net force on each charge?<br />
⊲ Problem 1.2<br />
You have three charges as shown.<br />
-<br />
+<br />
a<br />
y (m)<br />
0.2<br />
0.1<br />
0.0<br />
- 0.1<br />
- 0.2<br />
- 0.3<br />
1.0 µC<br />
3.0 µC<br />
0.2 0.4 0.6 0.8 1.0<br />
-2.0 µC<br />
x (m)<br />
What is the net force on each charge?<br />
§ 1.3 Electric Field<br />
Definition: Electric Field<br />
If a particle with charge q a is placed at a point in space and F a is<br />
the net electric force on this particle due to all other charges, then<br />
the electric field at that point in space is the ratio of the force on<br />
the particle and the charge of the particle.<br />
⃗E = ⃗ F a<br />
q a<br />
You can think of the electric field as the force per charge in the<br />
same way that pressure is the force per area. It is important to understand<br />
the following points about the electric field:<br />
• The electric field does not depend on the test charge q a in any way.<br />
• The electric field represents the effect of all the other charges.<br />
• The electric field is different at each point in space.<br />
We will now work a specific example in order to demonstrate these<br />
three properties of the electric field. It is important to follow the details<br />
of the computation in this example very closely since there is much to<br />
learn from it. Consider the configuration of two charges shown below.
12 Electric Field 1.3<br />
0.4<br />
0.2<br />
y (m)<br />
r a<br />
r b<br />
2.0 nC<br />
r c<br />
-3.0 nC<br />
0.0<br />
0.0<br />
0.2 0.4 0.6 0.8 1.0<br />
We calculated this configuration in a previous example where the test<br />
charge q a was at the point ⃗r a = (0.2î + 0.4ĵ)m and we found that the<br />
force at this point was<br />
x (m)<br />
⃗F a = q a (37.5 î − 42.5 ĵ) N C<br />
so that the electric field at this point is<br />
F<br />
⃗E(0.2m, 0.4m) = ⃗ a<br />
= (37.5 î − 42.5 ĵ) N q a<br />
C<br />
We can find the electric field at other points ⃗r a as well.<br />
⃗E(⃗r a ) = ⃗ F a<br />
q a<br />
= ⃗ F ab<br />
q a<br />
+ ⃗ F ac<br />
q a<br />
= q b ⃗r a − ⃗r b<br />
4πɛ 0 |⃗r a − ⃗r b | 3 + ⃗r a − ⃗r c<br />
4πɛ 0 |⃗r a − ⃗r c | 3<br />
We will stop at this point to notice that the charge q a has already<br />
dropped out of the equation. So that we can write the electric field at<br />
an arbitrary location ⃗r as<br />
⃗E(⃗r) =<br />
q b ⃗r − ⃗r b<br />
4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c<br />
4πɛ 0 |⃗r − ⃗r c | 3<br />
This observation will be used after this example is finished. Let us now<br />
find the electric field at various other points. Let ⃗r = xî + yĵ so that<br />
the electric field at ⃗r is<br />
⃗E(x, y) =<br />
q b ⃗r − ⃗r b<br />
4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c<br />
4πɛ 0 |⃗r − ⃗r c | 3<br />
(x − 0.8)î + (y − 0.5)ĵ<br />
= 18 N C<br />
[(x − 0.8) 2 + (y − 0.5) 2 ] − 27 (x − 0.7)î + (y − 0.2)ĵ<br />
3/2 N<br />
C<br />
[(x − 0.7) 2 + (y − 0.2) 2 ] 3/2<br />
Evaluating this at a few locations we find the following.<br />
q c<br />
⃗E(0, 0.4) = (22î − 17ĵ) N C<br />
⃗E(0, 0.3) = (28î − 14ĵ) N C<br />
⃗E(0, 0.1) = (33î − 2ĵ) N C<br />
⃗E(0, 0.0) = (32î + 3ĵ) N C<br />
⃗E(0, 0.2) = (32î − 9ĵ) N C<br />
⃗E(0, −0.1) = (28î + 8ĵ) N C<br />
The fields at these six points are graphed as the six bold arrows on<br />
the left side of the following plot. The field is graphed at many other
1.3 Electric Field 13<br />
points a well.<br />
+2<br />
-3<br />
This graph gives some idea of how the field changes as you move around<br />
in the space. The graph is a little confusing in the region where the<br />
arrows cross each other.<br />
This happens wherever the field is strong.<br />
Because of this there is another kind of graph that is usually used to<br />
map out the electric field. Take a look at the following graph, which<br />
has both styles of maps together.
14 Electric Field 1.3<br />
+2<br />
-3<br />
The second style has curved field lines that follow the direction of the<br />
electric field. Check to see that the arrows are parallel to the field lines<br />
at the base of the arrow. This type of graph clearly shows the direction<br />
of the field. You can get an idea of the magnitude of the field by how<br />
close the field lines are to each other: the field is strong where the field<br />
lines are close together. Here is a graph showing just the field lines.
1.3 Electric Field 15<br />
Look back now to the middle of the example, where we observed<br />
that in the calculation of the electric field the test charge q a had<br />
dropped out of the equation. We observed, at that time, that the<br />
electric field at any location ⃗r in space could be written out as<br />
⃗E(⃗r) =<br />
q b ⃗r − ⃗r b<br />
4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c<br />
4πɛ 0 |⃗r − ⃗r c | 3<br />
⃗r−⃗r s<br />
|⃗r−⃗r s| 3<br />
We see the contribution of each source charge is of the form<br />
qs<br />
4πɛ 0<br />
and that the net field is the sum of the contributions of each source<br />
charge. This is so because the net force is the vector sum of the individual<br />
force. Following this idea we would arrive at these two theorems.<br />
Theorem: Electric Field due to a Point Charge<br />
The electric field at the location ⃗r due to a point charge q s at the<br />
location ⃗r s is given by the following expression.<br />
⃗E(⃗r) =<br />
q s ⃗r − ⃗r s<br />
4πɛ 0 |⃗r − ⃗r s | 3<br />
The magnitude of the electric field due to a point charge at a<br />
distance r from the point charge is<br />
E(r) =<br />
q 1<br />
4πɛ 0 r 2<br />
Theorem: Superposition Theorem<br />
The electric field due to a collection of point charges is the sum of<br />
the fields of the individual charges.<br />
⃗E(⃗r) = ∑ q n ⃗r − ⃗r n<br />
4πɛ<br />
n 0 |⃗r − ⃗r n | 3<br />
⊲ Problem 1.3<br />
A charge q 1 = +q is placed at the location ⃗r 1 = 0î + aĵ and a second<br />
charge q 2 = −q is placed at ⃗r 2 = 0î − aĵ.<br />
(a) Write out the electric field at an arbitrary location on the x-axis:<br />
⃗r = xî + 0ĵ.<br />
(b) Sketch a map of the electric field lines, in the first quadrant. Here<br />
is a table that gives the angle ( from the positive x-axis) of the electric<br />
field at different locations in the first quadrant.
16 Electric Field 1.5<br />
x/a = 0.0 0.5 1.0 1.5 2.0 2.5 3.0<br />
y/a = 0.0 -90 -90 -90 -90 -90 -90 -90<br />
0.5 -90 -54 -48 -52 -57 -61 -65<br />
1.0 45 -3 -11 -20 -29 -36 -42<br />
1.5 90 42 19 5 -7 -16 -23<br />
2.0 90 61 39 23 11 1 -8<br />
2.5 90 69 51 36 24 13 5<br />
3.0 90 74 58 45 33 24 15<br />
§ 1.4 Vector and Scalar Fields<br />
Definition: Scalar Field<br />
A scalar field is an entity that has a magnitude at each point in<br />
space.<br />
The potential energy is a scalar field, the quantity U = mgy gives<br />
the amount of potential energy at each point (⃗r = xî+yĵ) in space.<br />
Definition: Vector Field<br />
A vector field is an entity that has a magnitude and direction at<br />
each point in space.<br />
Gravity is a vector field, the constant g = 9.8 m s<br />
is the magnitude<br />
2<br />
of the field near the surface of the earth, and the direction is toward the<br />
earth. If you move away from the surface of the earth the gravitational<br />
field decreases in strength. The gravitational field is similar to the<br />
electric field in the sense that they both give the force on a body. The<br />
gravitational field gives the force on a massive body.<br />
⃗F G = m⃗g<br />
While, the electric field gives the force on a charged body.<br />
⃗F E = q ⃗ E<br />
Definition: Uniform Field<br />
A uniform vector field is a vector field that has the same magnitude<br />
and direction at all points in space. A uniform scalar field is a<br />
scalar field that has the same value at all points in space.
1.5 Continuous Charge Distributions 17<br />
§ 1.5 Continuous Charge Distributions<br />
In principle one could compute the electric field in any situation<br />
by summing the fields of each proton and electron in the system. In<br />
practice this is not done because there are too many electrons and<br />
protons. In this section we will see how to compute the electric field<br />
when there are too many charges to count.<br />
Suppose that we have a string that has been rubbed on a cat until<br />
the string has built up a charge Q that is spread uniformly over the<br />
length of the string. We then take the string and stretch it out in a<br />
straight line. We wish to calculate the electric field due to the string.<br />
For convenience assume that the string is stretched out along the x-axis<br />
from x = −L/2 to x = L/2.<br />
One way of finding the electric field is to conceptually break the<br />
string into short little sections, say N of them. Each of the sections<br />
will be a length dx = L/N, and carry a charge dq = Q/N. We can<br />
number the sections from 1 to N, and then the n’th section will be<br />
at the position ⃗r n = x n î (x n = nL/N − L/2). As long as we break<br />
the string into enough sections so that dx is small, we can treat each<br />
section as a point charge and then we can compute the electric field as<br />
a sum of N point charges<br />
⃗E(⃗r) = ∑ q n ⃗r − ⃗r n<br />
4πɛ<br />
n 0 |⃗r − ⃗r n | 3 = ∑ dq ⃗r − ⃗r n<br />
4πɛ<br />
n 0 |⃗r − ⃗r n | 3<br />
Keep in mind when you look at this formula that the vector ⃗r n points<br />
to the charge q n . With this formula and the aid of a computer it<br />
is possible to compute the electric field for nearly any distribution of<br />
charge that you can imagine.<br />
It is also possible, in some cases, to find a closed form solution,<br />
without using a computer. If we take the limit as N goes to infinity,<br />
the sum becomes an integral and the electric field can be written in the<br />
form<br />
∫<br />
dq ⃗r − ⃗r s<br />
⃗E(⃗r) =<br />
4πɛ 0 |⃗r − ⃗r s | 3<br />
Here again we imagine that we break the object into many small pieces<br />
of charge dq, the vector ⃗r s points toward dq, ranging over all dq’s and
18 Electric Field 1.5<br />
that we “sum” over all the pieces.<br />
Let us evaluate this integral for the case of the string that we<br />
introduced earlier. First we need to relate dq to dx. The entire length<br />
L of the string has a charge Q so that the charge per length is λ = Q/L.<br />
If the charge is uniformly spread over the string we expect the charge<br />
per length to be the same everywhere so that<br />
dq<br />
dx = Q = λ −→ dq = λdx<br />
L<br />
With this, and the fact that ⃗r s = x s î, we can write<br />
∫<br />
dq ⃗r − ⃗r s<br />
⃗E(⃗r) =<br />
4πɛ 0 |⃗r − ⃗r s | 3<br />
∫ L/2<br />
λdx s ⃗r − x s î<br />
=<br />
−L/2 4πɛ 0 |⃗r − x s î| 3<br />
If we write the field point as ⃗r = xî + yĵ, and then do the integration,<br />
we find that<br />
⎡<br />
⎤<br />
⃗E(x, y) =<br />
λ ⎣<br />
1<br />
1<br />
− √<br />
⎦ î<br />
4πɛ 0<br />
√y 2 + (x − L 2 )2 y 2 + (x + L 2 )2<br />
⎡<br />
⎤<br />
+ λ 1 x −<br />
⎣−<br />
L 2<br />
x + L 2<br />
+ √<br />
⎦ ĵ<br />
4πɛ 0 y<br />
√y 2 + (x − L 2 )2 y 2 + (x + L 2 )2<br />
⊲ Problem 1.4<br />
Suppose that you have a circular hoop of radius R with a net charge<br />
Q spread uniform around the hoop. Compute the electric field at a<br />
distance z along the axis of the hoop?
1.6 Gauss’s Law 19<br />
§ 1.6 Gauss’s Law<br />
We have seen that the electric field due to any charge distribution<br />
can be computed. From this relationship between the electric field and<br />
the charge distribution, another one can be derived.<br />
First we will define a new quantity, the electric flux through a<br />
surface. To get an idea of what the flux represents first consider the<br />
following story. You wish to catch some butterflies but you don’t have<br />
time to chase them around, so you just set up your butterfly net on<br />
a pole. The butterflies are migrating south for the winter. They are<br />
flying by your house heading in a southerly direction, so you orient the<br />
net so that it faces north. The number of butterflies you catch should<br />
be proportional to both the density of butterflies in the air and the<br />
area of the mouth of the net. The number of butterflies caught will<br />
also depend on the orientation of the net relative to the direction the<br />
butterflies are moving. For example, if the butterflies end up flying to<br />
the south-west instead of directly south, you will not catch as many<br />
since the net was not facing the optimal direction.<br />
The electric flux is similar, it is the amount of electric field that<br />
“passes through” a surface. There are three things that determine the<br />
quantity of electric flux: the area of the surface, the magnitude of the<br />
electric field and the orientation of the field relative to the surface.<br />
To write this out clearly, we need to have a way to mathematically<br />
represent the orientation of a surface. We will define a vector area ⃗ A<br />
as the vector that has a magnitude equal to the area of the surface and<br />
has a direction that is normal to the surface. A good picture to keep<br />
in mind is a thumbtack,<br />
A<br />
A<br />
the nail part of the tack is the vector area and the flat part of the<br />
tack is the surface. This ends up being the best way to use a vector to<br />
represent a surface. Now we can define the electric flux.
20 Electric Field 1.6<br />
Definition: Electric Flux<br />
The electric flux (φ e ) through a surface is<br />
φ e = ⃗ E · ⃗A<br />
if the field is uniform over the surface. If the field is not uniform<br />
then one must integrate over the surface,<br />
∫<br />
φ e = ⃗E · dA ⃗<br />
We see that the dot product represents the dependence of the flux<br />
on the relative orientation of the surface and the field. When thinking<br />
of the flux integral over a surface it can be helpful to imagine gluing<br />
tacks to the surface with the nail part sticking out, so that you end up<br />
with a spiky covering of the surface. Each tack represents one small<br />
surface element dA ⃗ and the integral is the sum of the flux’s through all<br />
of the small surface element.<br />
Now we can state the theorem that relates the electric field to the<br />
charge density.<br />
Theorem: Gauss’s Law<br />
The electric flux out of any closed surface is proportional to the<br />
total charge enclosed within the surface.<br />
∮<br />
⃗E · dA ⃗ = Q in<br />
ɛ 0<br />
Note that the integral of the electric field over the surface does<br />
not depend on the charge density outside the surface in any way. For<br />
example if there is no charge inside the surface then the integral must<br />
be zero.<br />
Example<br />
We can use Gauss’s law in order to find the electric field strength.<br />
Here is an example of how this can be done. The result of this example<br />
is also generally useful.
1.6 Gauss’s Law 21<br />
Suppose that you have a block of material which has charges inside<br />
that move freely. We will show later that in side the block the electric<br />
field is zero and that outside but near the block the field is normal to<br />
the surface. Also we will see later that there is no net charge inside the<br />
block, but that on the surface there can be a surface charge density.<br />
Let us form a Gaussian surface in the shape of a small tin can that is<br />
half inside and half outside the material, with the can oriented so that<br />
the ends of the can are parallel to the surface of the material.<br />
Since the field is zero inside the material we know that the flux is zero<br />
through the half of the can that is inside the material. We also know<br />
that the flux is zero through the sides of the can because the field is<br />
normal to the surface of the material. So we see that the only flux<br />
through the surface of the can is the flux through the top of the can.<br />
Assuming that the can is small enough so that the field is uniform over<br />
the can, then the flux through the top is<br />
∫<br />
⃗E · dA ⃗ = E ⃗ · ⃗A = EA<br />
top<br />
where A is the cross sectional area of the can. We have now computed<br />
the left hand side of Gauss’s law. The right hand side is the charge<br />
inside the gaussian surface. Since the charge is only on the surface we<br />
can find the charge inside the can as the area of the surface that is<br />
inside the can A times the surface charge density σ: so Q in = Aσ. And<br />
∮<br />
∫<br />
bottom<br />
⃗E · ⃗ dA = Q in<br />
ɛ 0<br />
∫<br />
∫<br />
⃗E · dA ⃗ + ⃗E · dA ⃗ + ⃗E · dA ⃗ = Q in<br />
sides<br />
top ɛ 0<br />
−→ 0 + 0 + EA = Aσ −→ E = σ ɛ 0 ɛ 0<br />
In the previous example, the determination of the flux was greatly
22 Electric Field 1.6<br />
simplified because we picked a the gaussian surface so that the electric<br />
field was either normal to the surface or in the plane of the surface.<br />
For example, the field was in the plane of the sides of the can in the<br />
previous example, thus the flux was zero through the sides. Further,<br />
in the previous example, the electric field was normal to the part of<br />
the gaussian surface that did not have a zero flux, in addition the field<br />
strength was uniform over this part of the surface. Thus the first thing<br />
to do, when you are using Gauss’s law to find the electric field strength,<br />
is to choose the gaussian surface so that the field is either parallel or<br />
normal to all parts of the surface.<br />
⊲ Problem 1.5<br />
Four closed surfaces are near three charges as shown.<br />
S 3<br />
S 4<br />
-Q<br />
+Q<br />
S 2<br />
-2Q<br />
S 1<br />
Find the electric flux through each surface.<br />
⊲ Problem 1.6<br />
A charge q is placed at one corner of a cube. What is the electric flux<br />
through each face of the cube.<br />
⊲ Problem 1.7<br />
Use Gauss’s law to show that the field strength at a distance r from a<br />
q 1<br />
point charge is<br />
4πɛ 0 r<br />
. 2<br />
⊲ Problem 1.8<br />
A solid sphere of radius R has a total charge of Q uniformly distributed<br />
throughout its volume. Calculate the magnitude of the electric field at<br />
a distance r from the center of the sphere. Be sure to consider the cases<br />
of r < R and r > R separately.<br />
⊲ Problem 1.9<br />
A sphere of radius a carries a volume charge density ρ = ρ 0 (r/a) 2 .<br />
Find the electric field inside and outside the sphere.<br />
⊲ Problem 1.10<br />
Use Gauss’s law to show that the electric field near a line charge with<br />
uniform charge density λ is given by E =<br />
λ<br />
2πɛ 0r<br />
where r is the distance<br />
from the line.
1.7 More Examples 23<br />
⊲ Problem 1.11<br />
Consider a long cylindrical charge distribution of radius R with a uniform<br />
charge density ρ. Find the electric field at a distance r from the<br />
axis for r < R.<br />
⊲ Problem 1.12<br />
A spherical shell of radius R carries a net charge of Q uniformly distributed<br />
over it’s surface.<br />
(a) Find the electric field strength at a point inside the shell.<br />
(b) Find the electric field strength at a point outside the shell.<br />
§ 1.7 More Examples<br />
Example<br />
Return to the pithballs in the example in section 1: assume the length<br />
of each string is 10cm. If the masses have the same net electric charge,<br />
what is the magnitude of the charge on each mass? Can you determine<br />
the sign of the net charge on each ball?<br />
Use the following diagram to determine the distances between the<br />
masses:<br />
L<br />
θ/2<br />
L = 10cm<br />
d<br />
d = 2L sin θ 2 = 2(10cm) sin 5◦ = 1.7cm.<br />
Assume each mass has a charge q, then using Coulomb’s law:<br />
F E = 1 q 2<br />
4πɛ 0 d 2 −→ q2 = F Ed 2<br />
1<br />
4πɛ 0<br />
.<br />
√<br />
(0.0034N)(.017m)<br />
−→ q =<br />
2<br />
9 × 10 9 N · m 2 /C 2 = 1.0 × 10−8 C<br />
The sign of the charge cannot be determined; all that can be said is<br />
the net charge on the masses have the same sign.<br />
Example<br />
Three particles with electric charge are attached to a meter stick, as<br />
shown. The value of Q is 1 × 10 −6 C (= 1µC). (a) What is the electric<br />
force on the center charge? (b) To what position could the center charge<br />
be moved so that the net electric force on it is zero?
24 Electric Field 1.7<br />
0.5m<br />
+3Q<br />
+Q +2Q<br />
(a) The electric forces on the center charge due to the other two charges<br />
are:<br />
The net force is<br />
F<br />
2Q<br />
Q<br />
F<br />
3Q<br />
F net = F 3Q − F 2Q<br />
= 1<br />
4πɛ 0<br />
=<br />
Q(3Q)<br />
(0.5m) 2 − 1<br />
4πɛ 0<br />
Q(2Q)<br />
(0.5m) 2<br />
(<br />
)<br />
( )<br />
9 × 10 9 N·m2<br />
3 − 2<br />
(1 × 10 −6 C) 2<br />
C 2 (0.5m) 2 = +3.6mN<br />
(b) There are three regions to consider: in between the 2Q and 3Q<br />
charges, outside of these charges to the left and outside of these charges<br />
to the right. Here are force diagrams for placing the charge Q outside<br />
of the two larger charges:<br />
F<br />
3Q<br />
F<br />
3Q<br />
F<br />
2Q<br />
+Q<br />
+3Q<br />
+2Q<br />
+Q<br />
F<br />
2Q<br />
Since the forces point in the same direction, there is no location outside<br />
of the two larger charges where the net electric force on the charge Q<br />
can be zero. For positions inside the two larger charges, the forces will<br />
point in opposite directions, as seen in part (a), and will cancel at the<br />
position where the forces have equal magnitudes. Let’s measure the<br />
location from the+3Q charge:<br />
F<br />
2Q<br />
F 3Q<br />
+3Q<br />
+2Q<br />
The net force on the charge +Q at a position x is<br />
F net = F 3Q − F 2Q<br />
= 1 Q(3Q)<br />
4πɛ 0 x 2 − 1 Q(2Q)<br />
4πɛ 0 (1 − x) 2<br />
x
1.7 More Examples 25<br />
The net force is to be zero:<br />
1 Q(3Q)<br />
4πɛ 0 x 2 − 1 Q(2Q)<br />
4πɛ 0 (1 − x) 2 = 0<br />
−→ 3(1 − x) 2 − 2x 2 = 0<br />
Solving the resulting quadratic equation yields two answers: x = 5.45m<br />
and x = 0.55m. The latter value is the answer, since we already argued<br />
that the position must lie between the two larger charges.<br />
⊙ Do This Now 1.4<br />
Analyze the configuration in the previous example, except replace the +3Q<br />
charge with a −3Q charge.<br />
(a) −0.18N (b)4.45m to the right of the +2Q charge.<br />
Example<br />
Four charges are located at the corners of a square as shown in the<br />
diagram below. A fifth charge is located at the center of the square.<br />
For the charge values indicated compute the net electric force on the<br />
charge at the center of the square.<br />
25cm<br />
+4μC<br />
+1μC<br />
+4μC<br />
Draw a force diagram:<br />
+4μC<br />
-1μC<br />
+2μC<br />
+1μC<br />
F +2<br />
45 o<br />
y<br />
x<br />
F -1<br />
F +1<br />
45 o<br />
F +4<br />
-1μC<br />
+2μC<br />
The distances between each corner charge and the center charge are<br />
the same:<br />
√ (a ) 2 ( a<br />
) √<br />
2 2<br />
d = + =<br />
2 2 2 a,
26 Electric Field 1.7<br />
where a = 25cm. To find the net force, add all the forces using vector<br />
addition:<br />
⃗F net = îF x + ĵF y ,<br />
where<br />
F x = (F +4 − F +2 − F +1 − F −1 ) sin(45 ◦ )<br />
F y = (−F +4 + F +2 − F +1 − F −1 ) cos(45 ◦ )<br />
Use Coulomb’s law to compute the magnitude of each force:<br />
F +4 = (9 × 109 )(4 × 10 −6 )(4 × 10 −6 )<br />
( √2 ) 2<br />
= 4.6N<br />
2 (.25)<br />
F +1 = F −1 = (9 × 109 )(1 × 10 −6 )(4 × 10 −6 )<br />
( √2<br />
2 (.25) ) 2<br />
= 1.15N<br />
F +2 = (9 × 109 )(2 × 10 −6 )(4 × 10 −6 )<br />
( √2<br />
2 (.25) ) 2<br />
= 2.3N<br />
Finally, compute the components of the net force:<br />
F x = (4.6N − 2.3N − 1.15N − 1.15N) sin(45 ◦ ) = 0<br />
F y = (−4.6N + 2.3N − 1.15N − 1.15N) cos(45 ◦ ) = −3.25N<br />
So, the net force points straight down with a magnitude of 3.25N.<br />
Example<br />
Consider the configuration of three charges along a line as shown below.<br />
Assuming the only forces acting are mutual electric forces between the<br />
charges, show that the configuration is in equilibrium, meaning that<br />
the net force on each charge is zero.<br />
a<br />
a<br />
+4q -q<br />
+4q<br />
Here is a force diagram for all the charges:<br />
+4q -q<br />
+4q<br />
The net force on the center charge must be zero, since the two forces<br />
are in opposite directions and caused by equal charges that are the<br />
same distance away. One down, two to go. Use the force law to write
1.7 More Examples 27<br />
out the net force on the leftmost charge:<br />
F L = 1 (4q)(q)<br />
4πɛ 0 a 2 − 1 (4q)(4q)<br />
4πɛ 0 (2a) 2<br />
= 1 (4q)(q)<br />
4πɛ 0 a 2 − 1 (4q)(q)<br />
4πɛ 0 a 2<br />
= 0<br />
So the net force on the leftmost charge is zero. Since the rightmost<br />
charge is the mirror image of the leftmost charge, it also has no net<br />
force acting on it. Thus, all three charges have a net force of zero acting<br />
on them for the configuration given, and the system is in equilibrium.<br />
Is the equilibrium stable?<br />
By stability we mean the following. If we move one of the charges<br />
a little bit, does the configuration return to the original equilibrium<br />
configuration, or does the configuration fall apart? Let’s move one of<br />
the charges a little bit and see what happens. Move the left most charge<br />
a small distance ɛ to the left. If the configuration is stable, then the<br />
net force on it must push it to the right. Check:<br />
a+ε<br />
a<br />
The net force is now:<br />
+4q -q<br />
+4q<br />
ε<br />
F L ′ = 1 (4q)(q)<br />
4πɛ 0 (a + ɛ) 2 − 1 (4q)(4q)<br />
4πɛ 0 (2a + ɛ) 2<br />
If this force points to the right, it will push the charge back from<br />
where it came and the equilibrium can be stable. If the force is to the<br />
left, the charge will be pushed away from the equilibrium position and<br />
the equilibrium can not be stable. With a little algebraic muscle the<br />
formula above for F L ′ can be rewritten:<br />
(<br />
)<br />
F L ′ =<br />
q2 1<br />
πɛ 0 a 2 (1 + ɛ − 1<br />
a )2 (1 + ɛ<br />
2a )2<br />
It is easy to see that this is a negative number since the denominator<br />
in the subtracted term is smallest, no matter how small ɛ is, as long<br />
as it is larger than zero. Thus, the force on the leftmost charge points<br />
to the left and the charge is pushed away from the original equilibrium<br />
position. The equilibrium is unstable.<br />
⊙ Do This Now 1.5
28 Electric Field 1.7<br />
Determine what the other two charges in the previous example do when the<br />
leftmost charge is moved to the left.<br />
They accelerate towards each other<br />
⊙ Do This Now 1.6<br />
An electron accelerates east due to an electric field. What direction does the<br />
electric field point?<br />
west<br />
Example<br />
A dust particle with mass 2µg has a net electric charge 3µC. The<br />
piece of dust is in a region of uniform electric field and is observed to<br />
accelerate at a rate of 180 m s 2 . What is the magnitude of the electric<br />
field in that region of space?<br />
The net force on the dust particle can be computed using Newton’s<br />
second law:<br />
F net = (2 × 10 −9 kg)(180 m s 2 ) = 3.6 × 10 −7 N.<br />
Using the definition of electric field and assuming no other forces act<br />
on the dust particle:<br />
E = F q = 3.6 × 10−7 N<br />
3 × 10 −6 C = 0.12 N C .<br />
Example<br />
If a proton is placed in the same electric field from the previous example,<br />
what is the resulting acceleration?<br />
Work backwards:<br />
F = qE = (1.6 × 10 −19 C)(0.12N/C) = 1.9 × 10 −20 N<br />
a = F m =<br />
1.9 × 10−20 N<br />
1.67 × 10 −27 kg = 1.15 × 107 m s 2<br />
Example<br />
Three charges lie along a line as shown below. What is the electric<br />
field at a position that is a horizontal distance a to the right of the −q<br />
charge?<br />
-4q +2q -q<br />
2a<br />
a<br />
Here is a diagram indicating the electric field due to each charge at the<br />
point:
1.7 More Examples 29<br />
-4q +2q -q<br />
E -1<br />
E -4<br />
E +2<br />
2a<br />
a<br />
a<br />
In the diagram, the electric field due to the −4q charge has been labeled<br />
E −4 , etc. The electric field at the point indicated is just the<br />
superposition of the individual electric fields:<br />
E = E +2 − E −1 − E −4 = 1 (2q)<br />
4πɛ 0 (2a) 2 − 1 q<br />
4πɛ 0 a 2 − 1 (4q)<br />
4πɛ 0 (4a) 2<br />
−→ E = − 3 q<br />
16πɛ 0 a 2 .<br />
Example<br />
Using the configuration of charges in the previous example, determine<br />
the electric field a vertical distance a above the +2q charge.<br />
First, let’s add a coordinate system, since this problem will involve<br />
vectors in two dimensions:<br />
y<br />
E +2<br />
(0, a)<br />
E -4<br />
E -1<br />
-4q θ 1<br />
+2q θ 2 -q<br />
x<br />
(-2a, 0) (0, 0) (a, 0)<br />
The angles θ 1 and θ 2 can be computed using some geometry:<br />
cos θ 1 = √ 2 sin θ 1 = √ 1<br />
5 5<br />
cos θ 2 = √ 1 sin θ 2 = √ 1<br />
2 2<br />
To compute the electric field at the point (0, a), we add the individual<br />
electric fields as vectors:
30 Electric Field 1.7<br />
E +2<br />
θ 1<br />
θ 2<br />
E -4<br />
E -1<br />
⃗E = î(−E −4 cos θ 1 + E −1 cos θ 2 )<br />
+ ĵ(−E −4 sin θ 1 − E −1 sin θ 2 + E +2 )<br />
Use Coulomb’s law to compute the magnitudes of the electric field<br />
contribution from each charge:<br />
E −4 = 1 (4q)<br />
4πɛ 0 ( √ 5a) = 1 4q<br />
2 4πɛ 0 5a 2<br />
E +2 = 1 (2q)<br />
4πɛ 0 a 2 = 1 2q<br />
4πɛ 0 a 2<br />
Finally,<br />
(<br />
⃗E = î − 1<br />
+ ĵ<br />
E −1 = 1<br />
4πɛ 0<br />
4πɛ 0<br />
(<br />
− 1<br />
4πɛ 0<br />
(q)<br />
( √ 2a) = 1<br />
2 4πɛ 0<br />
q 2 √5<br />
5a 2 + 1<br />
4πɛ 0<br />
4q 1 √5<br />
5a 2 − 1<br />
4πɛ 0<br />
q<br />
2a 2<br />
)<br />
q 1 √2<br />
2a 2<br />
q 1 √2<br />
2a 2 + 1 )<br />
2q<br />
4πɛ 0 a 2<br />
−→ E ⃗ = 1 q<br />
(−î(0.36) + ĵ(1.29))<br />
4πɛ 0 a2 Flux example:<br />
Example<br />
A uniform electric field passes through a 1cm×1cm square in the xy<br />
plane. The electric field is ⃗ E = (5 N C )î + (3 N C )ˆk. What is the electric<br />
flux through the square?<br />
z<br />
y<br />
x<br />
Since the electric field is constant: ∫<br />
φ E = ⃗E · dA ⃗ = E ⃗ · ⃗A.
1.7 More Examples 31<br />
The area vector points in the ˆk direction, so<br />
(<br />
)<br />
φ E = (5 N C )î + (3 N C )ˆk · (0.01m) 2ˆk = 0.03<br />
N·m 2<br />
C<br />
Gauss’s Law Examples:<br />
Example<br />
A 1.5 × 10 −9 C point charge is located at the center of a cylinder. The<br />
electric flux through the sides of the cylinder is known to be 100 N·m2<br />
C .<br />
What is the electric flux through one of the endcaps?<br />
By Gauss’s law:<br />
∫<br />
sides<br />
−→<br />
⃗E · ⃗A +<br />
100 N·m2<br />
C<br />
∫<br />
caps<br />
∫<br />
∮<br />
caps<br />
+ ∫<br />
caps<br />
⃗E · ⃗A = Q in<br />
⃗E · ⃗A =<br />
⃗E · ⃗A = 169 N·m2<br />
C<br />
ɛ 0<br />
⃗E · ⃗A = 169 N·m2<br />
C<br />
1.5 × 10−9 C<br />
8.85 × 10 −12 C 2<br />
N·m 2<br />
N·m2<br />
− 100<br />
C<br />
= 69 N·m2<br />
C<br />
This is the flux through both caps. Since the charge is located symmetrically<br />
with respect to the two endcaps, the flux through one cap is<br />
just half this: 34.5 N·m2<br />
C .<br />
Example<br />
A large, thin rectangular plate has a uniform charge density σ distributed<br />
over its area:<br />
σ<br />
Assuming that the plate is so large that the electric field above and<br />
below the plate is uniform, compute the magnitude of the electric field.
32 Electric Field 1.8<br />
With the assumptions above, symmetry tells us that the electric field<br />
will point in a direction perpendicular to the surface; for a positive<br />
charge it will point away from the plate:<br />
E<br />
A<br />
σ<br />
E<br />
In the figure a Gaussian surface is indicated. Take it to be a small<br />
cylinder with cross-sectional area A. The electric flux in non-zero only<br />
through the endcaps:<br />
∮<br />
∫<br />
∫<br />
∫<br />
⃗E · dA ⃗ = E · dA ⃗ + E · dA ⃗ + E · dA<br />
⃗<br />
upper cap<br />
sides<br />
= EA + 0 + EA = 2EA<br />
Using Gauss’s law:<br />
2EA = σA<br />
ɛ 0<br />
−→ E =<br />
σ<br />
2ɛ 0<br />
.<br />
lower cap<br />
§ 1.8 Homework<br />
⊲ Problem 1.13<br />
y<br />
7.0 µC<br />
x<br />
0.50 m<br />
0.50 m<br />
2.0 µC<br />
0.50 m<br />
−4.0 µC<br />
Three point charges are located at the corners of an equilateral triangle<br />
as shown. Calculate the net electric force on the 7.0µC charge.
1.8 Homework 33<br />
⊲ Problem 1.14<br />
Richard Feynman said that if two people were at arm’s length from<br />
each other and each person had 1% more electrons than protons, the<br />
force of repulsion between them would be about equal to the “weight”<br />
of the entire Earth. Was Feynman exaggerating?<br />
⊲ Problem 1.15<br />
A point charge q is at the position (x 0 , y 0 ). Find the electric field at<br />
an arbitrary point (x, y) due to the charge q.<br />
⊲ Problem 1.16<br />
<strong>Two</strong> 2.0 µC charges are located on the x axis at x = 1.0m and at<br />
x = −1.0m.<br />
(a) Determine the electric field on the y axis at y = 0.5.m.<br />
(b) Calculate the electric force on a −3.0µC charged place at this point.<br />
⊲ Problem 1.17<br />
A uniformly charged insulating rod of length 14 cm is bent into the<br />
shape of a semicircle. If the rod has a total charge of −7.5µC, find<br />
the magnitude and direction of the electric field at the center of the<br />
semicircle.<br />
⊲ Problem 1.18<br />
An electron moves at a speed of 3 × 10 6 m s<br />
into a uniform electric field<br />
of magnitude 1000 N/C. The field is parallel to the electron’s velocity<br />
and acts to decelerate the electron. How far does the electron travel<br />
before it is brought to rest?<br />
⊲ Problem 1.19<br />
A proton moves at 4.5 × 10 5 m s<br />
in the horizontal direction. It enters a<br />
uniform electric field of 9.6 × 10 3 N/C directed downward. Ignore any<br />
gravitational effects and find the time it takes the proton to travel 5.0<br />
cm horizontally, its vertical displacement after it has traveled 5.0 cm<br />
horizontally, and the horizontal and vertical components of its velocity<br />
after it has traveled 5.0 cm horizontally.<br />
⊲ Problem 1.20<br />
y<br />
x<br />
θ<br />
E
34 Electric Field 1.8<br />
A charged ball of mass 1.00 g is suspended on a light string in the<br />
presence of a uniform electric field as shown. When ⃗ E = (3.00î +<br />
5.00ĵ) × 10 5 N/C, the ball is in equilibrium at θ = 37.0 ◦ .<br />
(a) Find the charge on the ball.<br />
(b) Find the tension in the string.<br />
⊲ Problem 1.21<br />
A triangular box is in a horizontal electric field of magnitude E =<br />
7.8 × 10 4 N/C as shown<br />
30 cm<br />
10 cm 60°<br />
E<br />
(a) Compute the electric flux through each face of the box.<br />
(b) Compute the net electric flux through the entire surface of the box.<br />
⊲ Problem 1.22<br />
The nose cone of a rocket is in a uniform electric field of magnitude E 0<br />
as shown.<br />
E<br />
E<br />
It is parabolic in cross section, of length d and of radius r. Compute<br />
the electric flux through the paraboloidal surface.<br />
⊲ Problem 1.23<br />
A charge Q is at the center of a cube of side L.<br />
(a) Find the flux through each face of the cube.<br />
(b) Find the flux through the entire surface of the cube.<br />
(c) Would these answers change if the charge was not at the center?<br />
⊲ Problem 1.24<br />
A charge Q is located just above the center of the flat face of a solid<br />
hemisphere of radius R.<br />
(a) What is the electric flux through the curved surface.<br />
(b) What is the electric flux through the flat surface.<br />
⊲ Problem 1.25<br />
Find the electric field at the origin if there is a charged rod that goes<br />
from x = a to x = b assuming that the charge density of the rod is<br />
λ = cx n .<br />
d
1.9 Summary 35<br />
§ 1.9 Summary<br />
Definitions<br />
Electric Field:<br />
Electric Flux:<br />
⃗E = ⃗ F a<br />
q a<br />
dφ e = E ⃗ · dA ⃗<br />
∫<br />
φ e = ⃗E · dA ⃗<br />
Facts<br />
Coulomb’s Law<br />
⃗F ab = 1 ˆr ab<br />
q a q b<br />
4πɛ 0 rab<br />
2 = 1 ⃗r ab<br />
q a q b<br />
4πɛ 0 rab<br />
3<br />
Theorems<br />
Electric Field of a Point Charge:<br />
⃗E(⃗r) = 1 q ⃗r − ⃗r s<br />
4πɛ 0 |⃗r − ⃗r s | 3<br />
where ⃗r points to the field point and ⃗r s points to the charge q.<br />
Electric Field of a Distribution of Charge:<br />
⃗E(⃗r) = 1 ∫<br />
dq ⃗r − ⃗r s<br />
4πɛ 0 |⃗r − ⃗r s | 3<br />
where ⃗r points to the field point and ⃗r s points to the charge element<br />
dq.<br />
Gauss’ Law: ∮<br />
⃗E · dA ⃗ = Q in<br />
ɛ 0<br />
Superposition Theorem: If a charge distribution ρ 1 produces an<br />
electric field ⃗ E 1 and the charge distribution ρ 2 produces an electric<br />
field ⃗ E 2 then if both charge distributions are present the electric field<br />
is ⃗ E = ⃗ E 1 + ⃗ E 2 .
36 Electric Potential 2.9
2.1 Electric Potential 37<br />
2 Electric Potential<br />
§ 2.1 Electric Potential<br />
The Coulomb force is a conservative force and thus we can consider<br />
the potential energy of a particle in an electric field that is created by<br />
a static charge distribution. Suppose that you have a charged particle<br />
in an electric field, and that you move the particle from point A to<br />
point B. As the particle is moved, the electric field will do work on the<br />
particle.<br />
W A→B =<br />
=<br />
= q<br />
∫ B<br />
A<br />
∫ B<br />
A<br />
∫ B<br />
A<br />
⃗F E · ⃗dr<br />
q ⃗ E · ⃗dr<br />
⃗E · ⃗dr<br />
Thus when the particle is moved from A to B the change in the electric<br />
potential energy is<br />
∫ B<br />
∆U = −W A→B = −q ⃗E · ⃗dr<br />
A<br />
and the change in potential energy per charge is<br />
∫ B<br />
∆U<br />
= − ⃗E ·<br />
q<br />
⃗dr.<br />
A<br />
Notice that the potential energy per charge does not depend on the<br />
test charge q, it only depends on the electric field.<br />
Definition: Electric Potential<br />
The electric potential, V , is the electric potential energy per charge.<br />
The electric potential is related to the electric field:<br />
∫ B<br />
V B − V A = ∆V = ∆U = −<br />
q<br />
A<br />
For a uniform electric field this simplifies to<br />
∆V = ∆U = −E q<br />
⃗ · ∆⃗r<br />
⃗E · ⃗dr
38 Electric Potential 2.1<br />
It is important to notice that the definition of electric potential<br />
only tells us how to find the difference (V B − V A ) between the electric<br />
potential at two different points, A and B . The value of the electric<br />
potential itself is not defined. This is not a problem because only the<br />
difference has physical significance. You may recall that this is true<br />
for potential energy also. The potential energy at ten meters above<br />
the surface of the earth is not defined, but the difference between the<br />
potential energy at ten meters and three meters is defined, ∆U =<br />
mg∆y. One can choose some point in space that is defined to be<br />
the zero of the electric potential. This point is sometimes called the<br />
ground. Then in reference to ground, all points have an absolute electric<br />
potential. This is like choosing the surface of the earth as the zero of<br />
gravitational potential energy.<br />
Often the electric potential is simply referred to as the potential.<br />
The units of electric potential is evidently the units of energy divided<br />
by the units of charge. This unit occurs frequently and has it’s own<br />
name, the Volt.<br />
1Volt =<br />
1Joule<br />
1Coulomb<br />
This unit is abbreviated as V. Be warned that this is a bad coincidence<br />
since the symbol used for the electric potential is V , which is very<br />
similar to V. This is not a serious problem, but if you mix up the<br />
symbols, you can end up doing silly things with the algebra.<br />
Example<br />
Suppose that there is a uniform electric field ⃗ E = (0.2 N C<br />
)î. The electric<br />
potential difference between the two points ⃗r A = (1.0m)î+(2.0m)ĵ and<br />
⃗r B = (5.0m)î + (−3.0m)ĵ is computed as follows.<br />
V B − V A = ∆V = − ⃗ E · ∆⃗r<br />
= −(0.2 N C<br />
)î · [⃗r B − ⃗r A ]<br />
= −(0.2 N C<br />
)î · [(4.0m)î + (−5.0m)ĵ]<br />
= −(0.2 N C<br />
)(4.0m) + 0 = −0.8V<br />
This example demonstrates that the electric potential decreases as<br />
one moves in the direction of the electric field (positive x in this case).<br />
This is a general rule: the electric field points from regions of high<br />
electric potential to regions of low electric potential.
2.1 Electric Potential 39<br />
Example<br />
Suppose that an electron moved from a region where the electric potential<br />
is 150 volts to a region where the electric potential is 100 volts.<br />
There is only the electric field acting on the electron. What would be<br />
the change in the kinetic energy of the electron? We know that the<br />
potential energy and electric potential are related: ∆V = ∆U/q.<br />
so that<br />
∆K + ∆U = 0<br />
∆K = −∆U = −q∆V = −(−e)(100V − 150V) = −8.0 × 10 −18 J<br />
In this example we see that negatively charged particles slow down<br />
when they go from a region of high electric potential to a region of low<br />
electric potential. A positively charged particles would speed up.<br />
Example<br />
Now let’s try an example where the field is not uniform.<br />
⃗E = α [ y 2 î + 2xyĵ ]<br />
If we move from the origin to the point (a, b) what is the change in the<br />
electric potential. First we need to pick a path from the starting point<br />
to the ending point. A straight line will do. Let ⃗r(t) = x(t)î + y(t)ĵ =<br />
atî + btĵ, we see that as the parameter t is varied from 0 to 1, that<br />
the vector ⃗r(t) points along the path from the origin to the final point<br />
(a, b): that is ⃗r(t) is the trajectory of a particle following our path. This<br />
function is called a parameterization of the path. Now we can compute<br />
the line integral ∫ E ⃗ · dr ⃗ since we can now write<br />
⃗dr = d⃗r<br />
dt dt = [ dx<br />
dt î + dy<br />
dt ĵ<br />
]<br />
dt = [aî + bĵ] dt
40 Electric Potential 2.1<br />
so that<br />
∫ B<br />
∆V = −<br />
= −<br />
= −<br />
= −α<br />
= −α<br />
= −α<br />
A<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
∫ 1<br />
0<br />
= −αab 2<br />
⃗E · ⃗dr<br />
⃗E(⃗r(t)) · d⃗r<br />
dt dt<br />
α [ y 2 î + 2xyĵ ] · [aî + bĵ] dt<br />
[ (bt)2î + 2(at)(bt)ĵ ] · [aî + bĵ] dt<br />
[<br />
(bt) 2 a + 2(at)(bt)b ] dt<br />
3ab 2 t 2 dt<br />
In general, to compute a line integral, one must first find a parameterization<br />
of the path.<br />
The following theorem gives a way of finding the electric field if<br />
the electric potential field is already know.<br />
Theorem: Electric Field from the Electric Potential<br />
[<br />
⃗E = −∇V ⃗ ∂V<br />
= −<br />
∂x î + ∂V<br />
∂y ĵ + ∂V ]<br />
∂z ˆk<br />
The symbol ⃗ ∇ represents the gradient operator ∂<br />
Thus the expression ⃗ ∇f is equal to ∂f<br />
∂x î + ∂f<br />
∂y ĵ + ∂f<br />
∂z ˆk.<br />
∂xî + ∂<br />
∂y ĵ + ∂ ∂z ˆk.<br />
Example<br />
The nonuniform electric field, ⃗ E = αy 2 î + 2αxyĵ, that we used in the<br />
previous example, has the electric potential V = −αxy 2 . Let us check<br />
the above theorem.<br />
⃗E = − ⃗ ∇V<br />
= ∂<br />
∂x αxy2 î + ∂ ∂y αxy2 ĵ + ∂ ∂z αxy2ˆk<br />
= αy 2 î + 2αxy ĵ + 0 ˆk OK
2.2 Equipotentials 41<br />
⊲ Problem 2.1<br />
You touch the terminals of a 9 Volt battery to your tongue. While it is<br />
touching your tongue a charge of 0.08 Coulombs passes through your<br />
tongue from one terminal to the other. How much energy is dissipated<br />
in your tongue, by the charged particles passing through it?<br />
⊲ Problem 2.2<br />
<strong>Two</strong> metal plates are held at an electric potential difference of 1000V.<br />
An electron is released from the plate that is at a lower electric potential.<br />
What is the speed of the electron when it reaches the other<br />
plate?<br />
⊲ Problem 2.3<br />
The electric field in a region is given by ⃗ E = ayî + axĵ, where a =<br />
2.0V/m 2 is a constant. What is the electric potential difference between<br />
the origin and the point x = 1.0m, y = 2.0m?<br />
⊲ Problem 2.4<br />
The electric potential in a region is given by V = axyz. What is the<br />
electric field?<br />
§ 2.2 Equipotentials<br />
Suppose that we have picked our zero for electric potential. Each<br />
point in space will now have it’s own value for electric potential. If we<br />
put a dot at every location that is at an electric potential of, say 4.3V,<br />
then the collection of all such dots will create a surface.<br />
Definition: Equipotential<br />
An equipotential is a surface on which the electric potential is constant.<br />
Below is a graph of a few equipotentials for the electric field produced<br />
by three point charges (front left is q = +1, back center is q = +2,<br />
and front right is q = −3).<br />
These 3D plots are difficult to generate and read, so often a section<br />
is cut through the surfaces, and only the cut edge of the surfaces are
42 Electric Potential 2.2<br />
shown. The following graph is a section through the equipotentials for<br />
the same three charges (there are a few more equipotentials drawn in<br />
this section graph).<br />
+2<br />
+1<br />
-3<br />
Below is a graph of the same equipotentials with a few electric field<br />
lines graphed also.<br />
+2<br />
+1<br />
-3<br />
Notice that while the field lines go from positive charges to negative<br />
charges, the equipotentials encircle charges. Also notice that wherever<br />
a field line crosses an equipoential, the field line is perpendicular to<br />
the equipotential. This must happen, as we will now show. Imagine<br />
that you move a charge a small distance ⃗ dr along the surface of an<br />
equipotential. The change in electric potential must be zero since the<br />
beginning and ending points are both on the same equipotential. But<br />
we also know that the change in electric potential is given by,<br />
dV = − ⃗ E · ⃗dr<br />
So since dV = 0, it must be the case that E ⃗ · ⃗dr = 0, and<br />
⃗E · ⃗dr = 0 −→ E ⃗ is perpendicular to dr ⃗<br />
since ⃗ E ≠ 0 and ⃗ dr ≠ 0.
2.3 Conductors in Equilibrium 43<br />
⊲ Problem 2.5<br />
(a) Sketch the equipotentials for two like charges.<br />
(b) Sketch the equipotentials for two opposite charges.<br />
§ 2.3 Conductors in Equilibrium<br />
A conductor allows it’s free charges to move through it with some<br />
amount of ease. So, if there is a force acting on a free charge then that<br />
charge will move. Thus if there is an electric field inside a conductor<br />
then the free charges in the conductor will move. In some situations,<br />
for example if the conductor is electrically insulated, the charges will<br />
eventually reach an equilibrium configuration in which they no longer<br />
move around. This makes sense, since if you apply a field to an isolated<br />
conductor the charges will move around to adjust to the new force but in<br />
the end they must settle down into the equilibrium configuration, which<br />
is the state with the lowest energy. But this tells us something about<br />
the electric field inside a conductor once it has reached equilibrium.<br />
Since the charges are not moving, there can be no electric field inside<br />
the conductor.<br />
Theorem: Conductor in Equilibrium: Field<br />
The electric field is zero inside a conductor at equilibrium.<br />
This in turn tells us that there can be no net charge inside the<br />
volume of a conductor if the conductor is in equilibrium. We can argue<br />
this as follows. Suppose that there was a net charge in some volume<br />
inside a conductor. By Gauss’s Law we know that there must be a<br />
net electric flux through the surface of this volume, but the flux is the<br />
integral of the electric field over the surface, so the electric field cannot<br />
be zero if there is a region with a net charge. But by the previous<br />
theorem we know that if there is an electric field inside a conductor then<br />
the conductor is not in equilibrium. Thus there can be a net charge<br />
inside a conductor only if the conductor is not at equilibrium.<br />
Theorem: Conductor in Equilibrium: Charge<br />
There is no net charge in any volume inside a conductor in equilibrium.<br />
If a charge is placed on a conductor it will reside on the<br />
surface of the conductor.<br />
We can arrive at one more very important result by using the fact<br />
that there is no field in a conductor in equilibrium. Since the field is<br />
zero, this means that ∆V = − ∫ ⃗ E · ⃗ dr = 0 for any path inside the<br />
conductor. This tells us that a conductor in equilibrium is all at the<br />
same electric potential.
44 Electric Potential 2.4<br />
Theorem: Conductor in Equilibrium: Potential<br />
All points in a conductor at equilibrium have the same electric<br />
potential.<br />
Theorem: Conductor in Equilibrium: Surface Field<br />
The field at the outside surface of a conductor in equilibrium is<br />
normal to the surface of the conductor. The magnitude of the<br />
field is E = σ ɛ 0<br />
where σ is the surface charge density.<br />
We can see that the previous theorem must be true by considering<br />
that the surface of the conductor is an equipotential, and noting that<br />
the electric field is normal to any equipotential surface.<br />
⊲ Problem 2.6<br />
There is a solid conductor with a cavity within it. Floating within this<br />
cavity there is a second conductor. This has been drawn below with a<br />
quarter of the outer conductor removed so that you can see the inner<br />
conductor.<br />
A total charge Q a is place on the inner conductor and a charge Q b is<br />
placed on the outer conductor.<br />
(a) What is the amount of charge on the inside surface of the outer<br />
conductor?<br />
(b) What is the amount of charge on the outer surface of the outer<br />
conductor?<br />
⊲ Problem 2.7<br />
A conducting spherical shell having an inner radius of 4.0 cm and outer<br />
radius of 5.0 cm carries a net charge of +10µC. If a +2.0µC point charge<br />
is placed a the center of this shell, determine the surface charge density<br />
on the inner and outer surfaces.
2.4 Capacitors 45<br />
§ 2.4 Capacitors<br />
Suppose that we place two conductors near each other. Now suppose<br />
that we remove a quantity of charge Q from one conductor and<br />
place it on the other. One conductor will end up with a charge +Q and<br />
the other will end up with a charge −Q. In addition an electric field<br />
will be created between the conductors.<br />
Since there is an electric field between the conductors, there will also<br />
be an electric potential difference between the two conductors.<br />
∆V = V + − V − = −<br />
∫ ⃗r+<br />
⃗r −<br />
⃗ E · ⃗ dr<br />
Because the electric field strength is proportional to the charge Q, the<br />
electric potential difference will also be proportional to Q.<br />
∆V ∝ Q<br />
Example<br />
Here is a specific example of this general result. Place two conducting<br />
plates parallel to each other as shown, and charge the top plate to a<br />
net charge Q and the other plate to a net charge −Q<br />
In a previous example it was shown that the electric field strength near<br />
a plate with uniform charge density σ is E = σ/2ɛ 0 . Between the<br />
plates the fields of the two plates are in the same direction (toward<br />
the negatively charged plate) so that the strength of the net field is<br />
twice the field strength of each plate alone. So the net field between<br />
the plates is E = σ/ɛ 0 . If the area of the plate is A then the charge
46 Electric Potential 2.4<br />
density is σ = Q/A so that we find,<br />
E =<br />
Q<br />
Aɛ 0<br />
As long as the separation between the plates d is much less than the<br />
width of the plates the electric field will be effectively uniform between<br />
the plates, so that the potential difference between the plates is<br />
∆V = V + − V − = − ⃗ E · ∆⃗r<br />
Since ∆⃗r points from the negative plate to the positive plate and E ⃗<br />
points the opposite direction, the dot product of these two vectors is<br />
negative the product of the magnitudes; E ⃗ · ∆⃗r = −Ed.<br />
∆V = −(−Ed) =<br />
Q d =<br />
d Q<br />
Aɛ 0 Aɛ 0<br />
We see then that the electric potential difference is proportional to the<br />
charge on the plates.<br />
Since the electric potential difference is proportional to the charge,<br />
the ratio is a constant.<br />
Q<br />
∆V<br />
Definition: Capacitance<br />
The capacitance, C, of two conductors is the ratio of the charge on<br />
the conductors to the electric potential difference.<br />
C =<br />
Q or Q = C ∆V<br />
∆V<br />
The unit of capacitance is a Farad and is equal to a Coulomb per<br />
Volt: F = C V .<br />
⊲ Problem 2.8<br />
A 6.0µF capacitor is connected to a 1.5 Volt battery so that a electric<br />
potential difference of 1.5 volts is maintained between the two conductors<br />
of the capacitor. What is the charge on each of the conductors?<br />
⊲ Problem 2.9<br />
Show that the capacitance of two parallel plates is C = ɛ 0 A/d, where<br />
d is the distance between the plates and A is the area of the plates.<br />
⊲ Problem 2.10<br />
Show that the capacitance of two concentric spherical conducting shells<br />
4πɛ<br />
is 0<br />
1/a−1/b<br />
, where a and b are the radii of the shells.<br />
⊲ Problem 2.11<br />
Coaxial cables are commonly used to carry high frequency signals. Your<br />
cable for your cable-TV is a coaxial cable. A coaxial cable consists of
2.5 Energy in an Electric Field 47<br />
a wire (radius a) surrounded by a cylindrical conducting shield (radius<br />
b) Show that the capacitance for a length L of coaxial cable is 2πɛ0L<br />
ln(b/a) .<br />
⊲ Problem 2.12<br />
Consider two long, parallel, and oppositely charged wires of radius a<br />
with their centers separated by a distance b. Assuming the charge<br />
is distributed uniformly on the surface of each wire, show that the<br />
capacitance per unit length of this pair of wires is<br />
C<br />
l = πɛ o<br />
ln( b a − 1)<br />
§ 2.5 Energy in an Electric Field<br />
Let us consider how much work must be done to charge a capacitor.<br />
Suppose that we have already moved an amount of charge q from the<br />
negative plate to the positive plate of the capacitor, so that the electric<br />
potential difference between the plates is ∆V = q/C. In order to move<br />
a little bit more charge dq to the positive plate, we need to do an<br />
amount of work<br />
dW = dU = dq ∆V = dq q C<br />
The total amount of work to bring the capacitor from a charge of q = 0<br />
to a charge of q = Q is<br />
U =<br />
∫ Q<br />
0<br />
dq q C = Q2<br />
2C<br />
=<br />
(C∆V )2<br />
2C<br />
= 1 2C(∆V )2<br />
Theorem: Energy Stored in a Capacitor<br />
A capacitor charged to an electric potential difference of V C stores<br />
an amount of energy<br />
U = 1 2 CV 2 C<br />
This energy is stored in the electric field that has been created<br />
between the plates. The energy density (energy per volume) is<br />
u = energy 1<br />
1 ɛ<br />
volume = 2C(∆V )2<br />
0A<br />
2 d<br />
=<br />
(Ed)2 = 1<br />
V<br />
A d<br />
2 ɛ 0E 2<br />
In the intermediate step in the above equation, the properties of a parallel<br />
plate capacitor was used, but the final result is true in general.<br />
Theorem: Energy Density of an Electric Field<br />
The electric field contains an amount of energy per volume u.<br />
u = 1 2 ɛ 0E 2
48 Electric Potential 2.6<br />
⊲ Problem 2.13<br />
A 120µF capacitor is charged to an electric potential difference of 100V.<br />
(a) How much energy is stored in the capacitor?<br />
(b) If the field strength in a capacitor becomes to great then the charge<br />
will jump across the gap between the plates. Assume that this break<br />
down occurs when the field strength reaches 3 × 10 6 V m. What is the<br />
minimum volume that is required for a 120µF capacitor to be able to<br />
hold an electric potential difference of 100V?<br />
§ 2.6 Electric Potential of a Point Charge<br />
We wish to find the electric potential for a point charge. The<br />
electric field strength around a point charge is<br />
E =<br />
q 1<br />
4πɛ 0 r 2<br />
where r is the distance from the point charge. The electric field is<br />
either pointed away or toward the charge depending on if the charge is<br />
positive or negative. Let us start with the positive charge.<br />
V b − V a = −<br />
∫ rb<br />
r a<br />
⃗ E · ⃗ dr<br />
If we let r b > r a then dr ⃗ is pointed outward. This is the same direction<br />
as E ⃗ so that E ⃗ · ⃗dr = E dr. Thus<br />
∫ rb<br />
V b − V a = − E dr = − q ∫ rb<br />
1<br />
r a<br />
4πɛ 0 r a<br />
r 2 dr = − q [− 1 ] rb<br />
4πɛ 0 r<br />
r a<br />
= q 1<br />
−<br />
q 1<br />
4πɛ 0 r b 4πɛ 0 r a<br />
Thus we see that the electric potential at a distance r from a point<br />
charge is<br />
V (r) =<br />
q 1<br />
4πɛ 0 r + constant<br />
For simplicity one normally chooses for the constant to be zero. Note<br />
that this choice for the constant implies that the zero of the electric<br />
potential is at the position r = ∞.<br />
Theorem: Electric Potential of a Point Charge<br />
V (r) =<br />
q 1<br />
4πɛ 0 r<br />
⊲ Problem 2.14<br />
Show that V (r) =<br />
q 1<br />
4πɛ 0 r<br />
is correct for a negative charge as well.
2.7 More Examples 49<br />
⊲ Problem 2.15<br />
An electron is released from a distance of 2.0 cm from a proton. How<br />
fast will the electron be going when it is 0.5 cm from the proton?<br />
(Assume that the proton does not move.)<br />
§ 2.7 More Examples<br />
Example<br />
An electric field is given by E ⃗ = (3<br />
C·m<br />
)x 2 î). What is the electric<br />
2<br />
potential difference between the points (1m, 0, 0) and (3m, 0, 0)? What<br />
is the minimum work needed to move a +6µC charge between the two<br />
points, starting from (1m, 0, 0)?<br />
Since the electric potential is path independent (it comes from a conservative<br />
force), we can integrate along any path we want; choose the<br />
x-axis:<br />
∫<br />
∫3m<br />
( )<br />
∆V = − ⃗E · (îdx) = − 3 N<br />
x 2 dx<br />
C·m 2<br />
( ) [<br />
−→ ∆V = − 3 N x 3 ] 3<br />
= −26 N·m<br />
C·m 2 3<br />
= −26V<br />
1<br />
C<br />
N<br />
1m<br />
The work done by an external agent will be<br />
W = −W E = q∆V = (6 × 10 −6 )(−26V) = −1.56 × 10 −4 J<br />
Example<br />
The figure below is a schematic representation of an electron “gun.”<br />
A potential difference is maintained between the left and right plates,<br />
with the right plate having a higher potential. By heating the left<br />
plate, an electron is “boiled” off the plate. The electron, starting from<br />
rest, then moves toward the right plate, directed toward a small hole in<br />
the plate so that it “shoots” out of the gun. If the potential difference<br />
between the plates is 9V, how fast will the electron be moving when it<br />
reaches the right plate?<br />
- +<br />
s = ?<br />
ΔV
50 Electric Potential 2.7<br />
The change in potential energy of the electron will be:<br />
Use conservation of energy:<br />
∆U = q∆V = (−e)∆V.<br />
∆U + ∆K = −e∆V + ( 1 2 m ev 2 − 0) = 0<br />
√ √<br />
2e∆V 2(1.6 × 10<br />
−→ v = =<br />
−19 C(+9V)<br />
m e 9.11 × 10 −31 kg<br />
= 1.78 × 10 6 m s<br />
Example<br />
Near the surface of the Earth, there is always a background electric<br />
field that averages 100N/C and points down. Assuming the Earth to be<br />
a perfect conductor, how much electric charge is stored on the Earth’s<br />
surface?<br />
Since the electric field points down, we know the Earth’s charge is negative.<br />
So, lets find the magnitude. Use the previous result to determine<br />
the surface charge density:<br />
E = σ ɛ 0<br />
−→ σ = Eɛ 0 .<br />
Multiply this by the Earth’s surface area to get the total charge:<br />
Q E = σ(4πR 2 E) = Eɛ 0 (4πR 2 E) = 453, 000C<br />
Example<br />
<strong>Two</strong> equal and opposite charges are located a distance a apart. Find<br />
the electric field and the electric potential at a) the point directly between<br />
the two charges and b) the point a distance a/2 directly above<br />
the point inbetween the two charges.<br />
+q -q<br />
The electric fields due to each charge point in the same direction at the<br />
center point c:<br />
a<br />
E +<br />
c<br />
+q -q<br />
E -
2.7 More Examples 51<br />
The resulting electric field at c is<br />
E c = E + + E −<br />
= 1 q<br />
4πɛ 0 ( a + 1<br />
2 )2 4πɛ 0<br />
q<br />
( a 2 )2<br />
= 2q<br />
πɛ 0 a 2<br />
To find the potential we will have to come up with a rule for what to<br />
do with more than a single point charge. As just illustrated, for the<br />
electric field due to a system of charges, the resulting field at any point<br />
is just the superposition of the electric fields due to all charges at that<br />
point:<br />
⃗E = ⃗ E 1 + ⃗ E 2 + ⃗ E 3 + · · ·<br />
This followed because the electric field came from a net force. Likewise,<br />
because the electric potential is defined using the electric field, at any<br />
point in space we will just add up the potentials due to all the individual<br />
charges to find the resulting potential:<br />
For our problem:<br />
V = V 1 + V 2 + V 3 + · · ·<br />
V c = V + + V −<br />
= 1 q<br />
4πɛ 0 ( a 2 ) + 1 (−q)<br />
4πɛ 0 ( a 2 )<br />
= 0<br />
Now repeat the analysis for the second point, labeled d:<br />
E +<br />
d<br />
45 o<br />
45 o<br />
y<br />
a/2<br />
E -<br />
x<br />
+q<br />
a/2<br />
a/2 -q<br />
The electric field is<br />
⃗E d = î(E + + E − ) cos 45 ◦ + ĵ(E + − E − ) sin 45 ◦<br />
Compute the magnitudes of the individual fields:<br />
E + = E − = 1 q<br />
4πɛ 0<br />
=<br />
q<br />
2πɛ 0 a 2<br />
( √ 2<br />
2 a)2
52 Electric Potential 2.7<br />
And,<br />
The potential is:<br />
⃗E d = î(2E + ) + ĵ(0)<br />
√ √<br />
q 2 2q<br />
E d = 2<br />
2πɛ 0 a 2 2 = 2πɛ 0 a 2<br />
V d = V + + V − = 1<br />
4πɛ 0<br />
(<br />
q<br />
( √ 2<br />
2<br />
)<br />
(−q)<br />
+<br />
a) ( √ 2<br />
2 a)<br />
= 0<br />
The potential is zero at both points! The electric field is not zero, and<br />
it points in the x direction at both points. You should be able to argue<br />
for any point lying on the line defined by the points c and d that the<br />
electric field always points in the x direction and the potential is always<br />
zero.<br />
⊙ Do This Now 2.1<br />
For the configuration in the previous example, compute the electric potential<br />
at a point that is distance a above the +q charge.<br />
q<br />
8πɛ0 a(2 −<br />
√<br />
2)<br />
Example<br />
A charge Q is distributed uniformly along a line L that lies along the<br />
x-axis. Compute the electric potential a distance x from the left end<br />
of the charge distribution that is on the x-axis. Take the potential at<br />
∞ to be zero.<br />
Locate the origin at the point we want to compute the electric potential<br />
(labeled P ):<br />
x<br />
P<br />
x'<br />
dx'<br />
Consider the very small section of length dx ′ of the line that is at x ′ .<br />
The amount of charge contained in this small section is<br />
dq ′ = Q L dx′ .<br />
Since the section is very small, we can treat it as a point charge and<br />
the potential it causes at P is<br />
dV ′ = 1 Q<br />
L dx′<br />
4πɛ 0 x ′ ,<br />
where we have used the potential for a point charge, which takes the<br />
potential at ∞ to be zero. To find the total electric potential at P we
2.8 Homework 53<br />
add up potential due to each small section making up the entire line:<br />
∫<br />
x+L ∫ Q<br />
V (x) = dV ′ 1<br />
L<br />
=<br />
dx′<br />
4πɛ 0 x ′<br />
x<br />
= Q ( ) x + L<br />
4πɛ 0 L ln x<br />
Let’s check that this gives the expected result for L → 0, which will<br />
result in Q being a point charge. We need the limit:<br />
( ) (<br />
x + L<br />
lim ln = lim ln 1 + L )<br />
≈ L<br />
L→0 x L→0 x x<br />
So<br />
lim V (x) = Q<br />
L→0 4πɛ 0 L · L<br />
x = Q<br />
4πɛ 0 x ,<br />
which is indeed the electric potential a distance x from a point charge<br />
Q.<br />
§ 2.8 Homework<br />
⊲ Problem 2.16<br />
Through what potential difference would an electron need to be accelerated<br />
for it to achieve a speed of 40% of the speed of light, starting<br />
from rest?<br />
⊲ Problem 2.17<br />
How much work is required to move one mole of electrons from a region<br />
where the electric potential is 9V to a region where the electric potential<br />
is -5V?<br />
⊲ Problem 2.18<br />
An electron moving along the x axis has an initial speed of 2.7×10 6 m s<br />
at<br />
the origin. Its speed is reduced to 1.4 × 10 5 m s<br />
at the point x = 2.0cm.<br />
Calculate the potential difference between the origin and this point.<br />
Which point is at the higher potential?<br />
⊲ Problem 2.19<br />
In Rutherford’s experiments alpha particles (charge +2e, mass 6.6 ×<br />
10 −27 kg) were fired at a gold nucleus (charge +79e). An alpha particle<br />
initially very far from the gold nucleus is fired at 2.0 × 10 7 m s<br />
directly<br />
toward the center of the nucleus. How close does the alpha particle get<br />
to this center before turning around?<br />
⊲ Problem 2.20<br />
Show that the amount of work required to assemble four identical<br />
charges Q at the corners of a square of side s is 5.41kQ 2 /s.
54 Electric Potential 2.8<br />
⊲ Problem 2.21<br />
In a certain region of space the electric potential is V = 5x−3x 2 y+2yz 2 .<br />
Find the expressions for the x, y, and z components of the electric field<br />
in this region. What is the magnitude of the field at the point P , which<br />
has coordinates (1, 0, −2)m?<br />
⊲ Problem 2.22<br />
A rod of length L lies along the x axis with its left end at the origin<br />
and has a nonuniform charge density λ = αx. What are the units of<br />
α? Calculate the electric potential at the point x = −b on the x axis.<br />
⊲ Problem 2.23<br />
A wire that has a uniform linear charge density λ is bent into the shape<br />
shown below. Find the electric potential at the center of the circular<br />
part.<br />
2R<br />
R<br />
2R<br />
⊲ Problem 2.24<br />
There is a point charge near by. You determine that the electric potential<br />
at your location (due to the point charge) is 3000V and the electric<br />
field strength is 500 V m.<br />
(a) How far away is the charge?<br />
(b) What is the value of the charge?<br />
⊲ Problem 2.25<br />
Three charges are placed on the x-axis. <strong>Two</strong> charges Q at x = d and<br />
x = −d, and a third charge −2Q at x = 0.<br />
(a) Show that the electric potential along the x-axis is V =<br />
2kQd2<br />
x(x 2 −d 2 ) .<br />
(b) Show that the electric field along the x-axis is E ⃗ = 2kQd2 (3x 2 −d 2 )<br />
x 2 (x 2 −d 2 )<br />
î. 2<br />
(c) In the limit that x ≫ d, show that the field is proportional to 1/x 4<br />
and that the potential is proportional to 1/x 3 .<br />
⊲ Problem 2.26<br />
You have drop of conductive fluid with a net charge of Q 0 and a radius<br />
r 0 . The electric field and electric potential at the surface of this drop<br />
(due to the charge on the drop) are E 0 and V 0 . <strong>Two</strong> such drops join<br />
together to form a larger drop.<br />
(a) What is the radius of the larger drop?<br />
(b) What is the surface charge density of the larger drop?<br />
(c) What is the electric field at the surface of the larger drop?<br />
(d) What is the electric potential at the surface of the larger drop?
2.8 Homework 55<br />
⊲ Problem 2.27<br />
How much work is required to bring a total charge Q to a spherical<br />
shell of radius R?<br />
⊲ Problem 2.28<br />
You have two concentric spherical shells of radius a and b. The smaller<br />
shell has a charge of q 1 = 10nC and the larger shell has a charge of<br />
q 2 = −15nC.<br />
a<br />
b<br />
(a) Find the electric field at all points in space.<br />
(b) Find the electric potential at all points in space.<br />
(c) Sketch the electric potential as a function of the distance from the<br />
center of the shells, for a = 0.15m and b = 0.30m.<br />
⊲ Problem 2.29<br />
<strong>Two</strong> conductors having net charges of +10.0µC and −10.0µC have a<br />
potential difference of 10.0V. Determine the capacitance of the system<br />
and the potential difference between the two conductors if the charges<br />
on each are increased to +100.0µC and −100.0µC.<br />
⊲ Problem 2.30<br />
<strong>Two</strong> conductors insulated from each other are charged by transferring<br />
electrons from one conductor to the other. After 1.6 × 10 12 electrons<br />
have been transferred, the potential difference between the conductors<br />
is 14V. What is the capacitance of the system?<br />
⊲ Problem 2.31<br />
Einstein showed that energy is associated with mass according to the<br />
famous relationship, E = mc 2 . Estimate the radius of an electron,<br />
assuming that its charge is distributed uniformly over the surface of a<br />
sphere of radius R and that the mass-energy of the electron is equal to<br />
the total energy stored in the resulting nonzero electric field between<br />
R and infinity.
56 Electric Potential 2.9<br />
§ 2.9 Summary<br />
Definitions<br />
Electric Potential:<br />
Capacitance:<br />
∆V = ∆U<br />
q<br />
C =<br />
Q<br />
∆V<br />
Equipotential: an equipotential is a surface over which the electric<br />
potential is a constant.<br />
Theorems<br />
dV = − ⃗ E · ⃗dr<br />
∆V = −<br />
∫ rf<br />
r i<br />
⃗ E · ⃗ dr<br />
[ ∂V<br />
⃗E = −<br />
∂x î + ∂V<br />
∂y ĵ + ∂V ]<br />
∂z ˆk<br />
≡ − ⃗ ∇V<br />
Conductors in Equilibrium:<br />
• Field - the field is zero in the volume of a conductor, and at the<br />
surface it is normal to the surface.<br />
• Charge - the net charge in the volume of a conductor is zero.<br />
• Potential - the potential in the volume of a conductor is constant.<br />
Electric Potential of a Point Charge:<br />
V = 1 q<br />
4πɛ o r<br />
Electric Potential of a Distributed Charge:<br />
V (⃗r) = 1 ∫<br />
dq<br />
4πɛ o |⃗r − ⃗r s |<br />
Energy Density of an Electric Field<br />
E = 1 2 ɛ oE 2<br />
Energy Stored on a Capacitor:<br />
U = 1 2 CV 2 C
3.2 Electric Current 57<br />
3 Circuits<br />
§ 3.1 Introduction<br />
When you turn on a flashlight, a flow of electrons is passed through<br />
the light bulb. Inside the light bulb is a small filament. As the electrons<br />
pass through the filament they loose energy and in the process the<br />
filament heats up. The filament gets so hot that it glows, this is how<br />
the bulb produces light.<br />
The electrons are supplied from the negative terminal of the battery<br />
and flow to the light bulb through a metal wire. After the electrons<br />
pass through the filament they are carried back to the positive terminal<br />
of the battery by another wire.<br />
A system of electrical devices connected with wires, such as the<br />
flashlight system, is called an electric circuit. Circuits are often represented<br />
in a wiring diagram. The following circuit diagram represents<br />
the flashlight system.<br />
+<br />
–<br />
In this diagram there are three elements and the wires that connect<br />
them. The element on the left is the battery, with the positive terminal<br />
being wider and marked with a + sign. The element on the top is the<br />
switch that turns the flashlight off and on. The element on the right is<br />
the light bulb.<br />
This chapter will be an investigation of electrical circuits and the<br />
common electrical devices from which circuits are built.<br />
§ 3.2 Electric Current<br />
A flow of charge, such as the charge flowing through the light bulb<br />
filament, is called an electric current.
58 Circuits 3.3<br />
Definition: Current<br />
The electric current, I, is defined to be the amount of charge that<br />
flows per time.<br />
I = dq<br />
dt<br />
The unit of current is coulombs per second. This<br />
combination of units is called the Ampere or Amp,<br />
abbreviated as just A: (1A = 1C<br />
1s ).<br />
Definition: Current Density<br />
Let A be a small area that is normal<br />
to the flow of current in a particular<br />
region. Let I be the current flowing<br />
through the area A. The current density,<br />
⃗J, is a vector in the direction of the flow<br />
of current. The magnitude of the current<br />
density is equal to the current per area.<br />
J = I A<br />
⊲ Problem 3.1<br />
A light bulb draws a current of 1.0 mA from a battery.<br />
(a) How many electrons pass through the bulb in 160 seconds?<br />
(b) The filament has a radius of 0.20 mm. What is the current density<br />
through the filament?<br />
§ 3.3 Ohm’s Law<br />
When an electric field is applied to a conductor the free charges in<br />
the conductor begin to move. If the conductor is isolated the charges<br />
will distribute themselves so that the field inside the conductor is zero<br />
and then the charges will cease to move. One the other hand, if the<br />
conductor is not isolated but part of a circuit, as the filament was in<br />
the flashlight circuit, then a continual flow of charge can be sustained,<br />
and the electric field in the conductor will not be zero.
3.3 Ohm’s Law 59<br />
Some materials, such a copper, allow charge to pass through it<br />
with very little resistance, and therefore it takes very little electric field<br />
to sustain a large current. Other materials, such a carbon, resist the<br />
flow of charge, and it requires more electric field to sustain the same<br />
current. Carbon is more resistive. This property is quantified in the<br />
following law.<br />
Fact: Ohm’s Law: Resistivity<br />
For many materials the current density and the electric field are<br />
proportional.<br />
⃗J = σ ⃗ E<br />
The constant σ is called the conductivity and depends on the material.<br />
The inverse of the conductivity is called the resistivity:<br />
ρ = 1/σ.<br />
Not all materials follow this relationship: those that do are called<br />
ohmic materials, those that do not are called non-ohmic.<br />
Here is a table with the resistivity of a few materials.<br />
Material Resistivity(×10 −8 Ω · m)<br />
Silver 1.6<br />
Copper 1.7<br />
Aluminum 2.8<br />
Tungsten 5.5<br />
Iron 10<br />
Nichrome 100<br />
Carbon 3500<br />
Silcon 64000000000<br />
Wood 1000000000000000<br />
Amber 50000000000000000000000<br />
Imagine a piece of conductive material, with two terminals connected<br />
to it, and with a current I passing into the material through<br />
terminal a and draining out through terminal b. Such a circuit element<br />
is called a Resistor.<br />
In order to sustain this current there will need to be an electric field<br />
in the conductor going from terminal a to terminal b. Thus there will
60 Circuits 3.4<br />
be an electric potential difference ∆V = V a − V b = − ∫ a<br />
E ⃗ · ⃗dr between<br />
b<br />
the two terminals. It can be shown that, regardless of the shape of<br />
the conductors, this electric potential difference is proportional to the<br />
current if the material is ohmic.<br />
Theorem: Ohm’s Law: Resistance<br />
∆V = IR<br />
Where R is called the resistance of the element.<br />
The unit of resistance is the ohm which is one volts per amp. The<br />
ohm is abbreviated as Ω, so that 1Ω = 1V/1A.<br />
⊲ Problem 3.2<br />
You have a wire with cross sectional area A and length L. Show that if<br />
the terminals are placed at the ends of this conductor that the resistance<br />
of this element is R = ρ L A .<br />
⊲ Problem 3.3<br />
You have a block of carbon, with sides of length a, 2a, and 3a. If<br />
terminals are placed on two parallel sides we can make a resistor with<br />
this block. We have three choices for the placement of the terminals,<br />
the sides that are a apart, 2a apart or 3a apart.<br />
(a) Which choice will produce the most resistance.<br />
(b) Which choice will produce the least resistance.<br />
§ 3.4 Electric Power<br />
Suppose that you have a circuit element with two terminals, that<br />
has a current I running through it and a potential difference ∆V between<br />
the terminals. In a time dt an amount of charge dq = I dt will<br />
pass through the element. All of that charge falls through the electric<br />
potential difference of ∆V so that the charge dq looses an amount of<br />
potential energy<br />
dU = dq ∆V = I dt ∆V −→ dU<br />
dt = I ∆V.<br />
So we see that the element dissipates a power P = I ∆V .<br />
Theorem: Electrical Power<br />
The power dissipated in a circuit element is equal to the product<br />
of the current through the element and the potential difference<br />
between the terminals of the element.<br />
P = I ∆V
3.5 Kirchhoff’s Rules 61<br />
⊲ Problem 3.4<br />
A 60 Watt light bulb is plugged into the wall receptacle which supplies<br />
an electric potential of 120 Volts. How much current runs through the<br />
bulb when you turn it on?<br />
⊲ Problem 3.5<br />
(a) Show that a resistor with a current I running through it has a power<br />
of P = I 2 R.<br />
(b) Show that a resistor with a voltage ∆V across it has a power of<br />
P = (∆V ) 2 /R.<br />
§ 3.5 Kirchhoff’s Rules<br />
There are two theorems that are very useful in analysing a circuit.<br />
The first theorem stems from the conservation of charge, that is, that<br />
charge is neither created nor destroyed in a circuit. Consider a junction<br />
where a number of elements come together.<br />
Since the current does not build up at the junction the sum of current<br />
going into the junction must be equal to the sum of the current going<br />
out of the junction. In the case pictured above: I 1 + I 4 = I 2 + I 3 .<br />
Theorem: Kirchhoff’s Junction Rule<br />
The sum of the currents into a junction is equal to the sum of the<br />
current out of a junction. ∑<br />
Iin = ∑ I out<br />
The second theorem stems from the conservation<br />
of energy. Since the electric potential is the potential<br />
energy per charge that is due to the electric field in<br />
the system, if a charge moves around a loop in a circuit<br />
and comes back to where it started it must be at<br />
the same electric potential as when it started. Thus<br />
if we add the electric potential differences of all the<br />
elements that are crossed as you go around any loop<br />
in the circuit the sum must be zero.
62 Circuits 3.6<br />
Theorem: Kirchhoff’s Loop Rule<br />
The sum of the electric potential differences around any closed<br />
loop in a circuit is zero.<br />
∑<br />
∆V = 0<br />
As an example consider the following circuit with five elements.<br />
+ –<br />
+ –<br />
ΔV 1<br />
ΔV 2<br />
+<br />
+<br />
ΔV 3<br />
ΔV 4<br />
–<br />
+ ΔV 5 –<br />
There are a number of loop rules that we could write down. First,<br />
consider the loop starting at the bottom left corner and then taking<br />
the shortest route going clockwise back to the bottom left corner. You<br />
cross elements 1 and 3, both time going from high to low potential.<br />
−∆V 1 − ∆V 3 = 0<br />
Now the equation cannot be satisfied by positive numbers so one of the<br />
∆V ’s must be negative. This is alright, the algebra will sort it out in<br />
the end.<br />
Now consider starting at the lower right corner and taking the<br />
shortest clockwise loop. We pass through elements 5, 3, 2, and 4.<br />
∆V 5 + ∆V 3 − ∆V 2 − ∆V 4 = 0<br />
Notice which ones are positive and which are negative.<br />
We could also do the big loop, starting at the lower left and going<br />
clockwise through elements 1, 2, 4, and 5.<br />
−∆V 1 − ∆V 2 − ∆V 4 + ∆V 5 = 0<br />
It is important to notice that this last equation could have been<br />
arrived at by combining the first two equations, so it has not really given<br />
us any new information. In general it does not help to write down more<br />
loop equations than there are “windows” in the circuit.<br />
§ 3.6 Resistors in Combination<br />
In a circuit where the same current runs through two resistors,<br />
those resistors are said to be in series. If we a circuit where the same<br />
electric potential is across two resistors, those resistors are said to be in<br />
–
3.6 Resistors in Combination 63<br />
parallel. The diagram below will help explain why these configurations<br />
are named in this way.<br />
Suppose that we have two resistors in series. Since they are in<br />
series they must carry the same current.<br />
I 1 = I 2 = I<br />
We also know that the potential difference across the pair is the sum<br />
of the potential differences across each individual resistor;<br />
∆V = ∆V 1 + ∆V 2<br />
= I 1 R 1 + I 2 R 2<br />
= IR 1 + IR 2<br />
= I(R 1 + R 2 )<br />
−→ R effective = R 1 + R 2<br />
We see that the pair of resistors in series still follow Ohm’s law, and<br />
that the pair act as a single resistor with an effective resistance of<br />
R effective = R 1 + R 2 .<br />
Series Resistors<br />
Parallel Resistors<br />
+<br />
–<br />
ΔV<br />
+<br />
ΔV 1<br />
–<br />
+<br />
ΔV 2<br />
–<br />
I 1<br />
I 2<br />
Now consider two resistors in parallel. Since they are in parallel<br />
the must have the same electric potential.<br />
∆V 1 = ∆V 2 = ∆V<br />
We also know from the junction rule that the net current going into the<br />
system is equal to the sum of the two currents going into the resistors.<br />
I = I 1 + I 2 = ∆V 1<br />
+ ∆V 2<br />
R 1 R<br />
[ 2<br />
1<br />
= ∆V + 1 ]<br />
R 1 R 2<br />
= ∆V<br />
R 1<br />
I<br />
+ ∆V<br />
R 2<br />
−→<br />
1<br />
R effective<br />
= 1 R 1<br />
+ 1 R 2
64 Circuits 3.7<br />
Theorem: Effective Resistance<br />
Series R = R 1 + R 2<br />
Parallel<br />
1<br />
R = 1 + 1 R 1 R 2<br />
§ 3.7 Capacitors in Combination<br />
Capacitors in series and parallel can also be treated as a single<br />
capacitor. The argument is much the same, but with charge rather<br />
than current.<br />
Suppose that you have two capacitors in series. Then the charges<br />
on each must be the same.<br />
Q 1 = Q 2 = Q<br />
While the potential drop across the pair is the sum of the potential<br />
drop across each<br />
∆V = ∆V 1 + ∆V 2<br />
+<br />
–<br />
−→<br />
Series Capacitors<br />
= Q 1<br />
+ Q 2<br />
= Q + Q C 1 C 2 C 1 C 2<br />
1<br />
C = ∆V<br />
Q = 1 + 1 C 1 C 2<br />
+<br />
Parallel Capacitors<br />
+Q<br />
ΔV 1<br />
-Q –<br />
+Q 1 +Q 2<br />
ΔV<br />
+Q +<br />
-Q<br />
ΔV 1<br />
-Q 2<br />
2<br />
-Q<br />
For capacitors in parallel the potential is the same.<br />
–<br />
∆V 1 = ∆V 2 = ∆V<br />
While the net charge that flows into the system is shared between the<br />
capacitors.<br />
Q = Q 1 + Q 2<br />
= C 1 ∆V 1 + C 2 ∆V 2 = C 1 ∆V + C 2 ∆V<br />
−→ C =<br />
Q<br />
∆V = C 1 + C 2
3.8 Capacitor Circuits 65<br />
Theorem: Effective Capacitance<br />
Series<br />
1<br />
C = 1 C 1<br />
+ 1 C 2<br />
Parallel C = C 1 + C 2<br />
§ 3.8 Capacitor Circuits<br />
Consider the following circuit, which is composed of a power supply<br />
with a fixed electric potential output of V S , a capacitor, a resistor, and<br />
a double pole switch.<br />
b<br />
+<br />
a<br />
V S +Q<br />
–<br />
I -Q<br />
+<br />
V C<br />
–<br />
–<br />
To begin with the switch is in position a. In this position all of<br />
the charge will drain from the capacitor. At the time t = 0 the switch<br />
is moved to position b. In this position the power supply will begin to<br />
fill the capacitor with charge. The current I flows into the capacitor.<br />
Thus the charge on the capacitor increases at the rate<br />
dQ<br />
dt = I<br />
But for a capacitor the charge is proportional to the electric potential<br />
across the capacitor Q = CV C so that<br />
I = dQ<br />
dt = C dV C<br />
dt<br />
Kirchhoff’s loop rule gives us the following equation.<br />
But by Ohm’s Law we have<br />
V R<br />
+<br />
V S − V C − V R = 0<br />
V R = RI = RC dV C<br />
dt<br />
Putting this into the loop rule equation we find<br />
V S − V C − RC dV C<br />
dt<br />
= 0
66 Circuits 3.8<br />
This is a differential equation that describes how the voltage on the<br />
capacitor changes with time, similar to how F = ma is a differential<br />
equation that describes how a particles position changes with time. In<br />
this particular case we can find the solution by noting that if we define<br />
a new variable, f(t) = V C − V S then df<br />
dt = dV C<br />
dt<br />
Plugging this into our differential equation we find<br />
−f − RC df<br />
dt = 0<br />
or<br />
df<br />
dt = − 1<br />
RC f<br />
since V S is a constant.<br />
This just says that the derivative of f is simply a constant times f.<br />
The only function that has this property is the exponential. So let us<br />
try a function of the form<br />
f = Ae αt −→ df<br />
dt = αAeαt = αf<br />
Comparing this with our differential equation we see that it is the same<br />
if<br />
α = − 1<br />
RC<br />
Thus<br />
V C − V S = f = Ae −t/RC<br />
−→ V C = V S + Ae −t/RC<br />
In order to determine the constant A we need to use the initial condition<br />
of the system. At t = 0 we know that there was no charge on the<br />
capacitor, thus the electric potential on the capacitor was zero at t = 0.<br />
Putting this into the solution we fine that<br />
0 = V C (0) = V S + Ae −0/RC = V S + A −→ A = −V S<br />
So the voltage on the capacitor is<br />
V C = V S (1 − e −t/RC )<br />
The combination RC is called the time constant of the circuit because it<br />
has units of time and gives the time scale for the change in the voltage:<br />
in a time RC the difference of the electric potential from it’s final value<br />
(V S ) decreases by a factor of e −1 ≈ 3/8.
3.9 More Examples 67<br />
Note that the equation for V C does not depend on R or C separately,<br />
but only on the combination RC.<br />
⊲ Problem 3.6<br />
Suppose that the capacitor circuit described above is assembled with<br />
a 10 Volt power supply, a 5.0kΩ resistor, and a 3.0µF capacitor. How<br />
long after the switch is put in position b, will it be before the capacitor<br />
is half charged (5.0V)?<br />
⊲ Problem 3.7<br />
Assume that the capacitor circuit described above is left with the switch<br />
in position b for a long time, so that the capacitor is fully charged to<br />
the voltage V S . The switch is now moved to position a. Show that<br />
V C = V S e −t/RC .<br />
§ 3.9 More Examples<br />
Example<br />
The largest part of the nerve cells in your body is called the “axon.”<br />
The axon is in the shape of a cylinder and has sections through which<br />
ions pass, sending electrical signals through the nerve. During a nerve<br />
impuse, for the cylindrical section of an axon indicated below, 10,000<br />
sodium ions (Na + ) pass through the surface of the cell membrane in 1<br />
m/s. What is the current and current density into the cell?<br />
Na +<br />
20μm<br />
Na +<br />
100μm
68 Circuits 3.9<br />
Each sodium ion has a charge +1.6 × 10 −19 C, so for 10,000 ions, ∆Q =<br />
10, 000 × (1.6 × 10 −19 C) = 1.6 × 10 −15 C. Since it takes 1ms:<br />
I = ∆Q<br />
∆t = 1.6 × 10−15 C<br />
= 1 × 10 −12 A = 1pA.<br />
.001s<br />
The ions flow through the sides of the cylinder, so the magnitude of<br />
the current density is<br />
J = I S = 1.6 × 10 −12 A<br />
(2π)(10µm)(500µm) = 5.1 × A 10−5 m 2 = 51µA m 2<br />
Example<br />
A copper wire, with a diameter of 1mm, has a current of 5A flowing<br />
through it. What is the electric field in the wire?<br />
Comment: How can E ≠ 0? Earlier it was stated that E = 0 inside a<br />
conductor, however this was for the case of electrostatics. In the case<br />
of a current flowing, since electric charges are not static, the electric<br />
field will be nonzero.<br />
For our wire, the current density is<br />
J = I A = 5A<br />
π(0.0005m) 2 = 6.4 × A 106 m 2<br />
From Ohm’s law:<br />
J = 1 ρ E<br />
−→ E = ρJ = (1.7 × 10 −8 Ωm)(6.4 × 10 6 A m 2 )<br />
= 0.11 Ω·A<br />
= 0.11 V m<br />
m<br />
Example<br />
A light bulb with a resistance of 3Ω is connected to a 9V battery. In<br />
one second how many electrons flow through the bulb?<br />
I<br />
R = 3Ω<br />
-<br />
+<br />
9V<br />
Use Ohm’s law to find the current:<br />
I = ∆V<br />
R<br />
= 9V<br />
3Ω = 3A
3.9 More Examples 69<br />
This mean that in one second, 3C of charge flows through the resistor.<br />
Since each electron (which will flow in a direction opposite the<br />
conventional positive current) carries 1.6 × 10 −19 C of charge:<br />
3A<br />
N =<br />
1.6 × 10 −19 C × (1s) = 1.9 × 1016 electrons<br />
Example<br />
A 20W light bulb is left on in your car’s interior. If the car’s 12V<br />
battery is fully charged and has a capacity of 200Amp-hours. How<br />
long will it take to completely discharge the battery (assuming the<br />
terminal voltage remains 12V throughout the discharge)?<br />
Use electric power P to compute the current that flows:<br />
P = I∆V −→ I = P<br />
∆V = 20W<br />
12V = 1.67A<br />
The amount of charge that will pass through the bulb in a time ∆t<br />
is ∆Q = 1.67A × ∆t. The 200 Amp-hour rating tells us how much<br />
charge can pass from the positive terminal to the negative terminal of<br />
the battery before discharging the battery, so<br />
∆Q200Ahr = (1.67A)∆t<br />
−→ ∆t = 200Ahr<br />
1.67A<br />
How much charge has flowed?<br />
= 120hr = 5days<br />
200Ahr = 200A(3600s) = (200 C )(3600s) = 720, 000C<br />
s<br />
Example<br />
Determine the currents in the three resistors of the following circuit:<br />
12V<br />
6Ω<br />
-<br />
+<br />
6V<br />
+<br />
-<br />
3Ω<br />
9Ω<br />
Let’s label the currents. We will do our best to indicate the correct<br />
directions of the current:
70 Circuits 3.9<br />
6Ω<br />
12V<br />
-<br />
+<br />
6V<br />
+<br />
-<br />
I 1<br />
I 2 3Ω I 3<br />
9Ω<br />
A<br />
We now apply the loop rules to the two loops indicated. In both cases,<br />
start from the point labeled A, moving in the direction indicated, summing<br />
the potential gains or losses across each circuit element. For batteries<br />
the potential increases from the negative to the positive terminal.<br />
For resistors the current flows from high to low potential. Applying the<br />
loop rules gives two equations, on for each loop:<br />
−(3Ω)I 2 − (6Ω)I 1 + 6V = 0<br />
−(3Ω)I 2 + 12V − (9Ω)I 3 = 0<br />
We also must apply current conservation. Choose the point labeled A:<br />
I 1 − I 2 + I 3 = 0<br />
Now we have 3 equations with three unknowns. Solving the linear set<br />
of equations yields:<br />
I 1 = 0.364A<br />
I 2 = 1.27A<br />
I 3 = 0.91A<br />
Check that these are correct by putting them back into the loop equations:<br />
−(3Ω)(1.27A) − (6Ω)(0.364A) + 6 −→ −6.0V + 6.0V = 0<br />
−(3Ω)(1.27A) + 12 − (9Ω)(0.91A) −→ −12.0V + 12.0V = 0<br />
Example<br />
Determine the currents in the resistors of the following circuit:
3.9 More Examples 71<br />
15Ω<br />
5Ω<br />
10Ω<br />
15V<br />
+<br />
-<br />
10V<br />
+<br />
-<br />
15Ω<br />
Label the currents, including best guesses for the directions:<br />
I 1<br />
I 2<br />
I 3<br />
15Ω<br />
5Ω<br />
10Ω<br />
+<br />
-<br />
15V<br />
+<br />
-<br />
10V<br />
I 1<br />
A<br />
15Ω<br />
B<br />
Apply Kirchoff’s loop rules. First to the lefthand loop, beginning at<br />
point A and moving clockwise:<br />
15V − (15Ω)I 1 − (5Ω)I 2 + 10V − (15Ω)I 1 = 0,<br />
then to the righthand loop, beginning at point B and moving counterclockwise:<br />
(10Ω)I 3 − (5Ω)I 2 + 10V = 0.<br />
Note that for a resistor, the conventional positive current flows from<br />
high to low potential, which determines the sign to use when moving<br />
from one side of a resistor to the other.<br />
Now apply Kirchoff’s junction rule at the point labeled B:<br />
I 2 + I 3 − I 1 = 0<br />
Simplifying, we have three equations with three unknowns:<br />
25 − 30I 1 − 5I 2 = 0<br />
10 + 10I 3 − 5I 2 = 0<br />
I 2 + I 3 − I 1 = 0<br />
To solve, let first eliminate I 3 :<br />
10 + 10I 3 − 5I 2 = 10 + 10(I 1 − I 2 ) − 5I 2 = 0
72 Circuits 3.9<br />
−→ 10 + 10I 1 − 15I 2 = 0<br />
Next multiply the first equation, of the three listed above, by -3:<br />
−→ −75 + 90I 1 + 15I 2 = 0.<br />
Add the two modified equations together:<br />
10 + 10I 1 − 15I 2 − 75 + 90I 1 + 15I 2 = 0<br />
−→ −65 + 100I 1 = 0<br />
−→ I 1 = .65A<br />
Using this value of I ! , we can solve for I 2 and I 3 :<br />
I 2 = 1.1A<br />
I 3 = −0.45A<br />
The negative value obtained for I 3 indicates that the direction for<br />
this current is opposite the direction that was chosen. So I 3 flows<br />
up through the 10Ω resistor, not down as indicated in our diagram.<br />
⊙ Do This Now 3.1<br />
Repeat the analysis for the circuit in the previous example, using the current<br />
directions shown below:<br />
I 1<br />
I 2<br />
I 3<br />
15Ω<br />
5Ω<br />
10Ω<br />
+<br />
-<br />
15V<br />
+<br />
-<br />
10V<br />
I 1<br />
A<br />
15Ω<br />
B<br />
Example<br />
You should not get any negative-valued currents.<br />
What is the effective resistance between the points A and B for the<br />
following circuit.<br />
R = 30Ω<br />
4<br />
A<br />
R = 15Ω<br />
2<br />
R = 10Ω<br />
1<br />
R = 4Ω<br />
3<br />
B
3.9 More Examples 73<br />
First combine R 1 and R 2 . Since they are in parallel:<br />
1<br />
= 1 + 1 = 1<br />
R 12 R 1 R 2 10Ω + 1<br />
15Ω −→ R 12 = 6Ω<br />
The circuit has been reduced to:<br />
R = 30Ω<br />
4<br />
A<br />
R = 6Ω<br />
12<br />
R = 4Ω<br />
3<br />
Now combine R 12 and R 3 , which are in series:<br />
This leaves the effective circuit:<br />
R 123 = R 12 + R 3 = 6Ω + 4Ω = 10Ω.<br />
R = 30Ω<br />
4<br />
B<br />
A<br />
R = 10Ω<br />
123<br />
B<br />
Finally, combine the last two resistances in parallel:<br />
1<br />
= 1 + 1 = 1<br />
R AB R 123 R 4 10Ω + 1<br />
30Ω −→ R 12 = 7.5Ω<br />
Example<br />
A 12V battery is connected across the points AB in the circuit from the<br />
previous example. What current flows through the R 3 = 4Ω resistor?<br />
First find the current that flows from A to B:<br />
I AB = V AB<br />
= 12V<br />
R AB 7.5Ω = 1.6A<br />
The incoming current splits between the 30Ω resistor and the effective<br />
resistance R 123 :<br />
A<br />
I AB<br />
I 4<br />
I 123<br />
R = 30Ω<br />
4<br />
R = 10Ω<br />
123<br />
B<br />
We can easily compute the current through the 30Ω resistor, since the<br />
potential difference across it is V AB :<br />
I 4 = V AB<br />
= 12V<br />
R 4 30Ω = 0.4A<br />
Using Kirchoff’s junction rule:<br />
I 123 = I AB − I 4 = 1.6A − 0.4A = 1.2A.
74 Circuits 3.10<br />
Because the R 3 = 4Ω resistor in connected in series with the parallel<br />
combination of R 1 and R 2 , the current through it is I 123 :<br />
−→ I 3 = 1.2A<br />
§ 3.10 Homework<br />
⊲ Problem 3.8<br />
In the Bohr model of the hydrogen atom, an electron in the lowest<br />
energy state follows a circular path, 5.29 × 10 −11 m from the proton.<br />
Show that the speed of the electron is 2.19×10 6 m s<br />
. What is the effective<br />
current associated with this orbiting electron?<br />
⊲ Problem 3.9<br />
In a particular cathode ray tube, the measured beam current is 20µA.<br />
How many electrons strike the tube screen every 1 30 seconds?<br />
⊲ Problem 3.10<br />
The quantity of charge q (in coulombs) passing through a surface of<br />
area 2.0cm 2 varies with time as q = 4t 3 + 5t + 6 where t is in seconds.<br />
What is the instantaneous current through the surface at t = 1.0s ?<br />
What is the value of the current density?<br />
⊲ Problem 3.11<br />
An electric current is given by I(t) = 100.0 sin(120πt), where I is in<br />
amperes and t is in seconds. What is the total charge carried by the<br />
current from t = 0 to t = 1/240s ?<br />
⊲ Problem 3.12<br />
Calculate the average drift speed of electrons traveling through a copper<br />
wire with a cross-sectional area of 1.00mm 2 when carrying a current<br />
of 1.00A. It is known that about one electron per atom of copper<br />
contributes to the current. The atomic weight of copper is 63.54 and<br />
its density is 8.92g/cm 3 .<br />
⊲ Problem 3.13<br />
Eighteen gauge wire has a diameter of 1.024mm. Calculate the resistance<br />
of 15.0m of 18-gauge copper wire at 20.0 ◦ C.<br />
⊲ Problem 3.14<br />
Suppose that a voltage surge produces 140V for a moment. By what<br />
percentatge will the ouput of a 120V, 100W lightbulb increase, assuming<br />
its resistance does not change?<br />
⊲ Problem 3.15<br />
<strong>Two</strong> cylindrical copper wires have the same mass. Wire A is twice as<br />
long as wire B. What is the ratio of their resistances?
3.10 Homework 75<br />
⊲ Problem 3.16<br />
Batteries are rated in ampere hours (A · hr), where a battery rated at<br />
1.0A · hr can produce a current of 1.0A for 1.0hr.<br />
(a) What is the total energy stored in a 12.0V battery rated at 55.0A·hr<br />
?<br />
(b) At $0.12 per kilowatt hour, what is the value of the electrical energy<br />
stored in this battery?<br />
⊲ Problem 3.17<br />
What is the required resistance of an immersion heater that will increase<br />
the temperature of a 1.5 kg of water from 10 ◦ C to 50 ◦ C in 10<br />
min while operating at 110V?<br />
⊲ Problem 3.18<br />
A battery with an emf of 12V and internal resistance of 0.90Ω is connected<br />
to a load resistor R.<br />
(a) If the current in the circuit is 1.4A, what is the value of R?<br />
(b) What power is dissipated in the internal resistance of the battery?<br />
⊲ Problem 3.19<br />
(a) Find the equivalent resistance between points a and b in the figure<br />
below.<br />
(b) If a potential difference of 34V is applied between points a and b,<br />
calculate the current in each resistor.<br />
7 Ω<br />
a<br />
4 Ω<br />
9 Ω<br />
b<br />
10 Ω<br />
⊲ Problem 3.20<br />
For the figure below find the current in the 20Ω resistor and the potential<br />
difference between points a and b.<br />
25 V<br />
10 Ω - +<br />
a 10 Ω b<br />
5.0 Ω 20 Ω<br />
5.0 Ω<br />
⊲ Problem 3.21<br />
Determine the equivalent capacitance for the capacitor network shown<br />
below. If the network is connected to a 12V battery, calculate the<br />
potential difference across each capacitor and the charge on each capacitor.
76 Circuits 3.10<br />
3.0 µF 6.0 µF<br />
2.0 µF<br />
⊲ Problem 3.22<br />
Four capacitors are connected as shown. Find the equivalent capacitance<br />
between points a and b. Calculate the charge on each capacitor<br />
in V ab = 15V.<br />
15 µF 3.0 µF<br />
20 µF<br />
6.0 µF<br />
⊲ Problem 3.23<br />
When two capacitors are connected in parallel, the equivalent capacitance<br />
is 4.00µF. If the same capacitors are reconnected in series, the<br />
equivalent capacitance is one-fourth the capacitance of one of the two<br />
capacitors. Determine the two capacitances.<br />
⊲ Problem 3.24<br />
For the system of capacitors shown below find the equivalent capacitance<br />
of the system, the potential across each capacitor, the charge on<br />
each capacitor, and the total energy stored by the group.<br />
3 µF 6 µF<br />
2 µF 4 µF<br />
90 V<br />
⊲ Problem 3.25<br />
Calculate the equivalent capacitance between the two points shown<br />
in the circuit below. Note that this is not a simple series or parallel<br />
combination.
3.10 Homework 77<br />
4 µF<br />
+ -<br />
2 µF<br />
+<br />
-<br />
8 µF<br />
+<br />
-<br />
+ -<br />
+<br />
-<br />
4 µF<br />
2 µF<br />
⊲ Problem 3.26<br />
If R = 1.0kΩ and E = 250V determine the direction and magnitude of<br />
the current in the horizontal wire between a and e.<br />
R c<br />
b<br />
2R d<br />
ε + -<br />
4R<br />
3R<br />
+<br />
-<br />
2ε<br />
a<br />
⊲ Problem 3.27<br />
Determine the current in each branch of the figure.<br />
3.0 Ω<br />
e<br />
5.0 Ω<br />
8.0 Ω<br />
1.0 Ω<br />
1.0 Ω<br />
4 V<br />
12 V<br />
⊲ Problem 3.28<br />
A 25-W light bulb is connected in series with a 100-W light bulb and<br />
a voltage V is placed across the combination. Which bulb is brighter?<br />
Explain.<br />
⊲ Problem 3.29<br />
An electric heater is rated at 1500W, a toaster at 750W, and an electric<br />
grill at 1000W. The three appliances are connected to a common 120V<br />
circuit. How much current does each draw? Is a 25A circuit sufficient<br />
in this situation?<br />
⊲ Problem 3.30<br />
An 8-foot extension cord has two 18-gauge copper wires, each having<br />
a diamter of 1.024mm. How much power does this cord dissipate when<br />
carrying a current of 1.0A? How much power does this cord dissipate<br />
when carrying a current of 10A?
78 Circuits 3.10<br />
⊲ Problem 3.31<br />
Because aluminum has a greater resistivity than copper, aluminum<br />
wires heat up more than copper wires when they carry the same currents.<br />
For this reason aluminum wire of the same gauge (diameter) is<br />
rated to carry a smaller current. What current in an aluminum wire<br />
will heat up the wire the same amount as a copper wire of the same<br />
gauge carrying 20 Amps?<br />
⊲ Problem 3.32<br />
Consider the following circuit.<br />
12 kΩ<br />
9.0 V<br />
R 2 = 15 kΩ<br />
10 µF<br />
3 kΩ<br />
The switch is closed and the capacitor is allowed to charge up.<br />
(a) How much charge will there be on the capactor when it is fully<br />
charged?<br />
(b) Now the switch is opened and the charge on the capacitor drains<br />
off. How long will it take for the capacitor to reach 1/5 of the charge<br />
it had at the moment the switch was opened?
3.11 Summary 79<br />
§ 3.11 Summary<br />
Definitions<br />
Electric Current:<br />
Current Density<br />
I = dq<br />
dt<br />
J = I A<br />
Facts<br />
Ohm’s Law: For many materials current density is proportional to<br />
the electric field.<br />
⃗J = σ ⃗ E<br />
with σ the conductivity of the material. ρ = 1/σ is called the resistivity.<br />
∆V = IR<br />
with R the resistance of the device.<br />
Theorems<br />
Electric Power:<br />
Kirchhoff’s Loop Rule:<br />
P = I∆V<br />
∑<br />
∆V = 0<br />
Kierchhoff’s Junction Rule:<br />
∑<br />
Iin = ∑ I out<br />
Resistors in Parallel:<br />
Resistors in Series:<br />
1<br />
R eff<br />
= 1 R 1<br />
+ 1 R 2<br />
Capacitors in Parallel:<br />
R eff = R 1 + R 2<br />
Capacitors in Series:<br />
C eff = C 1 + C 2<br />
1<br />
C eff<br />
= 1 C 1<br />
+ 1 C 2
80 Magnetic Fields 4.11
4.2 Magnetic Force on a Current 81<br />
4 Magnetic Fields<br />
§ 4.1 Magnetic Field<br />
We have all played with magnets at some point. Magnets will<br />
attract some types of metals and can attract and repel other magnets.<br />
This seems similar to electric forces. Magnetic forces can indeed be<br />
explained in terms of magnetic fields, but the connection between the<br />
magnetic field and the magnetic force is significantly different from<br />
that for electric fields and force. The magnetic field is similar to the<br />
electric field in that they are both vector fields. We use the symbol B ⃗ to<br />
represent the magnetic field. If we put a compass needle in a magnetic<br />
field, the field will turn the needle until it is aligned with the direction<br />
of the field. At this point we will use the following simple operational<br />
definition of the magnetic field. The direction of the magnetic field<br />
is the direction that a compass needle points. The magnitude of the<br />
magnetic field is related to how strong the force is that aligns the needle<br />
with the direction of the field.<br />
A magnetic field does apply a force to charged particles. The<br />
magnetic force on the particle has the following properties:<br />
• F ∝ q<br />
• F ∝ v<br />
• F ∝ B<br />
• F ⃗ is perpendicular to both ⃗v and B. ⃗ These can be combined into the<br />
following succinct formula for the force.<br />
Fact: Magnetic Force<br />
A particle with charge q and velocity ⃗v in a magnetic field ⃗ B will<br />
feel a force<br />
⃗F = q ⃗v × ⃗ B<br />
Notice that there is no force if the velocity is parallel to the magnetic<br />
field. Also notice that the magnetic force is zero on a stationary<br />
particle.
82 Magnetic Fields 4.2<br />
We can determine the dimensions of the magnetic<br />
field from the force equation:<br />
F = qvB −→ B = F qv<br />
The unit of magnetic field is the Tesla, abbreviated as<br />
just T.<br />
Newton<br />
Tesla =<br />
Coulomb · meter/second = Kilogram<br />
Coulomb · second<br />
§ 4.2 Magnetic Force on a Current<br />
If a current carrying wire is placed in a magnetic field it will experience<br />
a magnetic force. The current in the wire is composed of many<br />
moving charges. Each of these charges experiences a magnetic force.<br />
The force on the wire is the sum of the forces on all the moving charges<br />
in the wire.<br />
Theorem: Magnetic Force on a Current<br />
Suppose that you have a wire that is carrying a current I. A small<br />
section of wire of length ⃗ dl, where the vector points in the direction<br />
of the current, will experience a force<br />
⃗ dF = I ⃗ dl × ⃗ B<br />
The force on a longer section of wire can be found by integrating<br />
over the length of the section.<br />
∫<br />
⃗F = I dl ⃗ × B ⃗<br />
If the magnetic field is uniform over the region containing the wire<br />
then the following result can be proved.<br />
Theorem: Force on a Current in a Uniform Field<br />
Let ⃗ ∆l be a vector that points from the beginning to the end of a<br />
section of wire carrying a current I. If that section is in a uniform<br />
magnetic field, the force on that section is<br />
⃗F = I ⃗ ∆l × ⃗ B<br />
⊲ Problem 4.1<br />
Consider a semicircular piece of wire or radius R in the first two quadrants<br />
of the x-y plane. The wire carries of current I in the counterclockwise<br />
direction.
4.3 Trajectories Under Magnetic Forces 83<br />
y<br />
dl<br />
R<br />
dθ<br />
θ<br />
x<br />
There is a uniform magnetic field in the y direction, B ⃗ = Bĵ. We<br />
wish to compute the net force on this section of wire without using the<br />
theorem F ⃗ = I∆l ⃗ × B. ⃗<br />
(a) If we break the semicircle into small sections, they will be small<br />
sections of arc, as pictured in the diagram above. If we take one of<br />
these, it will be at a position θ, and will subtend an angle dθ. Show<br />
that dl ⃗ = (− sin θî + cos θĵ)Rdθ<br />
(b) With this result you can now compute the integral F ⃗ = ∫ I dl ⃗ × B. ⃗<br />
Show that the net force on the semicircle is −I2RBˆk.<br />
(c) Show that this is the same answer you get when you apply the<br />
theorem F ⃗ = I∆l ⃗ × B. ⃗<br />
⊲ Problem 4.2<br />
Consider the rectangular current loop pictured below.<br />
w<br />
I<br />
B<br />
B<br />
The field is uniform and in the direction indicated in the diagram.<br />
Show that the torque about the axis indicated by the dotted line is<br />
IAB where A is the area of the loop.<br />
§ 4.3 Trajectories Under Magnetic Forces<br />
Because the magnetic force is perpendicular to the velocity of the<br />
particle, the magnetic force can not do work on the particle. Suppose<br />
that a particle moves a small distance ⃗ dr. The work done by a force ⃗ F<br />
as the particle moves is<br />
dW = ⃗ F · ⃗dr<br />
But ⃗ dr = ⃗v dt so<br />
dW = ⃗ F · ⃗v dt<br />
h
84 Magnetic Fields 4.3<br />
For a magnetic force, ⃗v and ⃗ F are prependicular so that ⃗ F ·⃗v = 0. Thus<br />
dW = 0 for a magnetic force.<br />
Theorem: Work by Magnetic Force<br />
The magnetic force does zero work. Thus the magnetic force can<br />
change the direction but not the speed of a particle.<br />
Because the magnetic force changes only the direction of a particle,<br />
a magnetic field is useful for steering charged particles, once they are<br />
already moving.<br />
Consider a particle that is in a region with a uniform magnetic<br />
field. Suppose that the particles initial velocity is perpendicular to the<br />
magnetic field. Since the magnetic force is always perpendicular to<br />
the direction of the magnetic field, the force will have no component<br />
parallel to the field. So, since the particle started with zero velocity<br />
parallel to the field it will continue to have zero velocity parallel to the<br />
field. But this means that the velocity will always be perpendicular to<br />
the magnetic field, so |⃗v × B| ⃗ = vB. Thus we know that the magnitude<br />
of the magnetic force is<br />
F ≡ | ⃗ F | = q|⃗v × ⃗ B| = qvB<br />
In addition we know by the previous theorem that the speed of the<br />
particle will be constant, v = v 0 . So that the magnitude of the force<br />
is constant. In summary, we see that the magnitude of the magnetic<br />
force on the particle is constant and perpendicular to the velocity, so<br />
that the acceleration of the particle is constant and perpendicular to<br />
the velocity. We have run into this situation before: In circular motion<br />
the acceleration is constant and always perpendicular to the velocity.<br />
We are lead by this observation to the following theorem.<br />
Theorem: Circular Trajectories<br />
If a particle is in a region of uniform magnetic field and entered<br />
the region with a velocity perpendicular to the magnetic field, then<br />
the particle will execute circular motion while in the region.<br />
One can relate the radius and velocity of the motion to the field<br />
strength and charge by employing Newton’s second law.<br />
F = ma<br />
−→ qvB = m v2<br />
r<br />
⊲ Problem 4.3<br />
One can use the circular motion of a charge particle to determine it’s<br />
mass if you already know it’s charge. Suppose that you send a particle
4.4 Lorentz Force 85<br />
with a charge 1.6 × 10 −19 C into a field 0.11mT with a speed of 3.83 ×<br />
10 5 m s<br />
. The radius of the resulting motion is 2cm. What is the mass of<br />
the particle?<br />
⊲ Problem 4.4<br />
A stream of charged particles<br />
enter a circular region of uniform<br />
magnetic field as shown (gray). The<br />
particles fan-out in the region and<br />
each particle exits along one of the<br />
six trajectories A through F . Pick<br />
the trajectories for particles with the<br />
following properties:<br />
(a) Which particles are positively charged?<br />
(b) If all the particles have the same mass and charge, which have the<br />
highest speed?<br />
(c) If all the particles have the same speed and charge, which have the<br />
highest mass?<br />
(d) If all the particles have the same mass and speed, which have the<br />
highest charge?<br />
⊲ Problem 4.5<br />
A particle is moving in a circular trajectory because of a magnetic field.<br />
Show that regardless of the velocity of the particle, it will take the same<br />
amount of time to complete one revolution. The fact that the time is<br />
a constant eased the development of the cyclotron, an early particle<br />
accelerator.<br />
§ 4.4 Lorentz Force<br />
If there is a magnetic as well as electric field then both forces act<br />
on a charge particle at the same time.<br />
Definition: Lorentz Force<br />
The combined electric and magnetic forces on a charge particle is<br />
called the Lorentz Force.<br />
⃗F = q( ⃗ E + ⃗v × ⃗ B)<br />
A<br />
F<br />
E<br />
B<br />
D<br />
C<br />
⊲ Problem 4.6<br />
There is a region with a uniform magnetic field ⃗ B = Bˆk and electric<br />
field ⃗ E = Eĵ. Particles with charge q are sent into this region with a<br />
velocity ⃗v = vî.
86 Magnetic Fields 4.5<br />
(a) Show that if v = E/B that the particles will go straight through<br />
the region without deflection.<br />
(b) Show that if v > E/B and q > 0 then the particles will be deflected<br />
in the negative y direction.<br />
(c) Show that if v > E/B and q < 0 then the particles will be deflected<br />
in the positive y direction.<br />
(d) Show that if v < E/B and q > 0 then the particles will be deflected<br />
in the positive y direction.<br />
A device setup with crossed magnetic and electric fields as in the<br />
previous problem is called a velocity selector. Since it sorts out the<br />
particles based on their speeds, if you select out just the ones that<br />
go straight ahead, you have selected particles with a particular speed<br />
v = E/B. By adjusting the field strengths you can choose what velocity<br />
you would like to select.<br />
§ 4.5 Hall Effect<br />
Let us look with more detail at the flow of current in a wire that<br />
is in a magnetic field. Suppose that we have a rectangular conductor<br />
with a current I running through it.<br />
w<br />
h<br />
If we consider the charges in the wire we see that there will be a<br />
magnetic force on the particles that will tend to deflect them to one<br />
face of the conductor.<br />
I<br />
-<br />
I<br />
But this force will be acting on all the free charges, so they will all shift<br />
upward as they move through the conductor. Keeping in mind that<br />
the positive charge in the conductor does not move, we see that this<br />
shift of the (negative) free charges will build up a net negative charge<br />
on one face of the conductor and net positive charge on the opposing<br />
face.<br />
-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
I<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
I<br />
- - - - - - - - - - - - - - - - - - -<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+<br />
+
4.5 Hall Effect 87<br />
But this charge separation will create an electric field, that is in the<br />
opposite direction to the magnetic force, thus the charge will continue<br />
to build up on the face of the conductor until the electric field builds<br />
up to the point where the electric force on the charges balances the<br />
magnetic force on the charges.<br />
- - - - - - - - - - - - - - - - - - -<br />
I<br />
E E E E E<br />
+ + + + + + + + + + + + + + + + + + +<br />
This will happen any time that a current carrying wire is in a magnetic<br />
field: the wire will spontaneously generate an electric field across the<br />
wire (not end to end but across). This effect, discovered by Edwin<br />
Hall, is called the Hall Effect. Because there is an electric field across<br />
the conductor there will also be an electric potential difference (called<br />
the Hall voltage), ∆V = Ew where w is the width of the conductor.<br />
But we know that in order for the electric force to cancel the magnetic<br />
force, we need E = vB. So we see that<br />
∆V = vBw −→ v = ∆V<br />
Bw<br />
Because it is relatively easy to measure the electric potential difference,<br />
the width of the conductor and the magnetic field strength, this type<br />
of device can be used to measure the velocity of the charge carriers.<br />
Knowing the velocity allows one to determine how many charge carriers<br />
there are per volume. In a time dt a quantity of charge dq = I dt passes<br />
out the end of the wire, and a number of electrons dn = dq/e = I dt/e<br />
passes out of the wire. But we can also say that the charges have moved<br />
a distance dx = v dt in this time, so that a section of charge v dt long<br />
has passed out of the wire. Thus a volume of charge dV = hw v dt<br />
has passed out of the wire (where h is the thickness of the wire and hw<br />
is the cross sectional area of the wire). So the number of conduction<br />
electrons per volume (η) is<br />
η = dn<br />
dV = I dt/e<br />
hw v dt = I<br />
ehw v<br />
The carrier density, η, depends on the type of material, not on the<br />
geometry of the material or the amount of current flowing through the<br />
material.<br />
The Hall effect can be used to make a magnetic field sensor. By<br />
using the equation v = ∆V<br />
Bw<br />
to eliminate the velocity from the equation<br />
η =<br />
I<br />
ehw v<br />
, and then solving for B we find that<br />
B = eηh ∆V<br />
I .<br />
The e, η and h are all constants for a given device, while ∆V and I are<br />
I
88 Magnetic Fields 4.6<br />
both easily measured.<br />
The Hall effect also allows us to determine if the charge carriers<br />
are positive or negative. In the analysis above we assumed that the<br />
charge carriers were negatively charged electrons. Consider what would<br />
happen if the charge carriers were positive instead. The positive charges<br />
would also be deflected by the magnetic field.<br />
I<br />
+<br />
I<br />
Which will build up a field in the opposite direction.<br />
I<br />
+ + + + + + + + + + + + + + + + + + +<br />
E E E E E<br />
- - - - - - - - - - - - - - - - - - -<br />
Thus we can determine from the sign of the Hall voltage, if the charge<br />
carriers are positive or negative. There are some types of semiconductors<br />
that have positive charge carriers, for example silicon with a little<br />
bit of aluminum mixed in. Semiconductors with positive charge carriers<br />
are called p-type semiconductors. Semiconductors with negative<br />
charge carriers are called n-type semiconductors.<br />
I<br />
§ 4.6 More Examples<br />
Example<br />
An electron is injected horizontally into a parallel plate capacitor with<br />
a velocity v = 4 × 10 6 m s : A, +Q<br />
-e<br />
v<br />
A, -Q<br />
The plates are square, with sides 20cm and the charge on the capacitor<br />
is 7.5µC. Describe what magnetic field is required such that the net<br />
force on the electron is zero between the capacitor plates. Ignore any<br />
edge effects.<br />
The electric field points down, since the top plate on the capacitor is<br />
positive. However, since the electron has a negative charge, the electric<br />
force is up. Thus, the force caused by the magnetic field must be down.<br />
Here is what the force diagram on the electron must look like
4.6 More Examples 89<br />
-e<br />
F E<br />
v<br />
F B<br />
= -e(v x B)<br />
Again, since the electron’s charge is negative, the magnetic force will<br />
point in the direction opposite the cross-product ⃗v × ⃗ B. So, ⃗ B must<br />
point into the page. The magnitude of B can be determined since the<br />
net force is to be zero:<br />
F B = F E<br />
−→ evB = eE<br />
Or, solving for B:<br />
B = E v<br />
For a parallel plate capacitor: E = σ/ɛ 0 , so<br />
B =<br />
σ Q<br />
ɛ 0 v = A<br />
ɛ 0 v<br />
7.5 × 10 −6 C<br />
−→ B =<br />
(0.2m) 2 (8.85 × 10 −12 C 2<br />
Nm<br />
)(4 × 10 6 m<br />
2 s ) = 5.3T<br />
So, ⃗ B must be perpendicular to the electric field, pointing into the page<br />
with a magnitude of 5.3T for the electon to have no net force on it.<br />
Example<br />
A proton is injected from a region with no magnetic field into a region<br />
with a strip of uniform magnetic field that is 10cm wide:<br />
+e<br />
v = ?<br />
L = 10cm<br />
B = 0.3T<br />
The magnetic field points into the page and has a magnitude of 0.3T.<br />
What is the maximum velocity the proton can have such that it does<br />
not make it across the 10cm magnetic-field region and exit through the<br />
other side?<br />
Upon entering the magnetic field, the proton will experience a force<br />
perpendicular the velocity, which will cause it to follow a circular path:
90 Magnetic Fields 4.6<br />
+e<br />
v = ?<br />
r<br />
L = 10cm<br />
B = 0.3T<br />
Let’s compute the radius of the path in terms of the velocity. Newton’s<br />
second law applied to the proton yields<br />
F B = m v2<br />
r<br />
evB = m v2<br />
r<br />
−→ r = mv<br />
eB<br />
So as v increases so does r. In order that the proton not exit through<br />
the other side of the magnetic field region we must have r < L;<br />
mv<br />
eB < L<br />
−→ v < eBL<br />
m<br />
= 2.9 × 106 m s<br />
Example<br />
A straight section of wire that is 0.5m long and carrying a current of<br />
I = 8A directed along the +x axis. One end of the wire is at the origin<br />
. The wire is in a magnetic field<br />
⃗B = ĵ<br />
(3 T m 2 )<br />
x 2 + ˆk<br />
What is the force on the wire?<br />
Look at a small section of the wire:<br />
y<br />
( )<br />
2 T x.<br />
m<br />
x<br />
dx<br />
x<br />
z
4.6 More Examples 91<br />
The force on this small section is:<br />
dF ⃗ = Id⃗x × B ⃗<br />
= (Idx)î ×<br />
( )<br />
ĵ<br />
(3 T m 2<br />
= (Idx)<br />
(ˆk(3x 2 ) − ĵ(2x)<br />
x 2 + ˆk<br />
(<br />
)<br />
2 T m<br />
) )<br />
x<br />
Integrate over the entire length of the wire to get the total force:<br />
0.5m ∫<br />
)<br />
⃗F = (Idx)<br />
(ˆk(3x 2 ) − ĵ(2x)<br />
0<br />
= I [ î(x 3 ) − ĵ(x 2 ) ] 0.5m<br />
0<br />
= (8A) [ (0.5) 3 î − (0.5) 2 ĵ ] [T · m]<br />
= (1N)î − (2N)ĵ<br />
Example<br />
An L-shaped section of wire, carryies a current I, and is in a constant<br />
magnetic field, as shown in the figure:<br />
L 1<br />
L 2<br />
I<br />
B<br />
In terms of the parameters shown, what is the direction and magnitude<br />
of the net magnetic force on the wire?<br />
Use the right hand rule to find the direction of the force on each segment:<br />
L 1<br />
L 2<br />
F 1<br />
F 2<br />
I<br />
B
92 Magnetic Fields 4.6<br />
The net force is<br />
⃗F = îF 1 − ĵF 2<br />
= îIL 1 B − ĵIL 2 B<br />
So the direction is given by<br />
( ) ( )<br />
θ = tan −1 IL2 B<br />
= tan −1 L2<br />
,<br />
IL 1 B<br />
L 1<br />
where θ is below the x-axis. The magnitude of the force is<br />
√<br />
F =<br />
√F1 2 + F 2 2 = IB L 2 1 + L2 2<br />
Example<br />
What is the net force on an a × b rectangular loop of wire carrying a<br />
current I that is in a uniform magnetic field perpendicular to the loop’s<br />
plane?<br />
Assume a direction for the magnetic field and use the right hand rule<br />
to draw the forces on each segment of the loop:<br />
a<br />
F 1<br />
b<br />
F 2<br />
F 3<br />
F 4<br />
B<br />
The net force on the loop is:<br />
⃗F = ⃗ F 1 + ⃗ F 2 + ⃗ F 3 + ⃗ F 4<br />
= î(F 2 − F 4 ) + ĵ(F 3 − F 1 )<br />
= î(IbB − IbB) + ĵ(IaB − IaB)<br />
= 0<br />
I<br />
Example<br />
Show that the magnetic force due to a uniform magnetic field for any<br />
closed current loop is zero.<br />
The force law states<br />
∮<br />
⃗F = I d ⃗ l × B ⃗
4.7 Homework 93<br />
Imagine the closed loop to be divided up into many d ⃗ l i , approximated<br />
by a many-sided polygon:<br />
B<br />
I<br />
So, the integral is just the limit of a sum over all of the small segments:<br />
∮<br />
d ⃗ l × ⃗ B = d ⃗ l 1 × ⃗ B + d ⃗ l 2 × ⃗ B + · · ·<br />
= (d ⃗ l 1 + d ⃗ l 2 + · · ·) × B ⃗ (∮ )<br />
= d ⃗ l × B ⃗<br />
⃗B is outside the integral because it is constant. The vector sum indicated<br />
by the integral in the last line must vanish since the vectors form<br />
a close polygon; graphical vector addition yields zero. Thus,<br />
∮<br />
⃗F = I d ⃗ l × B ⃗<br />
(∮ )<br />
= I d ⃗ l × B ⃗<br />
= 0<br />
§ 4.7 Homework<br />
⊲ Problem 4.7<br />
Consider an electron near the equator. In which direction does it tend<br />
to deflect by the magnetic field of the earth if its velocity is directed<br />
(a) downward, (b) northward, (c) westward, or (d) southeastward?<br />
⊲ Problem 4.8<br />
An electron moving along the positive x axis perpendicular to a magnetic<br />
field experiences a magnetic deflection in the negative y direction.<br />
What is the direction of the magnetic field?<br />
⊲ Problem 4.9<br />
An electron in a uniform electric and magnetic field has a velocity of<br />
1.2×10 4 m s in the positive x direction and an acceleration of 2.0×1012 m s 2
94 Magnetic Fields 4.7<br />
in the positive z direction. If the electric field has a strength of 20N/C<br />
in the positive z direction, what is the magnetic field in the region?<br />
⊲ Problem 4.10<br />
A duck flying due north at 15 m s<br />
passes over Atlanta, where the Earth’s<br />
magnetic field is 5.0 × 10 −5 T in a direction 60 ◦ below the horizontal<br />
line running north and south. If the duck has a net positive charge of<br />
0.040µC, what is the magnetic force acting on it?<br />
⊲ Problem 4.11<br />
A proton moves with a velocity of v = (2î − 4ĵ + ˆk) m s<br />
in a region in<br />
which the magnetic field is B ⃗ = (î + 2ĵ − 3ˆk)T. What is the magnitude<br />
of the magnetic force this charge experiences?<br />
⊲ Problem 4.12<br />
Show that the work done by the magnetic force on a charged particle<br />
moving in a magnetic field is zero for any displacement of the particle.<br />
⊲ Problem 4.13<br />
Below is shown a cube (40cm on each edge) in a magnetic field with a<br />
wire carrying a current I = 5.0A over the surface of the cube. If there<br />
is a magnetic field B ⃗ = 0.020Tĵ. What is the magnitude and direction<br />
of the magnetic force on each straight segment of the current loop?<br />
y<br />
B<br />
d<br />
a<br />
I<br />
z<br />
c<br />
b<br />
x<br />
⊲ Problem 4.14<br />
A singly charged positive ion has a mass of 3.20×10 −26 kg. After being<br />
accelerated through a potential difference of 833V, the ion enters a<br />
magnetic field of 0.920T along a direction perpendicular to the direction<br />
of the field. Calculate the radius of the path of the ion in the field.<br />
Find the velocity of the particle first then use the circular motion<br />
result.<br />
⊲ Problem 4.15<br />
A proton (charge +e, mass m p ), a deuteron (charge +e, mass 2m p ),<br />
and an alpha particle, (charge +2e, mass 4m p ) are accelerated through<br />
a common potential difference, V . The particles enter a uniform magnetic<br />
field B, in a direction perpendicular to B. The particles move in
4.7 Homework 95<br />
circular orbits. Write the radii of the orbits of the deuteron, r d , and<br />
the alpha particle, r α , in terms of the radius of the protons orbit r p .<br />
⊲ Problem 4.16<br />
Indicate the initial direction of the deflection of the charged particles<br />
as they enter the magnetic fields shown below.<br />
(a)<br />
(b)<br />
+ -<br />
(c)<br />
(d)<br />
+<br />
⊲ Problem 4.17<br />
A proton moves through a uniform electric field ⃗ E = 50ĵV/m and a<br />
uniform magnetic field ⃗ B = (0.20î + 0.30ĵ + 0.40ˆk)T. Determine the<br />
acceleration of the proton when it has a velocity ⃗v = 200î m s .<br />
+
96 Magnetic Fields 4.8<br />
§ 4.8 Summary<br />
Definitions<br />
Facts<br />
Force on a moving charge:<br />
⃗F = q⃗v × ⃗ B<br />
Force on a piece of wire carrying a current I:<br />
⃗ dF = I ⃗ dl × ⃗ B<br />
Theorems
5.1 Sources of Magnetic Field 97<br />
5 Sources of Magnetic Field<br />
§ 5.1 Sources of Magnetic Field<br />
So far we have not considered how magnetic fields are created. As<br />
you may know it is possible to create an electromagnet by wrapping<br />
a wire around a nail and running a current through the wire. The<br />
precise way in which a current produces a magnetic field is captured in<br />
the following law.<br />
Fact: Biot-Savart Law<br />
If you have a wire carrying a current I, a small section of the wire<br />
⃗dl will produce a magnetic field<br />
dB ⃗ = µ 0I ⃗dl × ˆr<br />
4π r 2<br />
where ⃗r is the vector that points from the location of the element<br />
⃗dl to the field point. The total field is the sum of the contributions<br />
of each element of the wire.<br />
This may look complicated but it represents the simple fact that<br />
the magnetic field wraps around the wire, and decreases in strength as<br />
you move away. In the figure below is depicted two magnetic field lines<br />
that are due to the current element ⃗ dl indicated.<br />
Note that the field warps around the extension of the current element<br />
in circles. This figure does not depict how the field strength changes<br />
with position. The direction the field wraps around the current is the<br />
same direction the fingers of your right hand will wrap around the wire<br />
if you grasp the wire with your thumb pointing in the direction of the<br />
current.
98 Sources of Magnetic Field 5.1<br />
In order to apply the Biot-Savart law one must first parameterize<br />
the curve that the wire follows. A parameterization for a curve is a<br />
vector function ⃗r(t) of one variable t, such that ⃗r(t) points from the<br />
origin to the curve and sweeps along the curve as the variable t is<br />
increased. Below is picture such a vector function at the values of t.<br />
The variable t is not the time, but it is sometimes helpful to think of<br />
it as the time, so that you can imagine ⃗r(t) is the position of some<br />
particle that is following the curve.<br />
As an example consider the following vector function of t.<br />
⃗r(t) = a cos tî + a sin tĵ<br />
This is a parameterization of a circle of radius a centered at the origin.<br />
There are other parameterizations of a circle. For example ⃗r(t) =<br />
a cos tî − a sin tĵ is also a parameterization of a circle, but as t increases<br />
the vector sweeps clockwise rather than counterclockwise.<br />
⊲ Problem 5.1<br />
For the parameterization of a circle ⃗r(t) = a cos tî + a sin tĵ, over what<br />
range would you need to vary t in order to have the vector sweep over<br />
the quarter circle in the second quadrant?<br />
Here are some other parameterizations.<br />
Straight Line :<br />
Helix :<br />
Spiral :<br />
Vertical Parabola :<br />
Horizontal Parabola :<br />
⃗r(t) = atî + btĵ + cˆk<br />
⃗r(t) = a cos tî + a sin tĵ + btˆk<br />
⃗r(t) = t cos tî + t sin tĵ<br />
⃗r(t) = ⃗a + btî + ct 2 ĵ<br />
⃗r(t) = ⃗a + bt 2 î + ctĵ
5.1 Sources of Magnetic Field 99<br />
⊲ Problem 5.2<br />
Write a parameterization for a parabola that goes through the three<br />
points (-1,0), (0,1) and (1,0).<br />
With the parameterization in hand we can proceed with a computation<br />
using the Biot-Savart law. First we notice that if we change t<br />
a little bit dt that the difference between ⃗r(t) and ⃗r(t + dt) is a short<br />
vector along the curve.<br />
So we see that the small section along the curve ⃗ dl that appears in the<br />
Biot-Savart law can be found from our parameterization. Further since<br />
we can write<br />
⃗dr = ⃗r(t + dt) − ⃗r(t) =<br />
⃗r(t + dt) − ⃗r(t) dr<br />
dt = ⃗ dt<br />
dt dt<br />
⃗dl = ⃗ dr<br />
dt dt<br />
This was the main point of the parameterization, now we can “add up”<br />
the contributions from all the section by integrating over the parameter<br />
t.<br />
We still need to do another step before we can write out the integral.<br />
We need to write out the vector that points from the current<br />
element ⃗ dl to the field point. So that we do not get the different vectors<br />
mixed up, let us call the vector that points from the origin to the field<br />
point ⃗r f , and let us call the parameterization of the current path ⃗r s (t).<br />
Then the vector that points from the current element to the field point<br />
can be written as<br />
⃗r = ⃗r f − ⃗r s (t)
100 Sources of Magnetic Field 5.1<br />
Now we can write out the Biot-Savart integral in terms of the<br />
parameterization.<br />
⃗B(r f ) = µ ∫<br />
0I ⃗dl × ˆr<br />
4π r 2<br />
= µ ∫<br />
0I ⃗dl × ⃗r<br />
4π r 3<br />
= µ ∫ drs ⃗<br />
0I<br />
dt<br />
× [⃗r f − ⃗r s (t)]<br />
4π |⃗r f − ⃗r s (t)| 3 dt<br />
This gives us a system by which we can find the field at any point due<br />
to any current that we can parameterize. Of course it is very rare that<br />
the integral has an analytic solution, but one can use a computer to<br />
evaluate the integral numerically once you have used this system to<br />
write out the integral.<br />
In order to see exactly what all this means let us do an example.<br />
Example<br />
Suppose that we have a current of 15 amps going through a section of<br />
wire that follows the curve below, which has the following parameterization.<br />
⃗r s (t) = (−t + t 3 )î + t 2 ĵ<br />
Where the distance is in meters, and t goes from -0.9 to 0.9.<br />
We want to find the field at the point ⃗r f = 0î + 1ĵ. In preparation for<br />
plugging into the Biot-Savart law we compute the following quantities.<br />
⃗r f − ⃗r s (t) = (t − t 3 )î + (1 − t 2 )ĵ = (1 − t 2 )(tî + 1ĵ)<br />
|⃗r f − ⃗r s (t)| 2 = (1 − t 2 ) 2 (t 2 + 1)<br />
|⃗r f − ⃗r s (t)| 3 = (1 − t 2 ) 3 (t 2 + 1) 3/2<br />
and<br />
⃗ dr s<br />
dt = (−1 + 3t2 )î + 2tĵ
5.2 Ampere’s Law 101<br />
and<br />
⃗ dr s<br />
dt × [⃗r f − ⃗r s (t)] = [ (−1 + 3t 2 )î + 2tĵ ] × [ (1 − t 2 )(tî + 1ĵ) ]<br />
= (1 − t 2 ) [ (−1 + 3t 2 )î + 2tĵ ] × [(tî + 1ĵ)]<br />
= (1 − t 2 ) [ (−1 + 3t 2 ) − 2t 2] ˆk<br />
= −(1 − t 2 ) 2ˆk<br />
Now putting this into the parameterized form of the Biot-Savart law<br />
we find:<br />
⃗B(⃗r f ) = µ 0I<br />
4π<br />
= µ 0I<br />
4π<br />
∫<br />
⃗ drs<br />
dt<br />
∫ 0.9<br />
= − µ 0I<br />
4π ˆk<br />
−0.9<br />
∫ 0.9<br />
× [⃗r f − ⃗r s (t)]<br />
|⃗r f − ⃗r s (t)| 3 dt<br />
−(1 − t 2 ) 2ˆk<br />
(1 − t 2 ) 3 (t 2 + 1)<br />
−0.9<br />
= − µ 0I<br />
4π ˆk [1.9367m −1 ]<br />
= −(2.90mT)ˆk<br />
3/2<br />
dt<br />
1<br />
dt<br />
(1 − t 2 )(t 2 + 1)<br />
3/2<br />
⊲ Problem 5.3<br />
Consider a straight section of wire along the x-axis that goes from x = a<br />
to the x = b. The wire carries a current I. What is the magnetic field<br />
at the position ⃗r f = yĵ.<br />
⊲ Problem 5.4<br />
A circular section of wire is carrying a current I counterclockwise. The<br />
arc of the wire subtends an angle of θ. The circle has a radius a. What<br />
is the magnetic field due to this section of wire at the center of the<br />
circle?<br />
!<br />
§ 5.2 Ampere’s Law<br />
In addition to the Biot-Savart law, there is another way of stating<br />
the relationship between a current and the magnetic field it produces.
102 Sources of Magnetic Field 5.2<br />
Fact: Ampere’s Law<br />
For any closed curve the line integral of the magnetic field around<br />
the curve is equal to the µ 0 times the net current through the<br />
surface enclosed by the curve.<br />
∮<br />
∫<br />
⃗B · ⃗dl = µ 0 I through = µ 0<br />
⃗J · dA ⃗<br />
Example<br />
It is not at all apparent that Ampere’s law is equivalent to the Biot-<br />
Savart law, but it is. One can prove the Biot-Savart law from Ampere’s<br />
Law and visa versa. While they are mathematically equivalent, they<br />
are useful in different situations. Ampere’s law is useful for abstract<br />
reasoning about fields, and for finding the field strength in highly symmetric<br />
configurations.<br />
It is important when applying Ampere’s law to keep in mind that<br />
the amperian loop does not correspond to anything physical. There<br />
does not need to be anything there in order for the law to work. You<br />
are free to choose the amperian loop to be any shape you like. Of<br />
course, as was the case with applying Gauss’s law, if you want to use<br />
the law to find the field strength you need to pick the loop correctly.<br />
In order to use Ampere’s Law to find the field strength, you need three<br />
things from the loop.<br />
• The loop must pass through the point at which you want to find the<br />
field strength and be parallel to the field at that point.<br />
• The loop must be either parallel or perpendicular to the field at all<br />
points on the loop.<br />
• In all regions where the loop is parallel to the field, the field must<br />
have the same strength.<br />
Ampere’s law can be used to find the field strength a distance r from<br />
a long straight wire. We will take our loop to be a circle that wraps<br />
around the wire and has a radius r.
5.2 Ampere’s Law 103<br />
This satisfies all three of the above requirements.<br />
parallel to the field at all points on the loop,<br />
⃗B · ⃗dl = Bdl<br />
and since B has the same value at all points on the loop,<br />
∫ ∫<br />
Bdl = B dl = B2πr<br />
But by Ampere’s Law this must also be equal to µ 0 I. Thus<br />
B2πr = µ 0 I −→ B = µ 0I<br />
2πr<br />
Since the loop is<br />
Example<br />
In the previous example we found the magnetic field outside of a long<br />
wire. Now let us find the magnetic field that is inside the wire itself.<br />
The method is much the same. We choose our loop to be a circle of<br />
radius r that wraps around the central axis of the wire, which we has<br />
a radius R. Since we want to find the field inside the wire we will be<br />
working with r < R. We still find the same results that B = µ0I<br />
2πr , but<br />
we must remember the current in this equation is the current that goes<br />
through our loop. This current is not the full current of the wire, since<br />
our loop does not enclose the entire wire. So we need to find the current<br />
through our loop of radius r. Let us assume that the current density<br />
in the wire is uniform. Then we can say that the total current in the<br />
wire is I = JπR 2 and the current through the loop is I through = Jπr 2 .<br />
Combining these two equations to eliminate J we find that<br />
So that<br />
B = µ 0I through<br />
2πr<br />
I through = r2<br />
R 2 I.<br />
= µ 0I r 2<br />
2πr R 2 = µ 0Ir<br />
2πR 2<br />
The following graph shows the magnetic field both inside and outside<br />
the conductor.
104 Sources of Magnetic Field 5.2<br />
Example<br />
In this example we will see how to deal with a current density that is<br />
not uniform. Suppose that we have a long wire with a current density<br />
that is greater at the center and drops off to zero at the edge: J(r) =<br />
3I<br />
πR<br />
(1 − r/R). We want to find the magnetic field and from Ampere’s<br />
2<br />
law we find B = µ0I through<br />
2πr<br />
. The difficulty is that since J is not uniform<br />
we cannot use I = JA to find I through . We need to use dI = JdA to<br />
find I through = ∫ JdA. This example will show how to do that.<br />
We need to pick the area elements<br />
so that the current density is uniform<br />
over the entire surface of each element.<br />
Since the current density is a function<br />
of r only, J is constant in regions that<br />
are circles around the center. So we will<br />
use a circular element dA, as pictured in<br />
gray in the figure. The figure shows a<br />
cross section of the wire.<br />
R<br />
dA<br />
r+dr<br />
r<br />
In the limit that dr is very small the area will be simply the length<br />
around the circular area element, 2πr times the width of the element<br />
dr so that dA = 2πr dr. Now we can compute the current inside a loop<br />
of radius r loop by integrating.<br />
∫ rloop<br />
∫ rloop<br />
3I<br />
I through = J dA =<br />
(1 − r/R)2πr dr<br />
πR2 0<br />
= I r2<br />
R 2 (3 − 2 r R<br />
−→ B = µ 0I through<br />
2πr<br />
)<br />
0<br />
= µ 0Ir<br />
2πR 2 (3 − 2 r R<br />
)<br />
⊲ Problem 5.5<br />
Consider a long wire where the current density is not uniform but<br />
instead increases as you approach the center of the wire, so that at a
5.4 More Examples 105<br />
distance r from the center the current density is J(r) =<br />
I<br />
2πRr . Note<br />
that this is not a very realistic current density but it is easy to work<br />
with. Find the magnetic field strength both inside and outside of this<br />
wire.<br />
§ 5.3 Force Between Parallel Wires<br />
Now that we know that the strength of the magnetic field produced<br />
by a long straight current carrying wire we can find the force between<br />
two parallel wires.<br />
In the figure above the field produced by current B in the region of<br />
current A is shown. Using the right hand rule we can see that the<br />
direction of the force on wire A is toward wire B. We can also see that<br />
the field and the current are perpendicular so that the magnitude of<br />
the force on a length L of wire A is<br />
F = I A LB = I A L µ 0I B<br />
2πd = µ 0I A I B L<br />
2πd<br />
where d is the distance between the wires.<br />
Theorem: Force Currents<br />
<strong>Two</strong> long parallel wires have a force per length between them of<br />
F<br />
L = µ 0I A I B<br />
2πd<br />
where d is the distance between the currents.<br />
⊲ Problem 5.6<br />
An electrical cable carries 45 amps in each of two wires that are a<br />
distance of 4mm apart. The currents are in opposites directions.<br />
(a) Is the force between the wires attractive or repulsive?<br />
(b) What is the force per length between the wires?<br />
§ 5.4 More Examples
106 Sources of Magnetic Field 5.4<br />
Example<br />
Finite Wire<br />
A straight wire of length a carries a current I. What is the magnetic<br />
field caused by the current at a point (x, y) measured from the center<br />
of the wire?<br />
What is B here?<br />
x<br />
y<br />
The wire is a straight line, which can be parametrized:<br />
⃗r s = (at)î,<br />
where t varies from − 1 2 to + 1 2<br />
. So we have:<br />
d⃗r s<br />
dt = aî<br />
We want to find the magnetic field at the position<br />
⃗r f = xî + yĵ,<br />
Collect the quantities needed for applying the Biot-Savart law:<br />
⃗r f − ⃗r s = (x − at)î + yĵ<br />
Using the Biot-Savart law:<br />
| ⃗r f − ⃗r s | = √ (x − at) 2 + y 2<br />
d⃗r s<br />
dt × (⃗r f − ⃗r s ) = (aî) × ((x − at)î + yĵ)<br />
⃗B(x, y) = µ 0I<br />
4π<br />
= µ 0I<br />
4π<br />
= ayˆk<br />
∫<br />
1<br />
∫2<br />
− 1 2<br />
= − µ 0Iy<br />
4π<br />
d⃗rs<br />
dt × [⃗r f − ⃗r s ]<br />
| ⃗r f − ⃗r s | 3 dt<br />
ayˆk<br />
dt<br />
((x − at) 2 + y 2 ) 3 2<br />
x−<br />
∫<br />
a 2<br />
x+ a 2<br />
du<br />
(u 2 + y 2 ) 3 2<br />
where I have substituted u = x − at into the last expression. Looking<br />
ˆk
5.4 More Examples 107<br />
up the integral:<br />
[<br />
] x+ a<br />
⃗B(x, y) = µ 2<br />
0Iy u<br />
4π y 2√ ˆk<br />
u 2 + y 2 x− a 2<br />
⎛<br />
⎞<br />
= µ 0I x +<br />
⎝<br />
a 2<br />
x − a 2<br />
√<br />
4πy (x )<br />
− √ ⎠<br />
+<br />
a 2 (x )<br />
ˆk<br />
2 + y<br />
2 −<br />
a 2<br />
2 + y<br />
2<br />
The resulting expression gives the magnetic field anywhere in the plane<br />
we’ve chosen as the xy-plane. What about points that lie out of this<br />
plane? Let’s rotate our diagram:<br />
Looking from the side<br />
Looking down the wire<br />
B points out<br />
(x, y)<br />
B<br />
(x, y)<br />
I<br />
I<br />
I points out<br />
Now imagine rotating the xy plane around the wire. Since the distances<br />
x and y remain the same, only the direction of ˆk will change. The<br />
following diagram shows how the position of the magnetic field will<br />
change for different points, that are equidistant from the line:<br />
I<br />
The magnetic field “rotates” around the wire. A quick method to<br />
determine which direction the rotation of the field goes is to use a<br />
modified right-hand rule: Stick your right thumb in the direction of<br />
the current and your fingers will point in the direction of the field’s<br />
rotation.<br />
Example<br />
Using Ampere’s Law we found that the magnetic field strength at a
108 Sources of Magnetic Field 5.4<br />
distance r from an infinite wire was µ0I<br />
2πy<br />
. Let us take the limit of a → ∞<br />
of the results we have for the finite wire, and see if we get the same<br />
results.<br />
⎛<br />
⎞<br />
µ 0 I x +<br />
B = lim ⎝<br />
a 2<br />
x − a 2<br />
√<br />
a→∞ 4πy (x )<br />
− √ ⎠<br />
+<br />
a 2<br />
2 + y<br />
2 (x −<br />
a<br />
2 )2 + y 2<br />
(<br />
= µ a<br />
0I<br />
2<br />
√<br />
4πy (<br />
a<br />
− − a 2<br />
√<br />
2 )2 (−<br />
a<br />
2 )2 )<br />
= µ 0I<br />
2πy<br />
OK<br />
Example<br />
A circular wire with radius a carryies a current I. Locate the circle<br />
in the xy-plane and compute the magnetic field it causes along a line<br />
through its center.<br />
Here is the geometry of the loop:<br />
z<br />
r f<br />
r s<br />
The path along the circle can be parameterized using t = [0, 2π]:<br />
We also have<br />
dl<br />
⃗r s = a cos t î + a sin t ĵ<br />
dr ⃗ s<br />
= −a sin t î + a cos t ĵ<br />
dt<br />
⃗r f = z ˆk<br />
⃗r f − ⃗r s = −a cos t î − a sin t ĵ + z ˆk<br />
√<br />
| ⃗r f − ⃗r s | = a 2 cos 2 t + a 2 sin 2 t + z 2<br />
= √ a 2 + z 2<br />
Compute the cross product needed for the Biot-Savart law:<br />
⃗ dr s<br />
dt × (⃗r f − ⃗r s ) = (−a sin tî + a cos tĵ) × (−a cos tî − a sin tĵ + zˆk)<br />
= az cos tî + az sin tĵ + (a 2 sin 2 t + a 2 cos 2 t)ˆk<br />
= az cos tî + az sin tĵ + a 2ˆk
5.4 More Examples 109<br />
Use the Biot-Savart law:<br />
⃗B = µ ∫<br />
0I<br />
4π<br />
= µ 0I<br />
4π<br />
∫2π<br />
0<br />
⃗ drs<br />
dt × [⃗r f − ⃗r s ]<br />
| ⃗r f − ⃗r s | 3 dt<br />
az cos tî + az sin tĵ + a 2ˆk dt<br />
(a 2 + z 2 ) 3 2<br />
Since the integrals of the cosine and sine vanish for the interval [0, 2π],<br />
the resulting magnetic field is:<br />
µ 0 Ia<br />
⃗B 2 ∫2π<br />
=<br />
ˆk dt<br />
4π (a 2 + z 2 ) 3 2<br />
−→<br />
⃗ B = µ 0I<br />
2<br />
0<br />
a 2<br />
(a 2 + z 2 ) 3 2<br />
ˆk<br />
Example<br />
A semi-circular shaped wire with radius a carryies a current I. What<br />
is the magnetic field caused at the center of the semi-circle’s diameter?<br />
B = ?<br />
a<br />
Divide the wire into small sections and compute the magentic field<br />
contribution due to one small section:<br />
I<br />
dθ<br />
dB<br />
dl<br />
The direction for all sections is up, by the right hand rule. Use the<br />
Biot-Savart law for the magnitude:<br />
d ⃗ B = µ 0<br />
4π<br />
= Iµ 0<br />
4π<br />
Id ⃗ l × ⃗r ′<br />
ˆk<br />
r ′ 3<br />
(dl)(a)<br />
ˆk,<br />
a 3
110 Sources of Magnetic Field 5.4<br />
since d ⃗ l and the radius vector to the point are perpendicular. The<br />
length of dl can be approximated by the length of the arc subtended<br />
by the angle ∆θ, so<br />
|dB| ⃗ = Iµ 0 (adθ)(a)<br />
4π a 3 ,<br />
= Iµ 0 dθ<br />
4π a<br />
To find the magnetic field due to the entire wire, integrate:<br />
B = Iµ 0<br />
4π<br />
∫ π<br />
0<br />
dθ<br />
a = Iµ 0<br />
4a<br />
Example<br />
<strong>Two</strong> sections of straight wire carrying electric current lie parallel to<br />
each other, a distance b apart:<br />
I 2<br />
b<br />
a<br />
I 1<br />
The leftmost ends of the wires are aligned. What is the force that<br />
current I 1 exerts on current I 2 ? Draw a coordinate system:<br />
y<br />
dx'<br />
x'<br />
I 2<br />
b<br />
I 1<br />
x<br />
x=-<br />
a<br />
2<br />
The z-axis points out of the page. From a previous example, we know<br />
that the magnetic field due to I 1 in the xy-plane is<br />
⎛<br />
⎞<br />
⃗B 2 (x, y) = µ 0I 1 x +<br />
⎝<br />
a 2<br />
x − a 2<br />
√<br />
4πy (x )<br />
− √ ⎠ ˆk<br />
+<br />
a 2<br />
2 + y<br />
2 (x −<br />
a<br />
2 )2 + y 2
5.5 Homework 111<br />
The magnetic force on the small segment, dx ′ indicated for I 2 is<br />
dF ⃗ 21 = I 2 (dx ′ î) × B ⃗ 2 (x ′ , b)<br />
⎛<br />
⎞<br />
= µ 0I 1 I 2<br />
x<br />
(î × ˆk) ⎝<br />
′ + a 2<br />
x ′ − a 2<br />
√<br />
4πb<br />
(x′ )<br />
− √ ⎠<br />
+ a 2 (x′ )<br />
dx ′<br />
2 + b<br />
2<br />
− a 2<br />
2 + b<br />
2<br />
⎛<br />
⎞<br />
= µ 0I 1 I 2<br />
4πb<br />
(−ĵ) x<br />
⎝<br />
′ + a 2<br />
x ′ − a 2<br />
√ (x′ )<br />
− √ ⎠<br />
+ a 2 (x′ )<br />
dx ′<br />
2 + b<br />
2<br />
− a 2<br />
2 + b<br />
2<br />
To compute the total force on I 2 , integrate over the length of the wire:<br />
⎛<br />
⎞<br />
⃗F 21 = −ĵ µ ∫<br />
0I 1 I 0<br />
2 x<br />
⎝<br />
′ + a 2<br />
x ′ − a 2<br />
√<br />
4πb (x′ )<br />
− √ ⎠<br />
+ a 2 (x′ )<br />
dx ′<br />
2 + b<br />
2<br />
− a 2<br />
2 + b<br />
2<br />
−a<br />
2<br />
The integrals can be done using simple substitution, and yield:<br />
⃗F 21 = − µ 0I 1 I 2<br />
(√<br />
a2 + b<br />
4πb<br />
2 − b)<br />
ĵ<br />
What is the force exerted on I 1 by I 2 ? Newton’s third law gives us the<br />
answer:<br />
⃗F 12 = µ 0I 1 I 2<br />
(√<br />
a2 + b<br />
4πb<br />
2 − b)<br />
ĵ<br />
Notice that the two wires attract each other. What would happen if<br />
one of the current’s directions was reversed?<br />
They would repel each other<br />
§ 5.5 Homework<br />
⊲ Problem 5.7<br />
Calculate the magnitude of the magnetic field at a point 100 cm from<br />
a long, thin conductor carrying a current of 1.0 A.<br />
⊲ Problem 5.8<br />
A wire in which there is a current of 5.00 A is to be formed into a<br />
circular loop of one turn. If the required value of the magnetic field at<br />
the center of the loop is 10.0µT, what is the required radius?<br />
⊲ Problem 5.9<br />
In Neils Bohr’s 1913 model of the hydrogen atom, an electron circles<br />
the proton at a distance of 5.3 × 10 −11 m with a speed of 2.2 × 10 6 m s .<br />
Compute the magnetic field strength that this motion produces at the<br />
location of the proton.
112 Sources of Magnetic Field 5.5<br />
⊲ Problem 5.10<br />
Determine the magnetic field at a point P located a distance x from<br />
the corner of an infinitely long wire bent at a right angle, as shown.<br />
The wire carries a steady current I.<br />
I<br />
x<br />
P<br />
⊲ Problem 5.11<br />
Consider the current-carrying loop shown below, formed of radial lines<br />
and segments of circles whose centers are at point P . Find the magnitude<br />
and direction of ⃗ B at P .<br />
I<br />
b<br />
⊲ Problem 5.12<br />
<strong>Two</strong> long, parallel conductors separated by 10.0 cm carry currents in<br />
the same direction. If I 1 = 5.00A and I 2 = 8.00A, what is the force<br />
per unit length exerted on each conductor by the other?<br />
⊲ Problem 5.13<br />
Imagine a long cylindrical wire of radius R that has a current density<br />
J(r) = J 0 (1 − r 2 /R 2 ) for r ≤ R and J(r) = 0 for r > R, where r is the<br />
distance from a point of interest to the central axis running along the<br />
length of the wire.<br />
(a) Find the resulting magnetic field inside and outside the wire.<br />
(b) Plot the magnitude of the magnetic field as a function of r.<br />
(c) Find the location where the magnetic field strength is a maximum,<br />
and the value of the maximum field.<br />
⊲ Problem 5.14<br />
Below is shown a cross-sectional view of a coaxial cable. The center<br />
conductor is surrounded by a rubber layer, which is surrounded by<br />
an outer conductor, which is surrounded by another rubber layer. The<br />
current in the inner conductor is 1.00 A out of the page. The current in<br />
the outer conductor is 3.00 A into the page. Determine the magnitude<br />
and direction of the magnetic field at the points indicated at a distance<br />
1mm and 3mm from the center.<br />
a<br />
θ<br />
P
5.5 Homework 113<br />
3.00 A<br />
1.00 A<br />
1 mm 1 mm 1 mm<br />
⊲ Problem 5.15<br />
A long cylindrical conductor of radius R carries a current I. The current<br />
density J is not uniform over the cross-section of the conductor but is a<br />
function of the radius J = br, where b is a constant. find the magnetic<br />
field inside and outside the conductor.
114 Sources of Magnetic Field 5.6<br />
§ 5.6 Summary<br />
Definitions<br />
Facts<br />
Field due to a section of current:<br />
dB ⃗ = µ 0I ⃗dl × ˆr<br />
4π r 2<br />
Theorems<br />
Ampere’s Law:<br />
∮<br />
⃗B · ⃗dl = µ 0<br />
∫<br />
⃗J · ⃗ dA
6.2 Faraday’s Law 115<br />
6 Time Varying Fields<br />
§ 6.1 Introduction<br />
Let us take a moment to review what we know about the means<br />
of producing electric and magnetic fields. We have found two means to<br />
produce fields.<br />
• Electric fields are produced when there are regions in space with a<br />
net electric charge. This is encapsulated in Gauss’s law.<br />
∮<br />
⃗E · dA ⃗ = 1 ∫<br />
ρ dV<br />
ɛ 0<br />
• Magnetic fields are produced when there is a flow of charge. This is<br />
encapsulated in Ampere’s law.<br />
∮<br />
∫<br />
⃗B · ⃗dl = µ 0<br />
⃗J · dA ⃗<br />
So far, we have investigated only steady state fields (fields that don’t<br />
change in time). A net electric charge and a steady flow of charge are<br />
the only means to produce steady state electric and magnetic fields.<br />
But these are not the only means of producing time varying fields. In<br />
this chapter we will investigate time varying electromagnetic systems.<br />
§ 6.2 Faraday’s Law<br />
Let us start by considering what will happen if we have some source<br />
of time varying magnetic field. It could for example be a wire that is<br />
carrying a current and the current is changing rapidly in time. Since<br />
the current is changing in time, the magnetic field that the current<br />
creates will also change with time.<br />
If we place a loop of wire into this time varying<br />
magnetic field, the time varying field will produce<br />
a current in the wire. This is called an induced<br />
current. Nikola Tesla was able to produce such<br />
strong induced currents that he was able to make<br />
a light bulb glow with the current. In the photo<br />
to the right Tesla is holding a light bulb which<br />
has no wires connected to it. The current driving<br />
the bulb is entirely an induced current, which is<br />
caused by the time varying magnetic field.
116 Time Varying Fields 6.2<br />
The induced current is caused by an electric field. This induced<br />
electric field is not like the other electric fields we have seen. This is a<br />
new type of electric field. The electric field is created by the changing<br />
magnetic field.<br />
The electric field wraps around the changing<br />
magnetic field somewhat like a static magnetic field<br />
wraps around a steady current. There are some additional<br />
subtleties due to the fact that the induced<br />
electric field is caushed by the change in the magnetic<br />
field. The relationship between the induced electric<br />
field and the changing magnetic field is stated in the<br />
following law of physics.<br />
Fact: Faraday’s Law<br />
The line integral of the electric field around any closed curve is<br />
equal to negative the rate of change of the magnetic flux through<br />
any surface bounded by the curve.<br />
∮<br />
⃗E · ⃗dl = − d ∫<br />
dt<br />
⃗B · ⃗ dA<br />
Notice that this is very similar to Ampere’s Law:<br />
∮<br />
∫<br />
⃗B · ⃗dl = µ 0<br />
⃗J · dA ⃗<br />
Ampere’s Law and Faraday’s Law both relate the line integral around<br />
closed curve to the flux of a vector field through the area enclosed by<br />
the curve. The difference is that Faraday’s Law has the time derivative<br />
of the flux.<br />
Another important and potentially confusing thing to notice, is<br />
that the line integral of the electric field ( ∫ ⃗ E · ⃗ dl) has so far been<br />
referred to as the electric potential difference between the end points<br />
of the curve. The confusing thing is that since it is a closed curve the<br />
electric potential difference must be zero, ∆V = 0, while Faraday’s Law<br />
tells us that the line integral of the induced electric field is not zero, but<br />
equal to the rate of change of the magnetic flux. So we learn from this<br />
that the force caused by the induced electric field is not a conservative<br />
force field. We also see that there is no electric potential associated with<br />
this electric field, ( ⃗ E ≠ − ⃗ ∇V ). To avoid confusion we will refer to the<br />
line integral of the induced electric field by its historical name, the<br />
electromotive force, or EMF. In equations the EMF is represented by<br />
the symbol E, so that E = ∮ ⃗ E · ⃗ dl. Note that the unit of EMF is the<br />
volt.
6.2 Faraday’s Law 117<br />
Definition: Magnetic Flux<br />
The integral of the magnetic field over the area of the loop is called<br />
the magnetic flux.<br />
∫<br />
φ m = ⃗B · dA ⃗<br />
Note that this is of the same form as the definition of the electric<br />
flux (φ e = ∫ E ⃗ · dA), ⃗ in addition you may have noticed that current<br />
is the flux of current density, (I = ∫ J ⃗ · dA). ⃗ With the definition of<br />
magnetic flux and EMF we can rewrite Faraday’s Law in a simpler<br />
looking form.<br />
Theorem: Faraday’s Law (alternate form)<br />
The induced EMF in a loop is equal to the negative rate of change<br />
of the magnetic flux through the loop.<br />
E = − dφ m<br />
dt<br />
Example<br />
Suppose that we are in a region where the magnetic field is uniform<br />
and increasing with time:<br />
⃗B = ⃗ B 0 + atî + btĵ + ctˆk.<br />
We place a loop of wire with an area of A, so that it lies flat in the x-y<br />
plane. We want to compute the induced EMF in the loop. Since the<br />
loop is in the x-y plane we know that in computing the magnetic flux<br />
( ∫ B ⃗ · dA), ⃗ that the area elements will all be pointing in the z direction:<br />
dA ⃗ = ˆk dA. So ∫ ∫<br />
φ m = ⃗B · dA ⃗ = ( B ⃗ 0 + atî + btĵ + ctˆk) · ˆk dA<br />
∫<br />
∫<br />
= (B 0z + ct) dA = (B 0z + ct) dA = (B 0z + ct)A<br />
So we can compute the induced EMF in the loop.<br />
E = − dφ m<br />
dt<br />
= − d dt [(B 0z + ct)A] = −cA<br />
What can we learn from this example? First we learn that only the<br />
component of the magnetic field that is normal to the surface of the loop<br />
contributes to the magnetic flux. Second we see that no component<br />
of the constant part of the field, ⃗ B 0 , contributes to the induced EMF,<br />
only the time varying part of the field induces an EMF.
118 Time Varying Fields 6.2<br />
Example<br />
Consider now a field that is<br />
⃗B = B 0 cos ωt ˆk<br />
Again we take the loop in the x-y plane.<br />
Because the field is uniform we again get the result that the flux is the<br />
product of the normal component of the field and the area of the loop.<br />
And thus that<br />
φ m = B 0 A cos ωt<br />
E = − dφ m<br />
dt<br />
= B 0 Aω sin ωt<br />
The previous example will be very helpful in understanding the<br />
relationship between the direction of the magnetic field and the direction<br />
of the induced electric field. As we said before the induced electric<br />
field wraps around the magnetic field, but we did not say in which direction<br />
it wraps itself. The above example will help us understand the<br />
direction.<br />
First notice that in the figure above a positive magnetic field is<br />
one that is in the positive z direction, while a positive electric field is<br />
one that is counterclockwise. (This choice of positive directions is in<br />
compliance with another right hand rule, that we will express as Lenz’s<br />
law in just a moment.) Now examine the graph of the electric and<br />
magnetic fields. Sometimes the fields have the same sign and sometimes<br />
they do not. Let us analyze the the graph in four quarters. In the first<br />
quarter notice that the electric field is counterclockwise, the magnetic<br />
field is positive, and the magnitude of the magnetic field is decreasing.<br />
This is represented in symbols in the first row of the table below. The<br />
other three quarters follow.<br />
• First Quarter: E and B ↑ and |B| ↘.<br />
• Second Quarter: E and B ↓ and |B| ↗.
6.3 Maxwell’s Extension of Ampere’s Law 119<br />
• Third Quarter: E and B ↓ and |B| ↘.<br />
• Fourth Quarter: E and B ↑ and |B| ↗.<br />
Here are the same four quarters in diagrams:<br />
Notice that if the magnitude of the flux is decreasing, the fields have<br />
the same sign. While if the magnitude of the flux is increasing, the<br />
fields have opposite sign. This observation is usually stated in terms of<br />
the current that is caused by the induced electric field if a loop of wire<br />
is placed in line with the electric field loop.<br />
Fact: Lenz’s Law<br />
The induced current creates a magnetic field that opposes the<br />
change in the magnetic flux.<br />
⊲ Problem 6.1<br />
Check to see that Lenz’s law is obeyed in all four quarters in the previous<br />
example.<br />
§ 6.3 Maxwell’s Extension of Ampere’s Law<br />
We have seen that a changing magnetic field causes an electric field.<br />
It ends up that a changing electric field will also cause a magnetic field.<br />
∮<br />
∫<br />
⃗B · ⃗dl d<br />
= µ 0 ɛ 0<br />
⃗E · dA<br />
dt<br />
⃗<br />
This magnetic field is added to the magnetic field produced by currents<br />
so that we arrive at an extension of Ampere’s law that makes it<br />
applicable to time varying fields.<br />
Fact: Ampere’s Law - Time Varying<br />
∮<br />
∫<br />
∫<br />
⃗B · ⃗dl = µ 0<br />
⃗J · dA ⃗ d<br />
+ µ 0 ɛ 0<br />
dt<br />
= µ 0 I through + µ 0 ɛ 0<br />
dφ e<br />
dt<br />
⃗E · ⃗ dA
120 Time Varying Fields 6.4<br />
dφ<br />
The quantity ɛ e 0 dt<br />
is sometimes called the displacement current<br />
because it acts like a current in Ampere’s Law.<br />
⊲ Problem 6.2<br />
You have a parallel plate capacitor with circular plates. The capacitor is<br />
being charged so that the electric field between the plates is increasing<br />
at a constant rate: E = at, with a = 4.5 × 10 10 V · m −1 s −1 . Use<br />
the extended version of Ampere’s Law to find the magnetic field at<br />
a distance r = 4.0cm from the center of the capacitor and half way<br />
between the plates. Assume that the field is circular about the axis of<br />
the capacitor.<br />
§ 6.4 Inductance<br />
Suppose that you have a loop of wire in a region where there is no<br />
source of magnetic field. If you run a current through the wire there<br />
will now be a magnetic field due to the current. This magnetic field will<br />
create a magnetic flux through the loop. Because the field produced<br />
is proportional to the current, the flux will also be proportional to the<br />
current.<br />
φ m = LI<br />
The proportionality constant L is called the self inductance of the loop.<br />
The induced EMF on the loop is the product of the inductance and the<br />
rate of change of the current.<br />
E = L dI<br />
dt<br />
Circuit elements are manufactured with specific values of inductance,<br />
and are called inductors. The SI unit of inductance is called the Henry,<br />
abbreviated as H. Inductors are essentially a coil of wire, sometimes<br />
wrapped around a core of material designed to increase the inductance.<br />
The sign of E has been chosen in such a way as to facilitate the application<br />
if Kirchhoff’s loop rule to inductors. It is the same convention<br />
that was used for the resistor.<br />
+ ε –<br />
+ ΔV –<br />
I<br />
For an inductor Lenz’s law can be interpreted as “the induced EMF<br />
across an inductor is in such a direction as to oppose the change in the<br />
current through the inductor”. For example, suppose that a current<br />
is flowing through an inductor, if you now try and reduce the current,<br />
an EMF will be generated in the inductor that will tend to keep the<br />
current going. Similarly if you try to increase the current an EMF will<br />
be generated that will tend to stop the current from increasing.<br />
I
6.4 Inductance 121<br />
Example<br />
Suppose that you have the circuit shown.<br />
By some means you have established a current<br />
in the circuit (perhaps it is an induced current<br />
caused by some external magnetic field) such<br />
+ ε –<br />
that at t = 0 the current is I 0 . At t = 0 the<br />
external cause of the current disappears. What<br />
will happen to the current? – ΔV +<br />
I<br />
Since the inductor opposes the change in the current the current<br />
will not stop abruptly. Instead it will decrease gradually, the inductor<br />
will drive the circuit for a while. Note that this implies that an inductor<br />
can store energy, much like a capacitor can. Let us see how we would<br />
do the circuit analysis of this system.<br />
Kirchhoff’s Loop rule give us that<br />
E + ∆V = 0<br />
−→ L dI<br />
dt + IR = 0<br />
−→ dI<br />
dt = −R L I<br />
−→ I(t) = I 0 e − R L t<br />
Thus we see that the current will decay exponentially.<br />
I<br />
⊲ Problem 6.3<br />
The circuit below is assembled with the switch open. At the time t = 0<br />
the switch is closed.<br />
+ ε –<br />
V S<br />
+<br />
–<br />
I<br />
– ΔV<br />
+<br />
(a) What is the current as a function of time?<br />
(b) What is the EMF as a function of time?<br />
(c) What is the voltage on the resistor as a function of time?<br />
(d) Check to see if Kirchhoff’s loop rule is satisfied at all times.<br />
I
122 Time Varying Fields 6.6<br />
(e) Sketch the graphs of all three functions.<br />
§ 6.5 Alternating Current<br />
Often the currents in a circuit are sinusoidal.<br />
I(t) = I 0 cos ωt<br />
A circuit with such a current is referred to as an alternating current<br />
circuit, or AC circuit, because the current alternates directions in the<br />
circuit. In contrast a circuit with constant current is referred to as a<br />
direct current circuit, or DC circuit.<br />
§ 6.6 AC Power and RMS Voltage<br />
The electrical power supplied by the power company is sinusoidal.<br />
The electric potential between the two sockets of an electrical outlet is<br />
oscillating.<br />
V (t) = V 0 cos ωt<br />
The voltage supplied in the U.S. is said to be 120 volts. If you examine<br />
the underside of an electrical appliance there will be printed something<br />
like<br />
60Hz - 120V - 7.0AMPS<br />
someplace near the UL listing mark. As you may have guessed the first<br />
mark indicate that the appliance is designed to operate on AC power<br />
that has a frequency of 60Hz. You might expect that the amplitude of<br />
the supplied electric potential, V 0 , would be 120V, and the amplitude<br />
of the current drawn by the appliance, I 0 , would be 7.0AMPS. This<br />
is not quite correct. These numbers are not the amplitudes but the<br />
amplitudes divided by √ 2, so that the actual amplitudes are greater<br />
by a factor of √ 2 than these values marked on the appliance. Thus V 0 =<br />
(120V) √ 2 = 170V and I 0 = (7.0A) √ 2=9.9A. In order to understand<br />
why this is so we need to consider the power dissipated in the appliance.<br />
Recall that power is the product of the voltage and the current.<br />
Suppose for simplicity that the appliance that we are using is a toaster.<br />
This is simple because, as an electrical circuit, a toaster is simply a<br />
resistor. So our entire system, including the AC source, is diagramed<br />
as follows.
6.7 AC Circuit Elements 123<br />
+<br />
V S–<br />
R<br />
Because the current oscillates in time, the power dissipated in the resistor<br />
also oscillates in time.<br />
P (t) = I(t)V (t)<br />
= I(t)I(t)R<br />
= [I(t)] 2 R<br />
= [I 0 cos ωt] 2 R<br />
= I 2 0 R cos 2 ωt<br />
Notice that the power oscillates between zero and I 2 0 R so that the<br />
average power supplied to the toaster is 1 2 I2 0 R.<br />
P avg = I2 0 R<br />
= I 0 I<br />
√ 0 R √2 = I 0 V<br />
√ √ 0<br />
= (7.0A)(120V)<br />
2 2 2 2<br />
Notice that for a sinusoidal signal the root mean square (RMS) value is<br />
the amplitude divided by √ 2. Thus we can learn two things from this<br />
example. First, the values that are printed on the toaster are the RMS<br />
values. Second, the product of the RMS current and RMS voltage gives<br />
the average power. So in this example the average power consumed by<br />
the appliance is P = (7.0A)(120V) = 840W.<br />
⊲ Problem 6.4<br />
A 60 watt light bulb has an average power output of 60 watts.<br />
(a) What is the peak power?<br />
(b) What is the resistance of the bulb?<br />
(c) Is this resistance the same as the resistance that is required to<br />
dissipate 60 watts when connected to a 120 volt DC source?<br />
§ 6.7 AC Circuit Elements<br />
In many ways an AC circuit can be analyzed using the same techniques<br />
we used for DC circuits. In fact inductors and capacitors become<br />
as simple as resistors. In an AC circuit, inductors and capacitors<br />
follow an adapted Ohm’s Law: the amplitude of the voltage on the
124 Time Varying Fields 6.7<br />
element is proportional to the amplitude of the current through the<br />
element, V = ZI, just like a resistor. The proportionality constant Z<br />
is called the impedance.<br />
Let us see how this works for an inductor.<br />
Suppose that the current through an inductor is I(t) = I 0 cos ωt.<br />
Then we know that the EMF (V L ) on the inductor is<br />
V L = L dI<br />
dt = −ωLI 0 sin ωt<br />
So that the amplitude of the voltage oscillation is ωLI 0 .<br />
V L0 = ωLI 0 = Z L I 0 with Z L ≡ ωL<br />
So we see that the amplitudes of the voltage and current are proportional.<br />
The constant Z L = ωL is the equivalent of the resistance for an<br />
inductor.<br />
Theorem: Impedance: Inductor<br />
In an AC circuit the amplitude of the voltage on an inductor is<br />
proportional to the amplitude of the current flowing through the<br />
inductor.<br />
V L0 = Z L I 0 with Z L ≡ ωL<br />
The impedance of an inductor is Z L = ωL.<br />
The actual voltage and current are not proportional, since one is<br />
a sine function and the other is the cosine function.<br />
V L<br />
The voltage on an inductor reaches the peak value one quarter of a cycle<br />
before the current does. For this reason the voltage on an inductor in<br />
an AC circuit is said to lead the current by a phase of 90 ◦ .<br />
A capacitor is similar.<br />
V C = 1 C Q = 1 ∫<br />
Idt = 1<br />
C ωC I 0 sin ωt<br />
1<br />
So that the amplitude of the voltage oscillation is<br />
ωC I 0.<br />
V C0 = 1<br />
ωC I 0 = Z C I 0 with Z C ≡ 1<br />
ωC<br />
So we see that the amplitudes of the voltage and current are proportional.<br />
The constant Z C = 1<br />
ωC<br />
is the equivalent of the resistance for<br />
an capacitor.
6.8 Phasor Diagrams 125<br />
Theorem: Impedance: Capacitor<br />
In an AC circuit the amplitude of the voltage on a capacitor is<br />
proportional to the amplitude of the current flowing through the<br />
capacitor.<br />
V C0 = Z C I 0 with Z C ≡ 1<br />
ωC<br />
The impedance of an capacitor is Z C = 1<br />
ωC .<br />
The voltage on a capacitor reaches the peak value one quarter of a cycle<br />
after the current does. For this reason the voltage on a capacitor in an<br />
AC circuit is said to follow the current by a phase of 90 ◦ .<br />
§ 6.8 Phasor Diagrams<br />
This relationship between the leading and following phases is often<br />
represented in a phasor diagraph. As you may recall an oscillation can<br />
be thought of as the horizontal component of a circular motion. So we<br />
can represent the current or voltage in and AC circuit as the horizontal<br />
component of a circular motion.<br />
The current is represented as a vector with constant length I 0 that is<br />
rotating in the counter clockwise direction. The real physical current<br />
is the projection of this vector onto the horizontal (Re) axis.<br />
One thing that is very nice about the phasor representation is that<br />
it allows us to clearly represent in a diagram the phase relationship between<br />
different quantities. Let us take the leading phase of the inductor
126 Time Varying Fields 6.8<br />
voltage as an example. The fact that the inductor voltage is one quarter<br />
of a cycle ahead of the current, means that the inductor voltage<br />
phasor is 90 ◦ ahead of the current phasor.<br />
If you imagine the phasors rotating you can see that the projection of<br />
the voltage, onto the horizontal axis, will reach a peak one quarter of<br />
a cycle before the projection of the current reaches its peak.<br />
Here is the phasor diagram for a capacitor.<br />
At this point the phasor diagram has only allowed us to represent<br />
what we already know. The phasor diagram is far more useful. For<br />
example the phasor diagram will alow us to use Kirchhoffs loop rule<br />
for AC circuits, as we will see in the following example. The idea of a<br />
phasor is useful in many areas of physics and engineering. We will see<br />
more examples in the next chapter.<br />
Example<br />
Suppose that we have a capacitor and a resistor in series. What is the<br />
effective impedance of the combination. First imagine that we connect<br />
the combination to an AC source.<br />
+<br />
+<br />
V S–<br />
C<br />
R<br />
+<br />
V R<br />
–<br />
V C<br />
–
6.8 Phasor Diagrams 127<br />
What we want to find is the ratio of the amplitude of the supply voltage<br />
and the amplitude of the current: V S0 /I 0 .<br />
Kirchhoff’s loop rule gives us that<br />
V S (t) − V C (t) − V R (t) = 0 −→ V S (t) = V C (t) + V R (t)<br />
In the phasor diagram this implies that the sum of the phasors for V C<br />
and V R must be equal to the phasor of the source voltage V S . Since<br />
the resistor phasor is parallel to the current phasor we know that the<br />
phasors for V C must follow the phasor for V R by 90 ◦ . Thus the sum<br />
(also V S ) forms the hypotenuse of a right triangle.<br />
We can find the phasor for the sum of the voltage on the resistor and<br />
capacitor by adding the individual phasors like we add vectors. Also<br />
notice from the circuit diagram that the sum of the voltages on the<br />
resistor and capacitor is equal to the voltage on the supply: V S =<br />
V R + V C .<br />
Since the lengths of the phasors are the amplitudes of the voltages we<br />
can use the pythagorian theorem to find that<br />
V 2 S 0<br />
= V 2 R 0<br />
+ V 2 C 0<br />
= (RI 0 ) 2 + (Z C I 0 ) 2<br />
−→ V S 2 0<br />
I0<br />
2 = R 2 + ZC<br />
2 √<br />
−→ Z eff = V √<br />
S 0<br />
= R<br />
I 2 + ZC 2 = R 2 + 1<br />
0 (ωC) 2<br />
We see that the impedances do not simply add when the components<br />
are in series. In this case the square of the impedance was the sum of<br />
the squares of parts. This is not a general rule. We can also find the<br />
phase angle (φ) between the supply voltage and the current.<br />
tan φ = −V C 0<br />
= −Z CI 0<br />
= −Z C<br />
V R0 RI 0 R = −1<br />
ωRC<br />
There is a very significant difference between the resistance of a<br />
resistor and the impedance of a capacitor or inductor: the impedance<br />
depends on the frequency. In the previous example we see that the
128 Time Varying Fields 6.8<br />
effective impedance of the system depends on ω. This can be a useful<br />
property, since sometimes we wish to treat different frequencies differently.<br />
For example in an audio system, there are big speakers for<br />
low frequencies and small speakers for high frequencies, but the signal<br />
coming from the amplifier to the speakers contains both high and low<br />
frequencies. Let us see how we can use a capacitor and resistor is series<br />
to split the signal so that the appropriate signal goes to each speaker.<br />
Example<br />
For the circuit in the previous example, let us find the voltage on the<br />
capacitor and resistor relative to the voltage from the source.<br />
G R (ω) ≡ V R 0<br />
= V R 0<br />
/I 0 R<br />
= √<br />
V S0 V S0 /I 0 R2 + 1/(ωC) = 1<br />
√ 2 1 + 1/(ωRC)<br />
2<br />
G C (ω) ≡ V C 0<br />
= V C 0<br />
/I 0 1/ωC<br />
= √<br />
V S0 V S0 /I 0 R2 + 1/(ωC) = 1<br />
√ 2 (ωRC)2 + 1<br />
These ratios are called the gain.<br />
So we see that the voltage on the capacitor is equal the the source<br />
voltage at low frequencies and drops to zero at high frequencies, while<br />
the voltage on the resistor does just the reverse. So we see that the<br />
signal from the source is sorted by this circuit, high frequencies to the<br />
capacitor and low frequencies to the resistor. Thus we can send the<br />
voltage from the capacitor to the large speakers and the voltage from<br />
the resistor to the small speakers.<br />
⊲ Problem 6.5<br />
Suppose that you build a RL circuit instead of the RC, that is you<br />
replace the capacitor with an inductor in the previous example.<br />
(a) Find the gain for the resistor and inductor.<br />
(b) Graph the gain for the resistor and inductor.<br />
(c) Which will pass low frequencies better than high frequencies.
6.9 Homework 129<br />
⊲ Problem 6.6<br />
Show that this LRC circuit √ has a maximum<br />
current when ω =<br />
1<br />
LC<br />
. This is<br />
called the resonance frequency.<br />
L<br />
+<br />
V S<br />
–<br />
R<br />
C<br />
§ 6.9 Homework<br />
⊲ Problem 6.7<br />
Consider the hemispherical closed surface<br />
of radius R as shown . If the hemisphere<br />
is in a uniform magnetic field that makes an<br />
angle θ with the vertical, calculate the magnetic<br />
flux through the flat surface S 1 . Calculate<br />
the flux through the hemisphical surface<br />
S 2 .<br />
S 1<br />
S 2<br />
θ<br />
B<br />
⊲ Problem 6.8<br />
A powerful electromagnet has a field of 1.6T and a cross-sectional area<br />
of 0.20m 2 . If we place a coil having 200 turns and a total resistance<br />
of 20Ω around the electromagnet and then turn off the power to the<br />
electromagnet in 20ms, what is the current induced in the coil?<br />
⊲ Problem 6.9<br />
A rectangular loop of area A is placed in a region where the magnetic<br />
field is perpendicular to the plane of the loop. The magnitude of the<br />
field is allowed to vary in time according to B = B o e −t/τ . What is the<br />
induced emf as a function of time?<br />
⊲ Problem 6.10<br />
A long, straight wire carries a current<br />
I = I o sin(ωt+δ) and lies in the plane of a rectangular<br />
loop of N turns as shown. Determine<br />
the emf induced in the loop by the magnetic<br />
field of the wire.<br />
I<br />
a<br />
b<br />
c<br />
⊲ Problem 6.11
130 Time Varying Fields 6.9<br />
A magnetic field directed into the page changes<br />
with time according to B = at 2 + b. The field has<br />
a circular cross-section of radius R. What are the<br />
magnitude and direction of the electric field at a<br />
distance r from the center of the field.<br />
R<br />
⊲ Problem 6.12<br />
For the situation in the previous problem if B = (2.0t 3 − 4.0t 2 + 0.80)T<br />
and R = 2.5cm. Calculate the magnitude and direction of the force<br />
exerted on an electron located at a distance r = 2R from the center of<br />
the field at the time t = 2.0s.<br />
⊲ Problem 6.13<br />
A circular coil enclosing an area of 100cm 2 is made of 200 turns of<br />
copper wire. Initially a 1.10T uniform magnetic field points perpendicularly<br />
upward through the plane of the coil. The direction of the<br />
field then reverses. During the time the field is changing its direction,<br />
how much charge flows through the coil if the resistance of the coil is<br />
R = 5.0Ω.<br />
⊲ Problem 6.14<br />
The rotating loop in an ac generator is a square 10cm on a side. It<br />
is rotated at 60 Hz in a uniform field of 0.80T. Calculate (a) the flux<br />
through the loop as a function of time, (b) the emf induced in the loop,<br />
(c) the current induced in the loop for a loop resistance of 1.0Ω, (d) the<br />
power dissipated in the loop, and (e) the torque that must be exerted<br />
to rotate the loop.<br />
⊲ Problem 6.15<br />
A solenoid has n turns per unit length, radius a and carries a current<br />
I.<br />
Long Solenoid<br />
a<br />
a<br />
b<br />
c<br />
Large loop of wire<br />
Small loop of wire<br />
(a) A large circular loop of radius b > a and N turns encircles the<br />
solenoid at a point far away from from the ends of the solenoid. Find<br />
the magnetic flux through the loop.
6.9 Homework 131<br />
(b) A small circular loop of N turns and radius c < b is completely<br />
inside the solenoid, far from its ends, with its axis parallel to that of<br />
the solenoid. Find the magnetic flux through this small loop.<br />
⊲ Problem 6.16<br />
The two circular loops shown below have their planes parallel to each<br />
other.<br />
A<br />
B<br />
I<br />
As viewed from A toward B, there is a counter-clockwise current in loop<br />
A. Give the direction of the induced current in loop B and determine<br />
whether the loops attract or repel each other if the current in loop A<br />
is (a) increasing and (b) decreasing.<br />
⊲ Problem 6.17<br />
A bar magnet moves with constant velocity along the axis of a loop as<br />
shown in the figure below.<br />
B<br />
S<br />
N<br />
v 0<br />
(a) Make a qualitative graph of the flux φ B through the loop as a<br />
function of time. Indicate the time t 1 when the magnet is halfway<br />
throught the loop.<br />
(b) Sketch a graph of the current I in the loop versus time, choosing I<br />
to be positive when it is counterclockwise as viewed from the left.<br />
⊲ Problem 6.18<br />
Given the circuit below, assume that the switch S has been closed for<br />
a long time so that steady state currents exist in the circuit.<br />
S<br />
10 Ω<br />
10 V<br />
100 Ω 2 Η
132 Time Varying Fields 6.9<br />
Ignore any resistance of the inductor L. (a) Find the battery current,<br />
the current in the 100 Ω resistor, and the current through the inductor.<br />
(b) Find the intial voltage across the inductor when switch S is opened.<br />
(c) Give the current as a function of time measured from the instant<br />
of opening the switch S.<br />
⊲ Problem 6.19<br />
A 2.00H inductor carries a steady current of 0.500A. When the switch<br />
in the circuit is thrown open, the current disappears in 10ms. What is<br />
the average induced emf in the inductor during this time?<br />
⊲ Problem 6.20<br />
A coiled telephone cord has 70 turns, a cross sectional diameter of 1.3<br />
cm, and an unstretched length of 60 cm. Determine an approximate<br />
value for the self inductance of the unstretched cord.<br />
⊲ Problem 6.21<br />
A 10.0mH inductor carries a current I = I max sin ωt, with I max = 5.00A<br />
and ω/2π = 60.0Hz. What is the back emf as a function of time?<br />
⊲ Problem 6.22<br />
The switch in the circuit below is closed at time t = 0. Find the current<br />
in the inductor and the current through the switch as functions of time<br />
if V = 10.0V, R = 4.00Ω and L = 1.00H.<br />
R 2R<br />
V<br />
R<br />
L<br />
S<br />
⊲ Problem 6.23<br />
Show that I = I 0 e −t/τ is a solution of the differential equation<br />
IR + L dI<br />
dt = 0<br />
where τ = L/R and I 0 is the current at t = 0.<br />
⊲ Problem 6.24<br />
An air-coil solenoid with 68 turns is 8.0 cm long and has a diameter of<br />
1.2 cm. How much energy is stored in its magnetic field when it carries<br />
a current of 0.77A.<br />
⊲ Problem 6.25<br />
The switch in the circuit below is connected to point a for a long time.<br />
After the switch is thrown to point b find (a) the frequency of oscillation<br />
in the LC circuit, (b) the maximum charge on the capacitor, (c) the
6.9 Homework 133<br />
maximum current in the inductor, and (d) the total energy stored in<br />
the circuit at time t.<br />
R a b<br />
S<br />
V<br />
C<br />
L<br />
⊲ Problem 6.26<br />
Show that the rms voltage of the pictured “sawtooth” wave is V 0 / √ 3.<br />
V<br />
V o<br />
-V o<br />
⊲ Problem 6.27<br />
What is the resistance of a light bulb that uses an average power of 75W<br />
when connected to a 60Hz power source having a maximum voltage of<br />
170 V? What is the resistance of a 100W bulb?<br />
⊲ Problem 6.28<br />
An inductor has an AC current of frequency 50Hz passing through<br />
it. The maximum voltage is 100V and the maximum current is 7.5A.<br />
What is the inductance of the inductor? The frequency is now changed<br />
to ω new while the voltage is held constant. The maximum current is<br />
now 2.5A. What is the angular frequency ω new ?<br />
⊲ Problem 6.29<br />
A 1.0mF capacitor is connected to a standard wall outlet. Determine<br />
the current in the capacitor at t = (1/180)s, assuming that at t = 0<br />
the energy stored on the capacitor is zero.<br />
⊲ Problem 6.30<br />
For what linear frequencies does a 22.0µF capacitor have a impedance<br />
below 175Ω? Over this same frequency range , what is the impedance<br />
of a 44.0µF?<br />
⊲ Problem 6.31<br />
A sinusoidal voltage v(t) = V max cos ωt is applied to a capacitor. Write<br />
an expression for the instantaneous charge on the capacitor. What is<br />
the instantaneous current in the circuit?<br />
⊲ Problem 6.32<br />
At what frequency does the inductive impedance of a 57µH inductor<br />
equal the capacitive impedance of a 57µF capacitor.<br />
t
134 Time Varying Fields 6.9<br />
⊲ Problem 6.33<br />
A series AC circuit contains the following components: R = 150Ω, L =<br />
250mH, C = 2.00µF, and a generator with V max = 120V operating at<br />
50.0Hz. Calculate the (a) inductive impedance, (b) capacitive impedance,<br />
(c) impedance, (d) maximum current, and (e) phase angle.<br />
⊲ Problem 6.34<br />
An RLC circuit consists of a 150Ω resistor, a 21µF capacitor, and a<br />
460mH inductor, connected in series with a 120V, 60Hz function generator.<br />
What is the phase angle between the current and the applied<br />
voltage? Which reaches its maximum earlier, the current or the voltage?<br />
⊲ Problem 6.35<br />
Calculate the resonance frequency of a series RLC circuit for which<br />
C = 8.40µF and L = 120mH.<br />
⊲ Problem 6.36<br />
An RLC circuit is used in a radio to tune into an FM station broadcasting<br />
at 99.7MHz. The resistance in the circuit is 12.0Ω, and the<br />
inductance is 1.40µH. What capacitance should be used?
6.10 Summary 135<br />
§ 6.10 Summary<br />
Definitions<br />
Facts<br />
Theorems<br />
Ampere’s Law:<br />
∮<br />
⃗B · ⃗dl = µ 0<br />
∫<br />
Faraday’s Law:<br />
∮<br />
∫<br />
⃗J · dA ⃗ d<br />
+ µ 0 ɛ 0<br />
dt<br />
⃗E · ⃗dl = − d dt<br />
∫<br />
⃗B · ⃗ dA<br />
⃗E · ⃗ dA<br />
AC Circuits<br />
Capacitor: The phase of the voltage across a capacitor is 90 ◦ behind<br />
the phase of the current through the capacitor. The amplitude of the<br />
voltage and current are related as follows.<br />
V C 0 = 1<br />
ωC I C 0<br />
Inductor: The phase of the voltage across an inductor is 90 ◦ ahead<br />
of the phase of the current through the inductor. The amplitude of the<br />
voltage and current are related as follows.<br />
V L0 = ωL I L0
136 Wave Optics 7.10
7.1 Maxwell Equations 137<br />
7 Wave Optics<br />
§ 7.1 Maxwell Equations<br />
Let us write down, in one place, all of the fundamental equations<br />
about the electric and magnetic fields.<br />
Gauss ′ s Law<br />
Faraday ′ s Law<br />
Ampere ′ s Law<br />
Gauss ′ s Law for B<br />
∮<br />
∮<br />
∮<br />
∮<br />
⃗E · dA ⃗ = 1 ∫<br />
ɛ 0<br />
⃗E · ⃗dl = − d dt<br />
∫<br />
⃗B · ⃗dl = µ 0<br />
⃗B · ⃗ dA = 0<br />
ρ dV<br />
∫<br />
⃗B · ⃗ dA<br />
⃗J · ⃗ dA + µ 0 ɛ 0<br />
d<br />
dt<br />
∫<br />
⃗E · ⃗ dA<br />
These equations as a group are known as the Maxwell Equations.<br />
The last equation, which we have not discussed before, can be<br />
understood to say that there is no magnetic equivalent to the electric<br />
charge. That is, that there are no magnetic charges, places where<br />
magnetic field lines begin or end, thus that magnetic field lines do not<br />
end. If you want to think of the magnetic field as a flowing fluid, with<br />
B the velocity of the fluid, then this law says that the fluid does not<br />
compress: B flows like water, when it spreads out the velocity decreases,<br />
when it goes through a narrow region it speeds up, the volume rate of<br />
flow is a constant.<br />
Ok, so the new law tells us that there are no magnetic charge, let<br />
us see what the Maxwell equations look like in a region where there is<br />
no electric charge (ρ = 0) and no current (J = 0).
138 Wave Optics 7.1<br />
Gauss ′ s Law<br />
Faraday ′ s Law<br />
Ampere ′ s Law<br />
a new Law<br />
∮<br />
∮<br />
∮<br />
∮<br />
⃗E · dA ⃗ = 0<br />
⃗E · ⃗dl = − d ∫<br />
⃗B · dA<br />
dt<br />
⃗<br />
∫<br />
⃗B · ⃗dl d<br />
= µ 0 ɛ 0<br />
⃗E · dA<br />
dt<br />
⃗<br />
⃗B · ⃗ dA = 0<br />
You can see that the equations are essentially the same in E and B.<br />
Investigating these equations led Maxwell to discover that in regions<br />
where there is no current or charge that there can be traveling waves<br />
as pictured below.<br />
x<br />
y<br />
B<br />
E<br />
E<br />
B<br />
z<br />
time increasing<br />
E<br />
B<br />
E<br />
B<br />
E<br />
B<br />
E<br />
B<br />
B<br />
E<br />
E<br />
B<br />
B<br />
E<br />
E<br />
B<br />
What is drawn is the magnetic field and electric field at various posi-
7.2 Describing Oscillations 139<br />
tions along the z-axis. Notice that the magnitude of the field oscillates,<br />
as one moves along the z-axis. This pattern of spacial oscillations of<br />
the field, travels to the right along the axis as time passes. Also notice<br />
that E and B are perpendicular, and that the direction of travel is<br />
perpendicular to both E and B.<br />
Maxwell also found that the equations predict the velocity at which<br />
these electromagnetic waves travel.<br />
√ 1<br />
v EM =<br />
µ 0 ɛ 0<br />
From the known values of µ 0 and ɛ 0 , Maxwell computed that the velocity<br />
of these EM waves is the same as the velocity of light. He<br />
concluded, rightly, that light and electromagnetic waves are the same<br />
thing. This was a huge leap in our understanding of light.<br />
Fact: Electromagnetic Waves and Light<br />
Light is an electromagnetic wave.<br />
⊲ Problem 7.1<br />
Show that 1/ √ µ 0 ɛ 0 = c.<br />
§ 7.2 Describing Oscillations<br />
We need some terminology to make it easier to talk about sinusoidal<br />
waves and oscillations. The central idea of a wave is encapsulated<br />
in the term phase of an oscillation. The word phase is used here in the<br />
same way as the word is used to describe the cycle of the moon: new<br />
moon, waxing crescent, first quarter, full moon, etc.<br />
In a system with a periodic cycle, the phase of the system describes<br />
what point in the cycle the system is currently occupying.<br />
Recall from the section on AC circuits that we can describe a<br />
sinusoidal oscillation as the horizontal component of a rotation. Thus<br />
the phase of a sinusoidal oscillation can be described by giving the angle<br />
of rotation, φ. When the phase is φ = 0, the oscillation is at its greatest<br />
positive extension. When the phase is φ = π/2 the oscillation is at zero<br />
and with a negative velocity. When the phase is φ = π the oscillation<br />
is at its greatest negative extension. When the phase is φ = 3π/2 the<br />
oscillation is zero but with a positive velocity. These and other phases<br />
of an oscillation are depicted in the figure below.
140 Wave Optics 7.2<br />
-A 0 A<br />
φ = 0<br />
φ = π/4<br />
φ = 2π/4<br />
φ = 3π/4<br />
φ = 4π/4<br />
φ = 5π/4<br />
φ = 6π/4<br />
φ = 7π/4<br />
φ = 8π/4<br />
The actual value of the displacement (x) is the amplitude of the oscillation<br />
times the cosine of the phase angle.<br />
Displacement = Amplitude × cos φ<br />
x = A cos φ
7.3 Describing Waves 141<br />
For a harmonic oscillator the phase increases at a constant rate:<br />
dφ<br />
dt = ω −→ φ = ωt + φ 0<br />
and so the displacement is<br />
x = A cos(ωt + φ 0 )<br />
⊲ Problem 7.2<br />
An oscillator has an amplitude of 3.2. At this instant the displacement<br />
of the oscillator is 1.4. What are the two possible phases of the oscillator<br />
at this instant?<br />
⊲ Problem 7.3<br />
You have a mass connected to a spring.<br />
0 x<br />
You start the mass oscillating in the following ways. What is the initial<br />
phase φ 0 in each case.<br />
(a) Stretch the spring to the right, at t = 0 release the mass.<br />
(b) Compress the spring to the left, at t = 0 release the mass.<br />
(c) At t = 0 strike the mass so that it begins moving to the left.<br />
(d) At t = 0 strike the mass so that it begins moving to the right.<br />
(e) At t = 0 the mass is at the position x 0 = 2.0m and has a velocity<br />
of v 0 = 3.0 m rad<br />
s<br />
(assume that ω = 5.0<br />
s ).<br />
⊲ Problem 7.4<br />
It takes a time of T = 0.025s in order for an oscillator to complete one<br />
cycle. What is the angular frequency (ω) of the oscillator?<br />
§ 7.3 Describing Waves<br />
Consider dropping a rock into a pool of still water. Ripples spread<br />
out from the point at which the rock enters the water. If you examine<br />
the motion of the water at one fixed point, the surface of the water<br />
moves up and down as successive waves move past. The height of<br />
the water at our fixed point is an oscillation. This is true at other<br />
points a well, at each location the height of the water oscillates. Since<br />
an oscillation is so simple, the only thing that can differ between one<br />
point and another is the amplitude of the oscillation and the initial<br />
phase φ 0 . Let us write the height of the water, at position ⃗r and time<br />
t, as y(⃗r, t). Since the oscillation at each point is a harmonic oscillation<br />
we can write:<br />
y(r, t) = A(⃗r) cos [ωt + φ 0 (⃗r)]
142 Wave Optics 7.3<br />
Where the amplitude A(⃗r) and initial phase φ 0 (⃗r) both depend on the<br />
position ⃗r.<br />
In the situation where the wave spreads out from a point source<br />
(such as a rock dropped in the water), the wave travels outward at a<br />
speed v. Because of this, the oscillation at a distance r from the source<br />
will be the same as the oscillation of the source, only delayed by the<br />
time (t d = r/v) it takes for the wave to travel the distance r from<br />
the source to the observation point. Thus the phase of the oscillation<br />
at a position r at a time t + t d , will be the same as the phase of the<br />
oscillation at the source (r = 0) at time t.<br />
φ(r, t + t d ) = φ(0, t)<br />
ω(t + t d ) + φ 0 (r) = ωt + φ 0 (0)<br />
φ 0 (r) = φ 0 (0) − ωt d<br />
φ 0 (r) = φ 0 (0) − ω r v<br />
φ 0 (r) = φ 0 (0) − kr with k ≡ ω v<br />
For convenience we usually pick our zero of time so that φ 0 (0) = 0,<br />
so that φ 0 (r) = −kr and y(r, t) = A(r) cos(ωt − kr).<br />
Theorem: Wave due to a Sinusoidal Point Source<br />
A general wave radiating from a point source can be written down<br />
mathematically as follows:<br />
ψ(r, t) = A(r) cos (ωt − kr)<br />
( 2π<br />
= A(r) cos<br />
T t − 2π )<br />
λ x<br />
where k = ω/v and T ≡ 2π ω and λ ≡ 2π k<br />
are the period and<br />
wavelength as described below.<br />
Definition: Wavelength<br />
If you freeze a sinusoidal wave at one point in time, the distance<br />
you must travel along the wave in order to go through a complete<br />
cycle of the oscillation, is called the wavelength. We use the symbol<br />
λ to represent the wavelength in equations.<br />
ψ<br />
λ<br />
x
7.3 Describing Waves 143<br />
Definition: Period and Frequency<br />
If you stand in one place and allow the wave to pass you, the<br />
time you must wait in order to be back at the same point in the<br />
oscillation as when you started is called the period. We use the<br />
symbol T to represent the period in equations.<br />
ψ<br />
T<br />
t<br />
The frequency is the inverse of the period.<br />
f = 1 T<br />
Note that the angular frequency is ω = 2πf. Both frequencies are<br />
really the same thing, they both measure the rate of the oscillation, it<br />
is just that one is in the units of radians and the other is in the units<br />
of cycles. So you can convert between units by using the fact that one<br />
seconds = radians cycles<br />
cycle seconds<br />
cycle is 2π radians: ω = radians<br />
= 2π f.<br />
Notice that the realtionship between the angular frequency and the<br />
wave number can be rewritten in terms of the wavelength and period.<br />
k = ω 2π/T<br />
−→ 2π/λ = −→ λ = vT<br />
v<br />
v<br />
So we can also think of the wavelength as the distance the wave travels<br />
in one period, or writing the equation as T = λ/v, we can think of the<br />
period as the time it takes to travel one wavelength.<br />
Let us summarize the relationships between the different parameters<br />
that describe a wave.<br />
Time scale:<br />
ω = 2π<br />
T = 2πf<br />
Length scale:<br />
k = 2π λ<br />
Relationship between time and length scales:<br />
k = ω v<br />
OR λ = vT OR λf = v
144 Wave Optics 7.4<br />
⊲ Problem 7.5<br />
For the following wave, graphed at seven different times, determine<br />
the amplitude, wavelength, period, frequency, angular frequency, wave<br />
number, and velocity. Write the function y(x, t) that describes the<br />
wave.
7.4 Electromagnetic Waves 145<br />
§ 7.4 Electromagnetic Waves<br />
With our eyes we are able to see electromagnetic (EM) waves that<br />
have wavelengths between about 400nm to 700nm. We experience light<br />
with different wavelengths as different colors, as indicated in the following<br />
table.<br />
Wavelength Frequency Perceived Color<br />
740 to 625 nm 405 to 480 THz red<br />
625 to 590 nm 480 to 510 THz orange<br />
590 to 565 nm 510 to 530 THz yellow<br />
565 to 520 nm 530 to 580 THz green<br />
520 to 500 nm 580 to 600 THz cyan<br />
500 to 430 nm 600 to 700 THz blue<br />
430 to 380 nm 700 to 790 THz violet<br />
Note that the frequency and wavelength of an EM wave are related by<br />
λf = c where c is the speed of light.<br />
But this is only a small range of EM spectrum.<br />
We see that there are a number of familiar items in the spectrum: x-<br />
rays, microwaves, and radio waves are all EM waves. The range of<br />
the spectrum just above and just below the visible range are called the<br />
ultraviolet (UV) and infrared (IR). The prefixes ultra (above) and infra<br />
(below) refer to the frequency not the wavelength.<br />
⊲ Problem 7.6<br />
Your microwave oven is filled with EM waves with a frequency of about<br />
3 GHz. What is the wavelength of the wave? The microwave oven heats<br />
up the objects in the oven, because the oscillating EM wave causes an<br />
oscillating electric force on the electric dipoles in the object, which<br />
causes the dipoles to oscillate. The frequency of a standard microwave
146 Wave Optics 7.5<br />
oven is tuned to a natural oscillation frequency of the water molecule.<br />
Note that a cellular phone uses about the same frequency as an oven, a<br />
cellular phone is a small microwave emitter, so you are slowly cooking<br />
your brain when you hold the phone to your head.<br />
⊲ Problem 7.7<br />
Compute the wavelength of an FM radio station that transmits at a<br />
frequency of 94.1 MHz.<br />
§ 7.5 Interference of Waves<br />
We have claimed that light is an EM wave because it has the same<br />
speed as an EM wave. We have not looked at the consequences of<br />
the fact that light is a wave. One of the principle characteristics of<br />
waves is their ability to interfere with each other when two or more of<br />
the waves are combined. The following example will demonstrate this<br />
interference.<br />
Consider the follow situation. You have a point source of waves,<br />
and near this source you have a reflecting surface.<br />
Source<br />
Detector<br />
Reflecting Surface<br />
The wave arrives at the detector by two different paths, a direct path<br />
and a reflected path. The wave travels a distance r 1 along the direct<br />
path and a distance r 2 along the reflected path. If the reflected path<br />
was blocked the oscillation at the detector would be just that due to the<br />
direct path, ψ 1 (t) = A 1 cos(ωt + φ 1 ) where φ 1 = −kr 1 . Similarly if the<br />
direct path is blocked the oscillation at the detector would be just that<br />
due to the reflected path, ψ 2 (t) = A 2 cos(ωt + φ 2 ) where φ 2 = −kr 2 .<br />
When both paths are open then the resultant wave at the detector will<br />
be the sum of the waves from each path<br />
ψ(t) = ψ 1 (t) + ψ 2 (t)<br />
= A 1 cos(ωt + φ 1 ) + A 2 cos(ωt + φ 2 )<br />
Using the phasor diagram for these two oscillations we can find the<br />
amplitude of their sum.
7.5 Interference of Waves 147<br />
I<br />
I<br />
A<br />
A 2<br />
The law of cosines gives the amplitude, A, in terms of the amplitudes<br />
φ 2<br />
A 1<br />
A 1<br />
A 2<br />
φ 2 - φ 1<br />
R<br />
φ 1<br />
R<br />
A 1 and A 2 .<br />
A 2 = A 2 1 + A 2 2 + 2A 1 A 2 cos(φ 2 − φ 1 )<br />
Theorem: Addition of <strong>Two</strong> Waves<br />
If two waves (with the same frequency) come together at a detector,<br />
the amplitude A of the resultant wave at the detector will be as<br />
follows.<br />
A 2 = A 2 1 + A 2 2 + 2A 1 A 2 cos(φ 2 − φ 1 )<br />
In general the power carried by a wave is proportional to the square<br />
of the amplitude of the wave, so this result can be rewritten in<br />
terms of the power.<br />
P = P 1 + P 2 + 2 √ P 1<br />
√<br />
P 2 cos(φ 2 − φ 1 )<br />
Below is graphed the power at a detector where two waves come<br />
together, The power of one wave is 25 mW and the power of the other<br />
waves is 4 mW.<br />
P (mW)<br />
0 π 2π 3π 4π<br />
φ 2 - φ 1<br />
Notice the following points:<br />
• The power at the detector can be less with both sources on than with<br />
just one of them on.<br />
• The maximum power occurs when the phase difference is an even<br />
multiple of π. The waves as said to be perfectly in-phase.
148 Wave Optics 7.6<br />
• The minimum power occurs when the phase difference is an odd<br />
multiple of π. The waves as said to be directly out-of-phase.<br />
• When the phase difference is an odd multiple of π/2, the power is<br />
equal to the sum of the powers of the individual signals.<br />
⊲ Problem 7.8<br />
Suppose that you are at home talking on your wireless phone, and you<br />
walk near your refrigerator. Since the refrigerator is made of metal<br />
it acts as a reflector for the microwave signal going to and from your<br />
phone. Because of this there are two paths between your phone and<br />
the transceiver (the base that phone sits in) that it is communicating<br />
with.<br />
x<br />
Phone<br />
Refrigerator<br />
Transceiver<br />
When you are a distance x from the refrigerator the signal that reflects<br />
from the refrigerator must travel a distance 2x further. Assuming that<br />
the reflected amplitude is one half of the direct amplitude, graph the<br />
power of the signal as a function of your distance from the refrigerator.<br />
Assume that the wavelength of the signal is 10cm. Are there places<br />
that it would be better to avoid?<br />
§ 7.6 Interferometer<br />
It is possible to build a device that measures very small motions<br />
by using interference.<br />
The device represented in the diagram below is called an interferometer.
7.7 Interference of <strong>Two</strong> Sources 149<br />
Source<br />
Mirror A<br />
Half-silvered<br />
mirror<br />
Detector<br />
Mirror B<br />
A beam of light strikes a half-silvered mirror. The half silvered mirror<br />
reflects half of the wave toward mirror A and transmits the other half<br />
toward mirror B. The part that strikes mirror A, is reflected back (to<br />
the right) toward the half-silvered mirror where it is again split, half<br />
passing through to the detector on the left and half being reflected back<br />
toward the source. The part of the beam that strikes mirror B is also<br />
split so that half reaches the detector.<br />
So we see that there are two path by which light can reach the<br />
detector, one path that has been reflected from mirror A, and another<br />
path that has been reflected from mirror B. By adjusting the location<br />
of mirror A we can adjust the path length of the light. In this way<br />
the relative phase of path A and B can be adjusted. Suppose that we<br />
have adjusted the path length so that the power at the detector is a<br />
maximum. Then φ A − φ B is an even multiple of π.<br />
Now suppose that we move mirror A a distance of λ 4<br />
to the left.<br />
This will increase the path length of by ∆r A = 2 λ 4 = λ 2<br />
since the path<br />
includes the extra bit of length twice, once on the way to the mirror and<br />
once on the way back from the mirror. But a change in the path length<br />
of ∆r A will change the phase by −k∆r A = − 2π λ<br />
λ 2 = −π. Thus φ A −φ B<br />
is now an odd multiple of π, and the power will be at a minimum. By<br />
observing the intensity of the light at the detector we can easily tell if<br />
it has gone from bright to dim (maximum to minimum). Thus we can<br />
easily detect when the mirror has been moved by λ/4. A common laser<br />
pointer has a wavelength of something like 650nm so one quarter of this<br />
is 162nm. So an interferometer can be used as a “ruler” with markings<br />
about 162nm apart. This is a very small distance, for comparison, the<br />
finest human hair is about 20,000 nm in diameter.
150 Wave Optics 7.8<br />
§ 7.7 Interference of <strong>Two</strong> Sources<br />
Suppose that we have two wave sources that are in phase with each<br />
other.<br />
Detector<br />
r A<br />
r B<br />
Source A<br />
Source B<br />
While the sources are in phase, the waves are not when they reach the<br />
detector because they must travel a different distance: at the detector<br />
the phase difference between the waves will be φ A −φ B = k(r B −r A ) =<br />
k ∆r = 2π ∆r<br />
λ<br />
. The minimum and maximum power occur when this<br />
phase difference is an odd and even multiple of π. That is when<br />
2π ∆r<br />
{ max m is even.<br />
λ = mπ min m is odd<br />
∆r = m λ { max m is even.<br />
2 min m is odd<br />
Theorem: Interference of <strong>Two</strong> Sources<br />
If you have two wave sources that are in phase with each other,<br />
the minimum and maximum of the power occur at points where<br />
the path difference is and odd (min) or even (max) multiple of half<br />
of the wavelength of the wave.<br />
∆r = m λ { max m is even.<br />
2 min m is odd<br />
⊲ Problem 7.9<br />
Draw two points that are 3 cm apart on a piece of paper.<br />
(a) Find all points on the paper that have ∆r = 0.<br />
(b) Find all points on the paper that have ∆r = 1cm.<br />
(c) Find all points on the paper that have ∆r = 2cm.<br />
(d) If λ = 1.0cm what are the locations of the maximum and minimum<br />
of power?<br />
(e) If λ = 2.0cm what are the locations of the maximum and minimum<br />
of power?
7.8 Far Field Approximation 151<br />
§ 7.8 Far Field Approximation<br />
In most cases the detector is placed far from the two sources, far<br />
in the sense that the distance from the sources to the detector r is<br />
much bigger than the distance between the sources, d. There is a<br />
useful approximation for the path difference that can be used when the<br />
detector is far from the source.<br />
First consider the case when the detector<br />
is not far, as pictured in the figure to<br />
Detector<br />
the right. Notice that if we make of a section<br />
of arc, with the center at the detector,<br />
Source<br />
and going through the closest source, and<br />
then consider the section of the path from<br />
the furthest source that is cut off by this<br />
arc. This cut off section (marked as ∆r in<br />
the figure) is equal to the path difference.<br />
Δr<br />
Also notice that the arc is perpendicular to<br />
Source<br />
the path.<br />
Now imagine moving the detector further from the sources, as<br />
pictured in the diagram below. The section of arc that is between the<br />
two paths, subtends a smaller and smaller angle, and thus this section<br />
of arc becomes closer and closer to a straight line. At the same time<br />
the grey region becomes a right triangle.<br />
θ<br />
Let us examining this right triangle<br />
more closely. We see from the diagram<br />
to the right that<br />
sin θ = ∆r −→ ∆r = d sin θ<br />
d<br />
This is a very useful result, that allows<br />
us to find the path difference from the<br />
distance between the sources and the angle<br />
to the detector.<br />
Combining this with our previous result, ∆r = m λ 2<br />
, we find the<br />
d<br />
θ<br />
Δr<br />
θ
152 Wave Optics 7.9<br />
angles for the minima and maxima.<br />
d sin θ m = m λ { max m is even.<br />
2 min m is odd<br />
⊲ Problem 7.10<br />
You set up a sheet of aluminum foil to block a microwave transmitter.<br />
You now punch two holes in the foil, so that the microwaves pass<br />
through the holes. The holes are 10 cm apart. You now place a microwave<br />
detector on a track that is parallel to the foil and 3 meters away.<br />
By sliding the detector along the track you observe that the response<br />
of the detector is as shown in the diagram. What is the wavelength of<br />
the microwave?<br />
Foil with holes<br />
x<br />
120cm<br />
80cm<br />
Source<br />
10 cm<br />
Detector Response<br />
as a function<br />
of position<br />
40cm<br />
0cm<br />
300 cm<br />
§ 7.9 Thin Film Interference<br />
The swirling colors that you see reflected in a<br />
soap bubble and in puddles in a parking lot are a<br />
result of interference. Let us start by considering<br />
the colors in the puddles. The reason that you<br />
only see this in a parking lot is because it only<br />
happens when there is a thin film of oil on the top<br />
of the water. Consider the diagram to the right,<br />
Oil t<br />
of the layers of a puddle. At the bottom we have<br />
the pavement of the parking lot. Next there is the<br />
pool of water. On top of the water is a thin layer of<br />
Water<br />
Pavement<br />
oil. The thickness of the oil has been exaggerated<br />
so that we can see what is happening.<br />
The light from the sun strikes the surface of the oil and some of it<br />
is reflected back toward the detector (your eye). The rest of the light
7.9 Thin Film Interference 153<br />
passes into the oil, were it proceeds until it reaches the water, at which<br />
point part of the light is reflected back toward the detector. The light<br />
that is not reflected passes into the water and eventually strikes the<br />
pavement, where it is mostly absorbed, since the pavement is black. In<br />
the end then, we have light from the sun reflected into the detector,<br />
by two paths, one path is reflected from the air-oil interface and the<br />
other path is reflected from the oil-water interface. If the light is nearly<br />
normal to the surface the path difference will be twice the thickness of<br />
the oil: ∆r = 2t. The light will be strongly reflected when the two<br />
paths are in-phase, that is when ∆r = m λ 2<br />
with m even. Thus in order<br />
for the light to be strongly reflected we need,<br />
2t = m λ 4t<br />
−→ λ =<br />
2<br />
m<br />
So only wavelengths that “match” the thickness of the oil will be reflected.<br />
This is why you see swirling colors, what you are seeing is the<br />
different thicknesses of the oil film, and for each thickness there is a<br />
particular color that gets reflected.<br />
There are a two complications to thin films that we need to consider.<br />
The first complication is that the wavelength of light changes when<br />
it passes into the oil. This is because the light slows down in the oil.<br />
Definition: Index of Refraction<br />
The index of refraction of an optical medium is the ratio of the<br />
speed of light in a vacuum and the speed of light in the medium.<br />
n = c v<br />
Let λ be the wavelength in a vacuum, then λf = c. Let λ ′ be the<br />
wavelength in the medium, then λ ′ f = v. Taking the ratio of these two<br />
equations we find<br />
λf<br />
λ ′ f = c v = n<br />
Solving for λ ′ we find the following result.<br />
Theorem: Wavelength in a Medium<br />
λ ′ = λ n
154 Wave Optics 7.10<br />
The reason that we care the wavelength has changed, is that when<br />
we use the result that ∆r = m λ 2<br />
we need to use the wavelength where<br />
the path difference ∆r occurs. In the case considered above, the extra<br />
bit of path occurs within the oil, so we need to use the wavelength in<br />
the oil.<br />
In order to understand the root cause of the second complication,<br />
consider a soap bubble again. If you make a soap bubble and watch it<br />
carefully, you will notice that just before it pops, it appears that the top<br />
of the bubble has a hole in it. There is not actually a hole, but the top<br />
of the bubble looks like it has a hole because when the bubble gets very<br />
thin, no light is reflected from the surface of the bubble. Let us consider<br />
what our prediction would be for a very thin bubble. From our work<br />
with the oil film we expect that we will get a maximum reflection when<br />
2t = mλ/2, and m is even. Recall also that m = 0 gives a maximum,<br />
so t ≈ 0 should give a maximum for all wavelengths. Thus we expect<br />
the soap bubble to reflect very well when the film of bubble juice gets<br />
very thin. The truth is just the opposite.<br />
The key to this puzzle is to understand that when a wave is reflected<br />
it can be inverted, that is the direction of the displacement can<br />
be reversed. But a reversal is the same as a π phase shift.<br />
Fact: Reflection Phase Shift<br />
When a wave is reflected at an interface between two media there<br />
will be a π phase shift in the reflected wave if the index of refraction<br />
of the incident medium is lower than the index of refraction of the<br />
reflecting medium. If the incident medium has a higher index,<br />
there is no phase shift.<br />
⊲ Problem 7.11<br />
Explain using the reflection phase shift why a thin soap bubble reflects<br />
no light.<br />
§ 7.10 Single Slit Diffraction<br />
If a wave is passed through a rectangular hole that is much narrower<br />
in one direction than the other we find that we get an interference<br />
pattern like we do with a two source. Such a hole is usually referred to<br />
as a slit since it is so narrow in one direction.
7.11 Homework 155<br />
Second Min<br />
First Min<br />
Source<br />
θ<br />
Central Max<br />
Fact: Single Slit Diffraction<br />
If light is passed through a single slit of width a, then the angle<br />
between the central maximum and the first minimum is given by<br />
a sin θ = λ<br />
§ 7.11 Homework<br />
⊲ Problem 7.12<br />
A pair of speakers is configured as shown, with a microphone placed directly<br />
in front of one of the speakers. Sound waves having a wavelength<br />
10 cm are coming from the speakers. The speakers are in phase with<br />
each other. What is the minimum distance d between the speakers so<br />
that no sound is heard at the microphone labeled P ?<br />
1 m<br />
P<br />
d<br />
⊲ Problem 7.13<br />
A material having an index of refraction of 1.30 is used to coat a piece<br />
of glass (n = 1.50). What should be the minimum thickness of this film<br />
to minimize reflection of 500nm light?<br />
⊲ Problem 7.14<br />
A soap bubble of index of refraction 1.33 strongly reflects both the red<br />
and the green components of white light. What film thickness allows<br />
this to happen? (In air, λ red = 700nm, λ green = 500nm.)
156 Wave Optics 7.11<br />
⊲ Problem 7.15<br />
A beam of 560nm light passes through two closely spaced glass plates, as<br />
shown below. For what minimum nonzero value of the plate separation<br />
d is the transmitted power a maximum?<br />
d<br />
⊲ Problem 7.16<br />
A pair of narrow parallel slits separated by 0.25mm is illuminated by<br />
green light (λ = 546.2nm). The interference pattern is observed on a<br />
screen 1.2m away from the plane of the slits. Calculate the distance<br />
from the central maximum to the first bright region on either side of<br />
the central maximum and between the first and second dark bands.<br />
⊲ Problem 7.17<br />
On a day when the speed of sound is 354 m s<br />
, a 2000Hz sound wave<br />
impinges on two slits 30.0cm apart. At what angle is the first maximum<br />
located? If the sound wave is replaced by 3.00cm microwaves, what<br />
slit separation gives the same angle for the first maximum? If the<br />
slit separation is 1.00µm, light of what frequency gives the same first<br />
maximum angle?<br />
⊲ Problem 7.18<br />
<strong>Two</strong> waves with amplitudes y 1 = A cos(ωt−kd 1 +δ) and y 2 = A cos(ωt−<br />
kd 2 ) interfere with each other. Prove that the resultant amplitude,<br />
y 1 + y 2 = y is<br />
( k<br />
y = 2A cos<br />
2 (d 2 − d 1 ) +<br />
2)<br />
δ (<br />
cos ωt − k 2 (d 1 + d 2 ) + δ )<br />
2<br />
Choose d 1 = d 2 = 0 and make plots of y 1 (t), y 2 (t), and y(t) for δ = π/8,<br />
π/4, π/2, and π.<br />
⊲ Problem 7.19<br />
<strong>Two</strong> slits are separated by a distance d. Light of wavelength λ comes<br />
from a far distant source and strikes the slits at an angle θ 1 as in the<br />
figure below. An interference maximum is formed on a screen that is<br />
far to the right of the slits. The maximum occurs at an angle θ 2 . Show<br />
that sin θ 2 − sin θ 1 = mλ/d, where m is an integer.
7.11 Homework 157<br />
θ 2<br />
d<br />
θ 1<br />
θ 2<br />
θ 1<br />
⊲ Problem 7.20<br />
A material having an index of refraction of 1.30 is used to coat a piece<br />
of glass (n = 1.50). What should be the minimum thickness of this film<br />
to minimize reflection of 500nm light?<br />
⊲ Problem 7.21<br />
A soap bubble of index of refraction 1.33 strongly reflects both the red<br />
and the green components of white light. What film thickness allows<br />
this to happen? (In air, λ red = 700nm, λ green = 500nm.)<br />
⊲ Problem 7.22<br />
Light of wavelength 600 nm is used to illuminate, at near zero angle<br />
of incidence, two glass plates that are stacked on top of each other.<br />
At one end the plates are separated slightly because a wire has been<br />
placed between them on this end. The wire is 0.5 mm in diameter.<br />
How many bright fringes appear along the total length of the plates,<br />
when the plates are viewed from the same side as the light source?<br />
⊲ Problem 7.23<br />
The double slit equation d sin θ = mλ and the equation for a single slit<br />
a sin θ = mλ are sometimes confused. For each equation, define the<br />
symbols and explain the equation’s application.<br />
⊲ Problem 7.24<br />
Light from a He-Ne laser (λ = 632.8nm) is incident on a single slit.<br />
What is the minimum width for which no diffraction minima are observed?<br />
⊲ Problem 7.25<br />
A screen is placed 50cm from a single slit, which is illuminated with<br />
690nm light. If the distance between the first and third minima in the<br />
diffraction pattern is 3.0mm, what is the width of the slit?<br />
⊲ Problem 7.26<br />
Light from an argon laser strikes a diffraction grating that has 5310<br />
lines/cm. The central and first order principal maximum are separated
158 Wave Optics 7.11<br />
by 0.488 m on a wall that is 1.72 m from the grating. Determine the<br />
wavelength of the laser light.<br />
⊲ Problem 7.27<br />
Paul Revere received information from a person in a church steeple that<br />
was 1.8 miles away, by the number of lanterns that were displayed:<br />
”One if by land and, two if by sea.” What would be the minimum<br />
separation between two lantern for Paul Revere to be able to distinguish<br />
them? Assume that the diameter of his pupils was 4.00mm and that<br />
the light had a wavelength of about 580nm.<br />
⊲ Problem 7.28<br />
Redo all the problems in the book starting from chapter 1, but add one<br />
to each constant. Add all of the numerical answers and subtract 1 for<br />
each non numerical answer. What number do you get?
7.12 Summary 159<br />
§ 7.12 Summary<br />
Definitions<br />
• Index of refraction: n = c v<br />
Theorems<br />
• A traveling sinusoidal wave can be represented by<br />
where ω = 2π<br />
T and k = 2π λ .<br />
• For a traveling wave: λf = v.<br />
y = A cos(ωt − kx)<br />
• Interference of two waves: If two waves are added together with<br />
powers P 1 and P 2 then the resultant power is<br />
P = P 1 + P 2 + 2 √ P 1 P 2 cos(φ 2 − φ 1 )<br />
• Maxima and minima of the power occur when ∆φ = φ 2 − φ 1 is an<br />
even and odd multiple of π, respectively.<br />
• The phase difference due to a path difference is ∆φ path = 2π ∆r<br />
λ<br />
• If a field point is far from two sources a distance d apart then the<br />
path difference is ∆r = d sin θ where θ is the angle between the path to<br />
the field point and the normal to the line connecting the two sources.<br />
• When a wave is reflected at an interface between two media there<br />
will be a π phase shift in the reflected wave if the index of refraction of<br />
the incident medium is lower than the that of the reflecting medium.<br />
• The wave length in a medium is longer than in a vaccum: λ ′ = λ n<br />
• If light is passed through a single slit of width a, then the angle<br />
between the central maximum and the first minimum is given by,<br />
a sin θ = λ
160 Geometric Optics 8.12
8.2 Reflection 161<br />
8 Geometric Optics<br />
§ 8.1 Short Wavelength Limit<br />
In the last section we saw the wave nature of light in the sections<br />
where we considered interference. In many situations these interference<br />
effects are small enough to be ignored. For example if light comes in<br />
through two windows in your house you will not find an unexpected<br />
dark spots in the room where the light from the two windows combines.<br />
Fact: Short Wavelength Limit<br />
When the wavelength of the light is much shorter than the size<br />
of the objects that the light is traveling through, then the wave<br />
nature of the light can be ignored.<br />
For example if you set up a candle and cast shadow of your hand<br />
on the wall, the shadow is a faithful replica of your hand, there is no<br />
interference pattern developed because of the interference between the<br />
light going between different fingers.<br />
In this short wavelength limit we can think of a light source as sending<br />
out rays of light that continue in a straight line unless something stops<br />
them. Because of this use of straight lines, this way of dealing with<br />
light is called Geometric Optics or Ray Optics.<br />
§ 8.2 Reflection<br />
Now consider what happens when a ray of light strikes a mirror.
162 Geometric Optics 8.3<br />
θ i<br />
θ r<br />
Mirror<br />
Fact: Law of reflection<br />
The incident and reflected angles are the same.<br />
θ i = θ r<br />
§ 8.3 Virtual Image<br />
Let us draw some of the other rays that come from the source.<br />
Notice that each ray obeys the law of reflection. This may look<br />
a little confusing, but if we trace the reflected ray back through the<br />
mirror, we see that they all converge on the same point.<br />
Object<br />
Virtual Image
8.4 Snell’s Law 163<br />
We see then that from the point of view of an observer of the reflected<br />
rays, it appears that the object is really behind the mirror, because all<br />
of the rays appear to come from that one point. Our brain makes the<br />
simplest conclusions and assumes that the rays come in a straight line,<br />
not in the bent line that they actually followed.<br />
This type of image is called a virtual image of the object since the<br />
light does not actually come from the image. We will see later, ways<br />
in which it is possible to construct a real image.<br />
§ 8.4 Snell’s Law<br />
When a ray of light strikes an interface between two different media,<br />
some of the light is reflected, and some of the light is transmitted<br />
through the interface.<br />
n i<br />
n t<br />
θ i<br />
θ r<br />
θ t<br />
Note that both the reflected and transmitted beams are deflected from<br />
their original direction. The angle at which the transmitted ray leaves<br />
the interface is given by Snell’s Law.<br />
Fact: Snell’s Law<br />
n t sin θ t = n i sin θ i<br />
We can use Snell’s law to explain a strange phenomenon that you<br />
may have noticed. If you view a fish tank from the corner, you can see<br />
objects in the tank through the front glass and the side glass at the<br />
same time. This is because of the bending of the light when it leaves<br />
the water. The effect is depicted in the diagram below.
164 Geometric Optics 8.5<br />
Eye<br />
Tank<br />
Front<br />
Front View<br />
Both View<br />
Side<br />
View<br />
Side<br />
The dark grey region is the part of the tank that can be seen from both<br />
the front and the side. You can actually see around the corner. This is<br />
really weird. So if you want a wider view from your window, you could<br />
just fill your room with water. Remember to also install the air tanks.<br />
§ 8.5 Virtual Image Caused by Refraction<br />
Consider a shiny coin on the bottom of a swimming pool.<br />
Air<br />
Water<br />
The light comes up from the coin and out into the air. When the light<br />
comes out it spreads out more. This looks similar to the rays reflected<br />
from a mirror. This similarity leads one to wonder if there is a virtual<br />
image of the coin, that is, is there a place where all the rays appear to<br />
originate. The following diagram traces the rays back into the water.
8.5 Virtual Image Caused by Refraction 165<br />
Air<br />
Water<br />
We see that they do not converge on a single point. The location of<br />
the image changes as you view it from different angles. When you get<br />
down low and view it from near the horizon the image is close to the<br />
surface of the water and also displaced horizontally from the object.<br />
As you move your view point upward the image moves away from you<br />
and deeper into the water. When you view from straight overhead the<br />
image is directly over the object and about 1/5 of the way up toward<br />
the surface.<br />
Air<br />
Water<br />
Image from horizon<br />
Image from 45°<br />
Image from directly above<br />
Object<br />
This is very different from the image in a flat mirror. The image in a<br />
flat mirror is in the same position, from any viewpoint in the room.<br />
⊲ Problem 8.1<br />
Consider the beam of light shown going through a prism made of glass<br />
with an index of refraction n.
166 Geometric Optics 8.6<br />
α<br />
β<br />
β<br />
Show that the net deflection of the beam (2β) is<br />
§ 8.6 Thin Lens Equation<br />
Imagine that we take a stack<br />
of prisms as shown in the diagram.<br />
Since the prisms farther<br />
from the middle have a steeper<br />
angle they bend the light ray<br />
more sharply. If we pick the angle<br />
of each prism correctly all<br />
of the rays will converge on the<br />
same point.<br />
2β = 2 arcsin(n sin α 2 ) − α<br />
Such a device was used in old light houses in order to focus the<br />
light on the horizon (where there are ships that want to see the light)<br />
rather than letting it spread out and go into the water or into the sky.<br />
Now imagine that you make<br />
the prisms in the middle thicker,<br />
without changing the angle of<br />
the faces. We would now end<br />
up with a smooth curved surface.<br />
Since the angle of faces was not<br />
changed the light would still converge<br />
on the single point. You<br />
can see that you get a lens. This<br />
type of lens is called a converging<br />
lens, since it bends the rays<br />
toward each other.<br />
focal length<br />
focal<br />
point<br />
The optical axis of the lens is a line that passes through the center<br />
of the lens, that is normal to the surface of the lens. A focus is a point<br />
where rays converge. The focus of rays that are parallel to the optical<br />
axis is called the principle focus or focal point. The distance between<br />
the lens and the focal point is called the focal length of the lens.<br />
Suppose that you set up your lens at sunrise, so that the rays are<br />
parallel to the optical axis and converge at the focal point. If you come
8.6 Thin Lens Equation 167<br />
back in an hour the sun will have moved up in the sky, and the rays will<br />
no longer be parallel to the optical axis. The rays will still converge at<br />
a point but it will not be the focal point of the lens. The new focus<br />
will be in the focal plane, a plane parallel to the plane of the lens and<br />
one focal length away from the lens.<br />
Focal Plane<br />
Optical Axis<br />
Notice that the ray that passes through the center of the lens is not<br />
deflected. This is a particularly nice ray for doing constructions of<br />
images, as we will see. In this case it allows us to find the location of<br />
the focus, since the focus is at the intersection of this straight line and<br />
the focal plane. Once we have found the focus, using the central ray,<br />
we can draw the other rays, because they must pass though the focus<br />
also.<br />
We can also reverse these diagrams: Light that comes from a point<br />
in the focal plane and strikes the lens leaves the lens parallel to the ray<br />
that passes through the center of the lens.<br />
We have seen what happens to groups of rays that are parallel to<br />
each other. Let us now see what happens to groups of rays that diverge<br />
from a point that is not in the focal plane. Suppose that you have light<br />
radiating from a point. The point is indicated by the fat arrow on<br />
the left, in the following diagram. We follow the light from this point<br />
source to it’s focus.
168 Geometric Optics 8.6<br />
In the diagram we are able to draw the three rays that are bolded,<br />
because they are, (top ray) parallel to the optical axis, (middle ray)<br />
through the center, and (bottom ray) through the focal point. These<br />
three rays are called the principle rays. From the three principle rays<br />
(we only needed two really) the location of the focus is determined.<br />
After the location of the focus is determined, the other faint rays are<br />
drawn so that they go through the focus.<br />
Notice that the rays of light actually do pass through the focus. If<br />
we place a piece of paper at that point the paper would be illuminated.<br />
This type of image is called a real image.<br />
x o<br />
x i<br />
y o<br />
y i<br />
f<br />
f<br />
The location and size of the image can be computed from the<br />
location and size of the object.<br />
Theorem: Thin Lens Equations<br />
With the dimensions as shown in the figure above,<br />
1<br />
x o<br />
+ 1 x i<br />
= 1 f<br />
Theorem: Magnification Equations<br />
With the dimensions as shown in the figure above,<br />
= − x i<br />
y o y i<br />
where y is negative if below the optical axis.<br />
x o
8.7 Virtual Image in a Converging Lens 169<br />
⊲ Problem 8.2<br />
Using geometric arguments, prove the thin lens equations.<br />
⊲ Problem 8.3<br />
You have a lens with a focal length of 100cm. You place an object at<br />
150cm from the lens. The object is 3cm tall.<br />
(a) Construct the image location using the principle rays.<br />
(b) Find the location of the image using the thin lens equation.<br />
(c) Find the height of the image from the magnification equation.<br />
§ 8.7 Virtual Image in a Converging Lens<br />
You do not get a real image for all positions of the object. When<br />
the object distance is smaller than the focal length, you get a virtual<br />
image. You have probably notice this effect before while using a magnifying<br />
glass: if you look at something far away through a magnifying<br />
glass, you will se the object upside down. If you move closer to the<br />
object at some point it flips over and appears right side up.<br />
Virtual Image<br />
Real Image<br />
image<br />
paper<br />
lens<br />
paper<br />
lens<br />
image<br />
In the photograph above you can notice a few things. In the photo on<br />
the left, the letters on the page are almost in-focus, this means that<br />
the image is not too far from the page. In contrast, in the photo on the<br />
right, the letters on the page are very out of focus, this tells us that<br />
the image is far from the page. This is verified by the ray diagrams.<br />
Here is a larger diagram, showing the construction of the virtual<br />
image, that corresponds to the photo on the left.
170 Geometric Optics 8.7<br />
-x i<br />
x o<br />
f f<br />
Notice that we still use the three principle rays. The topmost ray in<br />
the figure above strikes the lens as if it came from the focal point. The<br />
dotted line going back to the focal point is an extension of the actual<br />
path of the light ray. Notice also that even though the ray constructed<br />
in this manner does not hit the lens (since the lens was not tall enough)<br />
we can still use the ray to construct the location of the image.<br />
The thin lens and magnification equations can still be used in this<br />
case, but in order to get the algebra to work out correctly you need to<br />
interpret a negative image distance (x i ) as being on the side of the lens<br />
where the light originates from.<br />
Example<br />
Suppose that in the figure above x 0 = 6cm, y 0 = 2.0cm and f = 9cm.<br />
Putting this into the thin lens equation we find<br />
1<br />
6cm + 1 = 1<br />
x i 9cm −→ 1 = 1<br />
x i 9cm − 1<br />
6cm = − 1<br />
18cm<br />
−→ x i = −18cm<br />
Now we can use the magnification equation to find the size of the image.<br />
y i<br />
= − y o<br />
−→ y i = − x i<br />
y o = − −18 (2cm) = 6.0cm<br />
x i x o<br />
x o 6<br />
⊲ Problem 8.4<br />
You have a lens with a focal length of 80cm. You place an object at<br />
60cm from the lens. The object is 3cm tall.<br />
(a) Construct the image location using the principle rays.<br />
(b) Find the location of the image using the thin lens equation.<br />
(c) Find the height of the image from the magnification equation.<br />
⊲ Problem 8.5<br />
For a converging lens with a focal length of f, over what range of object<br />
distances is the image:<br />
(a) real?
8.8 Diverging Lenses 171<br />
(b) virtual?<br />
(c) upright?<br />
(d) inverted?<br />
(e) larger than the object?<br />
§ 8.8 Diverging Lenses<br />
Now consider what happens when parallel rays strike a lens that<br />
is thiner in the middle. The rays diverge from the optical axis as if<br />
they came from the focal point on the incident side. This type of lens<br />
is called a Diverging Lens. This is drawn in the figure below.<br />
-f<br />
We might also consider what happens to rays that are converging toward<br />
the focal point on the far side of the lens, these rays exit the lens<br />
parallel to the optical axis, as indicated in the diagram below.<br />
-f<br />
We can, as before, construct the location and size of the image, by<br />
using three principle rays. The three rays are, (1) a ray that leaves the<br />
object parallel to the optical axis, (2) a ray going through the center of<br />
the lens, (3) a ray that leaves the object headed toward the focal point<br />
on the far side of the lens. This has been done in the figure below.
172 Geometric Optics 8.9<br />
x o<br />
-x i<br />
-f -f<br />
We can also find the location and size of the image by using the thin<br />
lens equation and the magnification equation. Once again a negative<br />
value for image distance x i , indicates that the image is on the same<br />
side of the lens as the source of the light. We must also use a negative<br />
value for the focal length of a diverging lens.<br />
Example<br />
Suppose that in the figure above x 0 = 20cm, y o = 8cm and f = −12cm.<br />
Putting this into the thin lens equation we find<br />
1<br />
20cm + 1 1<br />
=<br />
x i −12cm −→ 1 = − 1<br />
x i 12cm − 1<br />
20cm<br />
−→ x i = −7.5cm<br />
Now we can use the magnification equation to find the size of the image.<br />
y i<br />
= − y o<br />
−→ y i = − x i<br />
y o = − −7.5 (8cm) = 3.0cm<br />
x i x o<br />
x o 20<br />
⊲ Problem 8.6<br />
You have a lens with a focal length of -100cm. You place an object at<br />
150cm from the lens. The object is 5cm tall.<br />
(a) Construct the image location using the principle rays.<br />
(b) Find the location of the image using the thin lens equation.<br />
(c) Find the height of the image from the magnification equation.<br />
⊲ Problem 8.7<br />
For a diverging lens, show that the object distance must be negative, in<br />
order for the image distance to be positive. In other words the object<br />
must be virtual, in order for the image to be real. We will see, in<br />
the next section, how it is possible for to have a virtual object to be<br />
negative.
8.10 Multi-Lens Optical Systems 173<br />
Example<br />
§ 8.9 Sign Conventions and Coordinates System<br />
We have used the coordinates<br />
(x o , y o ) for the object and the coordinates<br />
(x i , y i ) for the image. The<br />
direction of positive y is the same<br />
for both coordinate systems. The<br />
direction of positive x was opposite.<br />
Light<br />
In<br />
A curved mirror is also a lens. The coordinate<br />
sign convention, stated as follows,<br />
covers both types of lenses.<br />
• The x-axis of the object coordinate system<br />
is in the opposite direction as the incoming<br />
light.<br />
• The x-axis of the image coordinate system<br />
is in the same direction as the outgoing light.<br />
y o<br />
y i<br />
Light<br />
x o<br />
x Out i<br />
Transmitting Lens<br />
Light<br />
In<br />
Light<br />
Out<br />
x o<br />
x i<br />
y o<br />
y i<br />
Reflecting Lens<br />
A concave mirror (as shown) is a converging lens, while a convex<br />
mirror is a diverging lens. The sign convention for focal lengths is the<br />
same for mirrors.<br />
§ 8.10 Multi-Lens Optical Systems<br />
Most optical systems are composed of more than one lens. We do<br />
not need more theory in order to work with a multi-lens system. The<br />
light passes from one lens to the next in sequence. In order to do a<br />
computation with a multi-lens system, we need only use the image of<br />
one lens as the object of the next lens.<br />
Suppose that you have a diverging lens, f = −12cm, followed by a<br />
converging lens, f ′ = 16cm. The lenses are 20 cm apart. You place an<br />
object that is 9cm tall at a distance of 24 cm from the diverging lens.<br />
We can construct the final image as follows.<br />
24 cm<br />
20 cm<br />
28 cm 37.3 cm<br />
Mirror<br />
Object 1<br />
8cm<br />
12cm<br />
Image 1<br />
Object 2<br />
12cm<br />
16cm<br />
Or we could compute the final image from the thin lens equation. Given<br />
f = −12cm and x o = 24cm we can compute<br />
1<br />
= 1 x i f − 1 1<br />
=<br />
x o −12cm − 1<br />
24cm = − 3<br />
24cm −→ x i = −8cm<br />
16cm<br />
Image 2
174 Geometric Optics 8.10<br />
So<br />
y i = − x i<br />
y o = − −8 (9cm) = 3cm<br />
x o 24<br />
The second lens is 20 cm past the first, so the object distance for the<br />
second lens is<br />
x ′ o = 20cm − x i = 20cm − (−8cm) = 28cm.<br />
From this we can compute the final image location<br />
1<br />
x ′ = 1<br />
i f ′ − 1 x ′ = 1<br />
o 16cm − 1<br />
28cm −→ x′ i = 37.33cm<br />
and<br />
y i ′ = − x′ i<br />
x ′ y o = − −37.3 (3cm) = 4cm<br />
o 28<br />
Notice that the second lens has formed a real image of the virtual image<br />
produced by the first lens.<br />
In the following example, the second lens has a virtual object.<br />
Example<br />
Suppose that you have a converging lens, f = 12cm, followed by a<br />
diverging lens, f ′ = −12cm. The lenses are 18 cm apart. You place an<br />
object that is 4cm tall at a distance of 24 cm from the converging lens.<br />
We can construct the final image as follows.<br />
24 cm<br />
18 cm<br />
Image 2<br />
Image 1<br />
Object 1<br />
12cm<br />
12cm<br />
12cm<br />
Object 2<br />
The top ray is not used to construct the first image, since it is not a<br />
principle ray of the first lens. The first image is constructed from the<br />
other two rays, and then the third ray is draw so that it goes to the<br />
first image and also through the center of the second lens, this makes<br />
it a principle ray of the second lens. Similarly the center ray is not a<br />
principle ray of the second lens, it is only used for constructing the first<br />
image. The bottom ray is a principle ray of both lenses.
8.11 Homework 175<br />
We can also compute the final image from the thin lens equation.<br />
Given f = 12cm and x o = 24cm we can compute<br />
1<br />
= 1 x i f − 1 = 1<br />
x o 12cm − 1<br />
24cm = 1<br />
24cm −→ x i = 24cm<br />
So<br />
y i = − x i<br />
y o = − 24 (4cm) = −4cm<br />
x o 24<br />
The second lens is 18 cm past the first, so the object distance for the<br />
second lens is<br />
x ′ o = 18cm − x i = 18cm − (24cm) = −6cm.<br />
From this we can compute the final image location<br />
1<br />
x ′ = 1<br />
i f ′ − 1 1<br />
x ′ =<br />
o −12cm − 1<br />
−6cm −→ x′ i = 12cm<br />
and<br />
y i ′ = − x′ i<br />
x ′ y o = − 12 (−4cm) = −8cm<br />
o −6<br />
Notice that the diverging lens has formed a real image. Also notice that<br />
the first lens would have formed a real image, but since the diverging<br />
lens enters the optical path before the image is formed, the first image<br />
does not actually get formed.<br />
§ 8.11 Homework<br />
⊲ Problem 8.8<br />
You are designing a movie projector. The film is 8mm wide, and you<br />
wish to project this film onto a screen that is 2.0 meters wide from a<br />
distance of 10 meters.<br />
(a) How far from the lens should the film be?<br />
(b) What should the focal length of the lens be?<br />
⊲ Problem 8.9<br />
An object is located 20cm to the left of a diverging lens having a focal<br />
length f = −32cm. Determine the location and magnification of the<br />
image. Construct a ray diagram for this arrangement.<br />
⊲ Problem 8.10<br />
An object is placed 40 cm in front of a lens with focal length +10 cm.<br />
Describe the image (i.e. location, magnification, etc.).<br />
⊲ Problem 8.11<br />
<strong>Two</strong> converging lenses, each of focal length 10 cm, are separated by<br />
35 cm. An object is 20 cm to the left of the first lens. (a) Find the
176 Geometric Optics 8.11<br />
position of the final image using both a ray diagram and the thin-lens<br />
equation. (b) Is the image real or virtual? Upright or inverted? (c)<br />
What is the overall magnification of the image?<br />
⊲ Problem 8.12<br />
An object is 15 cm in front of a positive lens of focal length 15 cm. A<br />
second negative lens of focal length -15 cm is 20 cm from the first lens.<br />
Find the final image and draw a ray diagram.<br />
⊲ Problem 8.13<br />
A concave mirror has a focal length of 40.0cm. Determine the object<br />
position for which the resulting image is upright and four times the size<br />
of the object.<br />
⊲ Problem 8.14<br />
A concave mirror has a radius of curvature of 60 cm, (f = 30cm).<br />
Calculate the image position and magnification of an object placed in<br />
front of the mirror at distances of 90 cm and 20 cm. Draw ray diagrams<br />
to obtain the image in each case.<br />
⊲ Problem 8.15<br />
Under what conditions will a concave mirror produce and erect image?<br />
A virtual image? An image smaller than the object? An image larger<br />
than the object? Repeat the exercise for a convex mirror.<br />
⊲ Problem 8.16<br />
A dentist wants a small mirror that will produce an upright image with<br />
a magnification of 5.5 when the mirror is located 2.1 cm from a tooth.<br />
(a) What should the radius of curvature of the mirror be? (b) Should<br />
it be concave or convex?
@ Summary 177<br />
§ 8.12 Summary<br />
Facts<br />
• Law of Reflection:<br />
θ i = θ r<br />
• Snell’s Law:<br />
n i sin θ i = n t sin θ t<br />
Theorems<br />
• Thin Lens Equation:<br />
1<br />
x o<br />
+ 1 x i<br />
= 1 f
178 Hints A<br />
A<br />
Hints<br />
1.1 Do the directions “by hand”. To make it easier to keep track of<br />
everything write all forces in terms of the quantity F 0 =<br />
1.2 Use the vector form of Coulomb’s law.<br />
1.3 The answer is E(x, ⃗ 0) = q<br />
4πɛ 0<br />
−2aĵ<br />
(x 2 +a 2 ) 3/2 .<br />
q2<br />
4πɛ 0a<br />
. 2<br />
1.4 Use polar coordinates, ⃗r s = R cos θî + R sin θĵ and dq = Q 2π dθ.<br />
1.5 Use Gauss’s Law.<br />
1.6 Do not attempt to integrate the flux over the surfaces. First,<br />
look carefully at the orientation of the electric field through each of the<br />
three faces that touch the charge. Second, because of the symmetry of<br />
the configuration the remaining three faces must have the same flux as<br />
each other. Third, notice that if you placed eight such cubes around<br />
the charge (forming a larger cube with the charge at the center), that<br />
all eight smaller cubes would have the same flux.<br />
1.7 Use a gaussian surface that is a sphere of radius r, centered on<br />
the charge.<br />
1.8 Use Gauss’s law. The gaussian surfaces are spheres. If r < R<br />
then the amount of charge inside the gaussian surface depends on r,<br />
show that the amount of charge inside is q in = Q r3<br />
R 3<br />
1.10 Assume that the field is directed straight out from the line. Let<br />
the Gaussian surface be a cylinder (like a tin can) with the line charge<br />
as the axis of the cylinder. Note that the flux through the ends of the<br />
can is zero because of the orientation relative to the field.<br />
1.11 The gaussian surface is a cylinder of radius r and length L.<br />
1.12 Recall that there can be no field inside the body of a conductor<br />
that is in static equilibrium.<br />
1.13 The geometry gives you the angle of the forces, but because<br />
the charges are not of the same size you must still deal with both<br />
components. Depending on your disposition, you might prefer using<br />
the vector form of Coulombs law.
A Hints 179<br />
1.14 Since this is only an estimate, assume that you are composed<br />
entirely of water.<br />
1.16 Be sure to add the parts as vectors.<br />
1.17 Let the charge elements dq be little sections of arc that subtend<br />
and angle dθ. Notice that the full charge is spread over a half a circle<br />
so that the charge density (charge per angle) is −7.5µC<br />
π<br />
, so that dq =<br />
−7.5µC<br />
π<br />
dθ.<br />
1.18 Use the work energy theorem. Recall that work is force time<br />
distance.<br />
1.19 This is like a projectile motion problem, but this time we have<br />
a constant electric force rather than an constant gravitational force.<br />
1.20 Start by making a free body diagram and be sure to include the<br />
force of gravity and the force of the string.<br />
1.21 A flux is negative if it is into the volume of the box and positive<br />
if the flux is out of the volume of the box.<br />
1.22 There is a very easy way to do this problem, by using the fact<br />
that there is no charge inside the volume of the nose cone.<br />
1.25 ⃗r = −xî<br />
2.1 Look at the definition of electric potential, how is the electric<br />
potential related to potential energy?<br />
2.2 Use the work energy theorem.<br />
2.3 Follow the example in the text for a nonuniform field.<br />
2.4 Follow the example in the text.<br />
2.6 Consider a closed surface that is totally within the outer conductor<br />
and surrounds the inside surface of the outer conductor, as indicated<br />
by the dotted line in the figure. Using the fact that there is no field<br />
inside a conductor argue that the electric flux through this surface is<br />
zero. From this use Gauss’s law to find the charge on the inside surface<br />
of the outer conductor. Next use the fact that the total charge on the<br />
outer conductor is equal to the sum of the charge on its two surfaces,<br />
to find the charge on the outside surface of the outer conductor.<br />
2.7 Use Gauss’s law. Remember that the electric field is zero inside<br />
the body of a conductor.<br />
2.8 Use the definition of capacitance.<br />
2.9 Use the fact that the electric field strength between the plates is<br />
both σ/ɛ 0 and ∆V/d, where σ is the charge density on the plates.<br />
2.10 Assume that the capacitor is charged so that the inside sphere<br />
has a charge Q and the outside sphere has a charge −Q. Use Gauss’s<br />
law to get the field strength between the shells.
180 Hints A<br />
2.11 First suppose that there is a charge Q on the length L of the<br />
central wire, and a charge −Q on the outer shield. Then use gauss’s<br />
law to show that the electric field strength between the wire and shield<br />
is E =<br />
Q 1<br />
2πLɛ 0 r<br />
. Next show that the potential difference between the<br />
wire and shield is ∆V =<br />
Q<br />
2πLɛ 0<br />
ln(b/a). Finally use the definition of<br />
capacitance.<br />
2.12 The field around a line charge has already been found, use this<br />
to find the electric potential around a single wire. Find the electric<br />
potential difference as you move from one to the other for each wire<br />
and then add. Once you have the electric potential difference between<br />
the wires you can find the capacitance.<br />
2.13 Since there is a maximum field strength there is also a maximum<br />
energy density, and the energy density is the energy stored divided by<br />
the volume of the capacitor.<br />
2.14 The change in the charge switches the direction of the electric<br />
field, which alters the dot product. Also q = −|q|.<br />
2.15 Use the work energy theorem.<br />
2.16 Use the work energy theorem.<br />
2.17 Use the work energy theorem.<br />
2.18 Use the work energy theorem.<br />
2.19 Use the work energy theorem.<br />
2.20 Assemble the particles one at a time. The work to bring the<br />
first particle in is zero since there is no field to do work against. The<br />
second particles feels the potential of the first as you bring it in. The<br />
third particle feels the potential of the first two, and so on.<br />
2.21 The field is the gradient of the potential.<br />
2.22 V = ∫ dq<br />
4πɛ . 0r<br />
2.23 Find the electric potential for each section first. You have actually<br />
done a problem like each of the sections before.<br />
2.24 Write out the field and electric potential of a point charge, then<br />
do some algebra.<br />
2.25 The potential is the sum of the three point potentials. The field<br />
can be found from the derivative of the potential.<br />
2.26 The answer will be in terms of the radius (r 0 ), field (E 0 ), charge<br />
density (σ 0 ), and potential (V 0 ) of the original drops. The charge is all<br />
on the surface of the drop.
A Hints 181<br />
2.27 Imagine that you build of the charge slowly. Suppose that you<br />
have already got a charge q accumulated and you want to bring in an<br />
amount dq more. How much work dW must you do? Once you have<br />
dW written out you can integrate to get W .<br />
2.28 Use Gauss’s law to find the field in all three regions. Then<br />
integrate from infinity inward to find the electric potential. You will<br />
need to split the integral into three regions because the form of the field<br />
changes each time you cross the surface of a shell.<br />
2.29 Look at the definition of capacitance.<br />
2.30 Look at the definition of capacitance.<br />
3.1 The amount of charge is equal to the number of particles times<br />
the charge per particle.<br />
3.2 You will need to use Ohm’s law, the definition of current density<br />
and the relationship between the electric field and the electric potential<br />
∆V = −E ∆x.<br />
3.3 Use the result that R = ρL/A.<br />
3.4 Use P = I ∆V .<br />
3.5 Look up the theorem on electrical power.<br />
3.6 Use the result of the example before this problem.<br />
3.7 The current and voltage on the resistor can be negative (and they<br />
are in this case).<br />
3.8 Assume that the centripetal acceleration of the electron is caused<br />
by the Coulomb force of the proton on the electron. Remember the<br />
definition of electric current.<br />
3.9 Remember that you know the charge of each electron, and that<br />
current is the amount of charge per time.<br />
3.10 Read the definition of current.<br />
3.11 q = ∫ dq = ∫ dq<br />
dt dt<br />
3.12 First compute how many electrons pass a particular point in the<br />
wire in one second. All of these electrons together fill a volume V of<br />
the wire. From the number of electrons you can compute the volume of<br />
electrons. Once you have the volume you can compute the length the<br />
electrons occupy in the wire, since you know the cross sectional area<br />
of the wire. But this length is the distance the electrons move in one<br />
second.<br />
3.13 Use the result that R = ρl/A.<br />
3.14 Look up electric power.
182 Hints A<br />
3.16 Compute the total charge that can flow from the battery. One<br />
kilowatt hour is a unit of energy 3600kJ.<br />
3.17 From the specific heat of water you can find the energy required<br />
and from this you can find the power and thus from the known voltage<br />
you can find the current and then resistance.<br />
3.18 The internal resistance acts like it is in series with the external<br />
resistor and and ideal voltage source.<br />
3.19 Reduce the system in stages. Look for pairs of resistors in the<br />
system that are in parallel or in series, combine this pair, then repeat<br />
until there is only one resistor left.<br />
3.23 Write out all of the equations and then solve.<br />
3.25 Connect the circuit to a voltage supply V S and using Kirchhoff’s<br />
rules for the resultant circuit you can find the charge on each capacitor.<br />
This will in turn allow you to compute the total charge drawn form the<br />
power supply and thus the effective capacitance of the system.<br />
3.26 Use Kirchhoff’s rules<br />
3.27 Use Kirchhoff’s rules<br />
3.28 The bulb that draws the most power is brighter.<br />
3.29 The appliances are connected in parallel to the voltage supply.<br />
3.30 Use the result of the other problem in which you computed the<br />
resistance of a 12 gauge copper wire.<br />
3.31 The amount of heating is proportional to the power dissipated<br />
by the wire per length.<br />
3.32 When the capacitor is fully charged, no current flows into it.<br />
4.1<br />
(a) Notice that ⃗ dl has length Rdθ and is perpendicular to the radius<br />
vector.<br />
4.2 Compute the force on each of the fours section of wire first. Recall<br />
that torque is ⃗τ = ⃗r × ⃗ F , where ⃗r points from the axis of rotation to<br />
the point at which the force is applied.<br />
4.5 For any radius it will take a time 2πm/qB<br />
4.6 Plug the given values into the Lorentz force formula and do the<br />
cross product.<br />
4.7 The magnetic field of the earth points roughly northward.<br />
4.8 Don’t forget that the electron is negatively charged.<br />
4.9 This question is asking you to relate acceleration and force so<br />
start with Newton’s second law.
A Hints 183<br />
4.12 Consider the vector nature of the definition of work and the<br />
direction of the velocity relative to the magnetic force.<br />
4.13 Write out the vectors ⃗ L for each line segment and then do the<br />
cross products.<br />
5.1 Try plotting points for various values of t to get a sense of the<br />
function.<br />
5.2 Start with ⃗r(t) = ⃗a + btî + ct 2 ĵ and find the value of the constants<br />
by demanding that the curve goes through the three given points.<br />
5.3 The parameterization is tî<br />
5.4 The parameterization is a cos tî + a sin tĵ.<br />
5.5 Since the current density is not uniform, I in = ∫ JdA.<br />
5.7 Use the result B = µ 0 I/2πr.<br />
5.8 Use the Biot-Savart law.<br />
5.9 Use the method of the previous problem, and also note that the<br />
current can be found from the electron speed and the radius of the<br />
orbit.<br />
5.10 First argue that half of the wire does not produce any field at<br />
the point of interest. Then use the Biot-Savart law on the other half.<br />
5.11 Argue that the radial lines do not contribute. Note that the<br />
field due to the two arcs are in opposite directions to each other.<br />
5.12 First find the field created by wire 1 at the location of wire 2.<br />
Then find the force that this field produces on wire 2.<br />
5.13 Use Ampere’s Law.<br />
5.14 Use Ampere’s Law.<br />
5.15 Since the current density is not uniform, I in = ∫ JdA.<br />
6.1 The induced current is in the same direction as the electric field.<br />
Determine the direction of the induced field in the center of the loop,<br />
from the known electric field direction. Compare the induced field with<br />
the existing field.<br />
6.2 Choose the Amperian loop to be a circle with radius r and going<br />
through the desired field point half way between the plates. Note that<br />
the current density between the plates is zero.<br />
6.3 Use Kirchhoff’s loop rule to write an equation in the current and<br />
the derivative of the current. Try a solution of the form I(t) = a+be −αt .<br />
6.4 The RMS voltage of the source in the a standard outlet is 120<br />
volts.<br />
6.5 Follow the example in the text for an RC circuit.
184 Hints A<br />
6.6 Use a phasor diagram to draw Kirchhoff’s loop rule, then find the<br />
frequency that minimizes the effective impedance of the RLC combination.<br />
6.8 Use Faraday’s law. Assume that the magnetic field decreases at<br />
a constant rate over the 20 ms that it goes from 1.6 T to zero, that is<br />
dB/dt = (−1.6T)/(20ms).<br />
6.10 Since the field strength decreases as you move further from the<br />
wire you will need to do an integral in order to compute the magnetic<br />
flux.<br />
6.11 Use Ampere’s Law.<br />
6.13 Consider that Q = ∫ Idt and I = V/R = E/R.<br />
7.2 Start from x = A cos φ.<br />
7.3 Take a look at the diagram of the oscillator that shows the phase<br />
angle at different points in the cycle. For the last part remember that<br />
v = dx/dt and consider the quantity v/x.<br />
7.4 Recall that the phase is φ = ωt + φ 0 so that ∆φ = ω∆t. Then<br />
consider by how much the phase must change in order for the oscillator<br />
to complete one cycle.<br />
7.8 First compute the phase difference. Remember that φ 1 = −kr 1<br />
and φ 2 = −kr 2 . Next use the addition of two waves theorem.<br />
7.9 The collection of points with the same ∆r will form a continuous<br />
line, just like the collection of points that are all 5cm from a single<br />
point makes a circle or radius 5 cm.<br />
7.10 Treat the two holes as two sources.<br />
7.11 The phase of each path is now −kr plus the reflection phase<br />
shift.<br />
7.12 Find the distance from the lower speaker to the point P as a<br />
function of the distance d.<br />
7.15 The path difference occurs in air.<br />
7.16 Use the far field approximation to find the path difference.<br />
7.18 Use the trig identities in the appendix.<br />
7.19 There is a path difference, d sin θ, on both sides of the screen.<br />
7.22 The air gap between the plates will be wedge shaped, because<br />
the thickness of the air gap increases as you move toward the wire,<br />
there will be an increasing phase difference between the two reflected<br />
beams. This leads to the fringes.<br />
7.24 As the slit gets narrower the angle of the first minima increases.<br />
There is a limit to the increase.
A Hints 185<br />
7.26 The distance from the central maximum to the first order maximum<br />
is the same as it would be for a double slit with the same distance<br />
between the two slits as there is between successive lines in the diffraction<br />
grating.<br />
8.1 Use geometry to show that<br />
α<br />
α/2<br />
β<br />
α/2<br />
Then use Snells law.<br />
8.2 Each straight line segment that crosses the optical axis, creates<br />
two similar triangles, one above the axis and one below the axis. Use<br />
the properties of similar triangles to get an equation for each of these<br />
pairs.<br />
8.5 Use the thin lens equation to find under what conditions the image<br />
distance is negative.<br />
8.7 Use the thin lens equation.<br />
8.8 Start with the magnification equation.
186 Index B<br />
B<br />
Index<br />
† Energy Density of an Electric Field, 48<br />
† Addition of <strong>Two</strong> Waves, 147<br />
† Ampere’s Law - Time Varying, 120<br />
† Ampere’s Law, 102<br />
† Biot-Savart Law, 97<br />
† Circular Trajectories, 84<br />
† Conductor in Equilibrium: Charge, 43<br />
† Conductor in Equilibrium: Field, 43<br />
† Conductor in Equilibrium: Potential, 44<br />
† Conductor in Equilibrium: Surface Field, 44<br />
† Coulomb’s Law in Vector Form, 8<br />
† Coulomb’s Law, 8<br />
† Effective Capacitance, 65<br />
† Effective Resistance, 64<br />
† Electric Charge, 4<br />
† Electric Field due to a Point Charge, 15<br />
† Electric Field from the Electric Potential, 40<br />
† Electric Potential of a Point Charge, 48<br />
† Electrical Power, 61<br />
† Electromagnetic Waves and Light, 139<br />
† Energy Stored in a Capacitor, 47<br />
† Faraday’s Law (alternate form), 117<br />
† Faraday’s Law, 116<br />
† Force Currents, 105<br />
† Force on a Current in a Uniform Field, 82<br />
† Gauss’s Law, 20<br />
† Impedance: Capacitor, 125<br />
† Impedance: Inductor, 124<br />
† Interference of <strong>Two</strong> Sources, 150<br />
† Kirchhoff’s Junction Rule, 61<br />
† Kirchhoff’s Loop Rule, 62<br />
† Law of reflection, 162<br />
† Lenz’s Law, 119<br />
† Magnetic Force on a Current, 82<br />
† Magnetic Force, 81<br />
† Magnification Equations, 169
B Index 187<br />
† Ohm’s Law: Resistance, 60<br />
† Ohm’s Law: Resistivity, 59<br />
† Reflection Phase Shift, 154<br />
† Short Wavelength Limit, 161<br />
† Single Slit Diffraction, 155<br />
† Snell’s Law, 163<br />
† Superposition Theorem, 15<br />
† Thin Lens Equations, 168<br />
† <strong>Two</strong> Types of Charge, 4<br />
† Wave due to a Sinusoidal Point Source, 142<br />
† Wavelength in a Medium, 154<br />
† Work by Magnetic Force, 84<br />
alternating current, 122<br />
capacitance, 46<br />
conductivity, 59<br />
conductor, 7<br />
converging lens, 166<br />
current density, 58<br />
current, 58<br />
cyclotron, 85<br />
direct current, 122<br />
displacement current, 120<br />
Diverging Lens, 171<br />
electric field, 10<br />
electric field, 11<br />
electric flux, 20<br />
electric potential, 37<br />
electromagnet, 97<br />
electromagnetic, 139<br />
electromotive force, 116<br />
elements, 57<br />
EMF, 116<br />
equipotential, 41<br />
Farad, 46<br />
field lines, 14<br />
focal length, 166<br />
focal plane, 167<br />
focal point, 166<br />
focus, 166<br />
frequency, 143<br />
gain, 128<br />
gradient, 40
188 Index B<br />
ground, 38<br />
Hall Effect, 87<br />
Henry, 120<br />
in parallel, 62<br />
in series, 62<br />
in-phase, 147<br />
index of refraction, 153<br />
inductance, 120<br />
infrared, 145<br />
Lorentz Force, 85<br />
magnetic flux, 117<br />
n-type, 88<br />
non-ohmic, 59<br />
ohm, 60<br />
ohmic, 59<br />
optical axis, 166<br />
out-of-phase, 148<br />
p-type, 88<br />
parameterization, 39<br />
parameterize, 98<br />
period, 143<br />
phase, 139<br />
phasor, 125<br />
principle focus, 166<br />
principle rays, 168<br />
real image, 168<br />
resistance, 60<br />
resistivity, 59<br />
Resistor, 59<br />
resonance, 129<br />
scalar field, 16<br />
steady state, 115<br />
Tesla, 82<br />
ultraviolet, 145<br />
uniform, 16<br />
vector field, 16<br />
velocity selector, 86<br />
wavelength, 142<br />
wiring diagram, 57
B Index 189
190 Useful Mathematics C<br />
C<br />
§ C.1 Trigonometry<br />
Useful Mathematics<br />
cos(a + b) = cos a cos b − sin a sin b<br />
sin(a + b) = sin a cos b + cos a sin b<br />
cos a + cos b = 2 cos b − a<br />
2<br />
cos a − cos b = 2 sin b − a<br />
2<br />
sin a + sin b = 2 cos b − a<br />
2<br />
cos b + a<br />
2<br />
sin b + a<br />
2<br />
sin b + a<br />
2<br />
§ C.2 Fundamental Derivatives and Integrals<br />
Basic Facts<br />
d<br />
dx [xp ] = px p−1<br />
d<br />
dx [eax ] = ae ax<br />
d<br />
dx [ln(x)] = 1 x<br />
∫<br />
d<br />
[sin(kx)] = k cos(kx)<br />
dx<br />
d<br />
dx<br />
[cos(kx)] = −k sin(kx)<br />
∫<br />
∫<br />
x p dx = xp+1<br />
∫<br />
p+1<br />
e ax dx = eax<br />
∫ 1<br />
x<br />
dx = ln(x)<br />
cos(kx) dx =<br />
sin(kx)<br />
k<br />
sin(kx) dx = −<br />
cos(kx)<br />
k<br />
a<br />
Divide and Conquer Rules<br />
Sum Rule:<br />
f(x) = a g(x) + b h(x) −→<br />
d<br />
Example:<br />
dx [3x2 + 4x 3 ] = 3 d<br />
dx [x2 ] + 4 d<br />
Product Rule:<br />
Example:<br />
2x sin(x).<br />
Chain Rule:<br />
Example:<br />
f(x) = g(x) h(x) −→<br />
df<br />
dx = a dg<br />
dx + bdh dx<br />
dx [x3 ] = 6x + 12x 2 .<br />
df<br />
dx = g(x)dh dx + dg<br />
dx h(x)<br />
d<br />
dx [x2 sin(x)] = x 2 d<br />
dx [sin(x)] + d<br />
dx [x2 ] sin(x) = x 2 cos(x) +<br />
d<br />
dx [sin(3x2 )] =<br />
df<br />
dx = dg dh<br />
dh dx<br />
d<br />
d[3x 2 ] [sin(3x2 )] d<br />
dx [3x2 ] = cos(3x 2 )6x<br />
f(x) = g(h(x)) −→
C Power Series Expansions 191<br />
The Fundamental Theorem of Calculus<br />
Recall that the distance traveled between time t 1 and time t 2 is<br />
equal to the area under the v versus t graph between these times.<br />
v(t)<br />
Area = ∫ t 2<br />
t 1<br />
v(t)dt<br />
t<br />
t 1 t 2<br />
But the distance traveled is also x(t 2 ) − x(t 1 ). So we find that<br />
∫ t2<br />
v(t)dt = x(t 2 ) − x(t 1 )<br />
t 1<br />
This works for any pair of functions where one is the derivative of the<br />
other. So in general<br />
IF<br />
f(t) = dg<br />
dt<br />
THEN<br />
∫ t2<br />
t 1<br />
f(t)dt = g(t 2 ) − g(t 1 )<br />
§ C.3 Power Series Expansions<br />
(1 + x) N =<br />
∞<br />
1<br />
1 − x = ∑<br />
x n = 1 + x + x 2 + x 3 + x 4 . . .<br />
sin(x) =<br />
n=0<br />
∞∑<br />
n=0<br />
cos(x) =<br />
e x =<br />
N∑<br />
n=0<br />
∞∑<br />
n=0<br />
(−1) n<br />
(2n + 1)! x2n+1 = x − 1 6 x3 + . . .<br />
∞∑<br />
n=0<br />
(−1) n<br />
(2n)! x2n = 1 − 1 2 x2 + . . .<br />
1<br />
n! xn = 1 + x + 1 2 x2 + 1 6 x3 + . . .<br />
N!<br />
N!<br />
n!(N − n)! xn = 1 + Nx +<br />
2!(N − 2)! x2 + · · ·
192 Physical Constants and Data D<br />
D<br />
Physical Constants and Data<br />
§ D.1 Index of Refraction<br />
(at λ=589.3 nm, from wikipedia.org)<br />
Material Index<br />
Vacuum<br />
1 (exactly)<br />
Helium 1.000036<br />
Air at STP 1.0002926<br />
carbon dioxide 1.00045<br />
water ice 1.31<br />
liquid water (20C) 1.333<br />
ethanol 1.36<br />
glycerine 1.47<br />
polycarbonate 1.59<br />
glass (typical) 1.5 to 1.9<br />
cubic zirconia 2.2<br />
diamond 2.4<br />
moissanite 2.7<br />
gallium phosphide 3.5<br />
gallium arsenide 3.9<br />
silicon 4.0<br />
§ D.2 Approximate Electrical Conductivity<br />
(from wikipedia.org)<br />
Material Ω −1 · m −1<br />
Silver 63.0 × 10 6<br />
Copper 59.6 × 10 6<br />
Gold 45.0 × 10 6<br />
Aluminium 37.8 × 10 6<br />
Brass 20.0 × 10 6<br />
Iron 10.0 × 10 6<br />
Bronze 7.0 × 10 6<br />
Lead 4.8 × 10 6<br />
Stainless Steel 1.4 × 10 6<br />
Seawater 5.0 × 10 0<br />
Drinking water 5.0 × 10 −3<br />
Deionized water 5.5 × 10 −6<br />
Glass<br />
∼ 10 −12<br />
Rubber<br />
∼ 10 −13
D Unit Prefixes 193<br />
§ D.3 Fundamental Constants<br />
speed of light c 2.99792458 × 10 8 m s<br />
(exact)<br />
Planck constant h 6.6260755(40) × 10 −34 J · s<br />
hc 1239.8424(93) eV · nm<br />
1240 eV · nm<br />
¯h 1.05457266(63) × 10 −34 J · s<br />
¯hc 197.327053(59) eV · nm<br />
fundamental charge e 1.60217733(49) × 10 −19 C<br />
mass of electron m e 9.1093897(54) × 10 −31 kg<br />
0.51099906(15) MeV/c 2<br />
mass of proton m p 1.672631(10) × 10 −27 kg<br />
938.27231(28) MeV/c 2<br />
Boltzman constant k 1.380658(12) × 10 −23 J/K<br />
8.617385(73) × 10 −5 eV/K<br />
Avogadro number N A 6.0221367(36) × 10 23<br />
permeability of free space µ 0 4π × 10 −7 N/A 2 (exact)<br />
permittivity of free space ɛ 0 1/(µ 0 c 2 ) (exact)<br />
8.854187817 × 10 −12 C 2 /N · m 2<br />
1<br />
4πɛ 0<br />
8.99 × 10 9 N · m 2 /C 2<br />
gravitational constant G 6.67259(85) × 10 −11 m 2 /kg · s<br />
§ D.4 Unit Conversions<br />
1 m 3.28 ft<br />
1.6 km 1 mile<br />
1 mile 5280 ft<br />
1 hp 746 W<br />
1 liter 1 × 10 −3 m 3<br />
1 gallon 3.79 liters<br />
1 atm 1.013 × 10 5 Pa<br />
1 J 0.239 cal<br />
1 kcal 4186 J<br />
§ D.5 Unit Prefixes<br />
f femto 10 −15<br />
p pico 10 −12<br />
n nano 10 −9<br />
µ micro 10 −6<br />
m milli 10 −3<br />
k kilo 10 3<br />
M mega 10 6<br />
G giga 10 9<br />
T tera 10 12