Introductory Physics Volume Two

1 

Introductory Physics 

Volume Two 

Saint Mary’s College 

Chris Ray 

Spring 2012 

cover art by Susana del Baño Ranner

2 Electric Field 1.0

1.1 Electric Charge 3 

Demo 

1 Electric Field 

§ 1.1 Electric Charge 

Sometimes when you pet a cat, the cat’s fur begins to stand up 

on end and if you then touch the cats nose, you and the cat are both 

shocked. Through petting, the cat has been charged with static electricity. 

The shock occurs when the cat discharges the built up static 

charge to your finger. She has not done this on purpose, it is simply 

the inevitable force of nature taking its course, so don’t get mad at the 

cat. Of course if you had not touched the cat’s nose she would have lost 

her charge more gradually through the air, so she might have reason to 

be mad at you for sucking all the charge from her nose. 

Let’s investigate static electricity further. 

Static Electricity: Rub a rubber rod with fur and hang it by a string. 

Rub a second rubber rod with fur and bring it near the first: 

Rubber Rods rubbed with fur 

Observe that the charge rods repel one another. Thus, for two rods 

of the same material rubbed in the same way we find there will be 

a repulsive force between them. Remove the second rubber rod, and 

replace it with a plexiglass rod that has been rubbed with the fur. 

Observe that the hanging rubber rod is attracted to the plexiglass rod. 

So it is possible by rubbing different objects with fur to cause either 

attractive or repulsive forces between the objects. 

The observations above can be explained by the existence of two 

types of “electric charge.” Objects with the same charge repel each 

other, while objects with different charges attract one another. The 

two types of charge, following the convention of Benjamin Franklin, 

are called positive and negative. Rubbing rubber with fur leaves the 

rubber with a negative charge, while rubbing plexiglass with fur leaves 

the plexiglass with a positive charge.

4 Electric Field 1.1 

Fact: Two Types of Charge 

From observations, such as the demonstration with the rubber rods 

and fur, we have the following rules: 

• There are two types of electric charge: + and -. 

• Objects with the same sign charge repel. 

• Object with oppositely signed charges attract 

⊙ Do This Now 1.1 

Your friend Albert claims that he has isolated a third type of electric charge 

in nature. He calls the three types of charge A, B and C. Explain how he 

would prove this to you. Hint: you might think about the rock-paper-scissors 

game, and make a table showing which pairs are attractive and which are 

repulsive. 

Does the fur rubbing against the rubber (plexiglass) create the 

negative (positive) electric charge? It turns out that the answer is 

no. Careful experiments show that whenever a negative electric charge 

shows up an equal amount of positive charge will be found somewhere 

else, and vice versa. The net charge in the universe remains zero. To 

see this for the rod-fur case, consider the following. Charge a rubber 

rod with fur, hang the rod and then bring the fur near the rod. The 

negatively charged rod will be attracted to the fur, showing that the 

fur has gained a positive charge. We can understand what is going 

on by considering the fundamental building blocks of normal matter: 

protons, neutrons and electrons. Protons have a positive charge +e, 

neutrons have no charge, and electrons have a negative charge −e. 

Uncharged objects contains an equal number of electrons and protons. 

A net electric charge is usually established on an object by adding 

or removing electrons. In the case of the fur and rod, electrons are 

removed form the fur and collected by the rod. This leaves the fur 

with a net positive charge (more protons than electrons) and the rod 

with a net negative charge (more electrons than protons). 


Explain why the cat’s fur stands on end when it is pet. 

Fact: Electric Charge 

Electric Charge is neither created nor destroyed. 

Demo 

Charging Without Rubbing: Hang two small metal balls as indicated 

below:

1.1 Electric Charge 5 

Then charge a rubber rod with fur, place it in contact with the metal 

balls, and then remove it. The two balls will be observed to repel each 

other. 

Here is how we can explain the demonstration. When the rubber rod 

comes in contact with the metal, some of the charge (which is negative 

since it is from a rubber rod rubbed with fur) moves over to the metal. 

This is because metals easily accept or give up electric charge; electric 

charge moves easily through metals. If we indicate the charges using 

negative signs, then we can represent what happens with the following 

picture: 

- - - 

- 

- 

- 

- - 

- - - - 

- 

- 

- 

- 

- 

- - 

- 

- 

- 

- 

- - 

- 

- 

- 

The charges move from the rubber rod onto the metal balls because 

of the repulsive forces between them. Once isolated, the two metal 

spheres repel each other because they both have a net charge of the 

same sign. 

Example 

Two equal masses, each with mass m = 4g, are electrically charged 

and hung using string as shown. If the masses hang in equilibrium at 

an angle θ = 10 ◦ , what is the magnitude of the electric force that each 

mass exerts on the other?


θ 

First draw a free body diagram for one of the masses: 

θ 

2 

F T 

F E 

mg 

Sum forces: 

∑ ⃗F = î(FT sin θ 2 − F E) + ĵ(F T cos θ 2 − mg) = 0 

This gives two equations: 

F E = F T sin θ 2 

F T cos θ 2 = mg 

Solving for the electric force F E gives 

F E = mg tan θ 2 = (.004kg)(9.8 m s 

) tan(5 ◦ ) = 3.4mN 

2 

Demo 

Moving Charges: Place an aluminum can on a table. Perviously we 

showed that we had two types or charge. If we produce either of these 

types of charge the can is attracted toward the charge. 

- - - 

- 

- - -

- 

- 

- 

- 

- 

1.2 Coulomb’s Law 7 

How can we understand this apparently new type of charge, a 

charge that is attracted to both positive and negative charges? The 

place to start is consider the aluminum of which the can is composed. 

Each aluminum atom is composed of an equal number of protons and 

electrons, so the net charge of the can is zero. But there is a difference 

between the protons and electrons. The aluminum atoms are locked 

together in a crystal lattice, with the protons locked together with the 

neutrons in the nucleus of the atoms. The protons cannot move. On the 

other hand some of the electrons are shared between all of the atoms, 

and move around freely. This is what makes aluminum a conductor, it 

has electrons that move freely through the bulk of the material. 

So when a positively charged rod is brought near the can the electrons 

are attracted toward the oppositely charged rod and the move 

within the metal a little bit toward the rod. This leaves the can with a 

net negative charge on the side toward the rod and a net positive charge 

on the side away from the rod. The net positive charge is because the 

protons did not move, so that when the electrons moved toward the 

other side of the can, the protons were left alone, unpaired. 

+ 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

can alone 

- 

+ 

- 

+ 

- 

+ 

+ 

- 

+ 

- 

+ 

+ 

+ 

- 

+ 

- 

- 

+ 

+ 

+ 

+ 

+ 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

+ 

- 

+ 

+ 

- 

+ 

- 

+ 

- 

- 

can next to 

+ rod 

- 

- 

- 

- 

- 

+ 

+ 

+ 

- 

+ 

- 

+ 

- 

+ 

- 

+ 

+ 

+ 

+ 

+ 

- 

- 

+ + + + 

- 

- 

- 

- 

+ 

showing 

net charge 

+ + + + 

One might think that the net force would still be zero once the electrons 

have redistributed themselves, since we end up with a net positive 

charge on one side that is repelled and a net negative charge on the 

other side that is attracted. But it ends up that the force between two 

charges decreases with distance, so that the repulsive force is less than 

the attractive force. 


Explain how a negatively charged rod also attracts the can. 

§ 1.2 Coulomb’s Law 

As we have seen there is a force between charged objects. After 

careful observation it was determined that the electrical force is


similar to the gravitational force. Recall that the gravitational force 

between two massive objects is inversely proportional to the square of 

the distance between the objects and proportional to the weight of each 

object. The electric force between two charged objects is also inversely 

proportional to the square of the distance between the objects and is 

proportional to the charge on each object. This observational fact is 

referred to as Coulomb’s Law. 

Fact: Coulomb’s Law 

The magnitude of the force between charges q a and q b that are a 

distance r apart is given by 

F = q aq b 1 

4πɛ o r 2 

1 

where 

4πɛ o 

= 8.987552 × 10 9 N · m 2 /C 2 ≈ 9.0 × 10 9 N · m 2 /C 2 . 

It has already been noted that the direction of the 

electric force is determined by the sign of the charges: 

opposites charges attract, like charges repel. 

Fact: Coulomb’s Law in Vector Form 

The magnitude and direction of the electric force can be combined 

into one vector expression of Coulomb’s Law as follows: 

⃗F ab = q aq b ⃗r a − ⃗r b 

4πɛ 0 |⃗r a − ⃗r b | 3 

where F ⃗ ab is the force on charge a due to charge b and ⃗r a and ⃗r b 

are the position vectors of the two charges. Note that if we let 

⃗r = ⃗r a − ⃗r b that we can write 

⃗F ab = q aq b 1 

4πɛ 0 r 2 ˆr 

The vectors in the above definition are pictured below. 

a 

r a 

r ab 

r b 

b 

Both vectors are from the origin of the coordinate system, which can 

be chosen for computational convenience.

1.2 Coulomb’s Law 9 

Example 

Consider three charges as shown. We want to know the net force on 

charge a due to charges b and c. 

0.4 

y (m) 

4.0 mC 

r b 

2.0 nC 

r a 

0.2 

r c 

-3.0 nC 

0.0 

0.0 

We find from the diagram that 

0.2 0.4 0.6 0.8 1.0 

x (m) 

⃗r a = (0.2î + 0.4ĵ)m and q a = 4.0 × 10 −3 C 

⃗r b = (0.8î + 0.5ĵ)m and q b = 2.0 × 10 −9 C 

⃗r c = (0.7î + 0.2ĵ)m and q c = −3.0 × 10 −9 C 

so that we can compute 

⃗r a − ⃗r b = (−0.6î − 0.1ĵ)m and q a q b = 8.0 × 10 −12 C 2 

⃗r a − ⃗r c = (−0.5î + 0.2ĵ)m and q a q c = −12.0 × 10 −12 C 2 

so that 

⃗F a = F ⃗ ab + F ⃗ ac 

= q aq b 

4πɛ 0 

⃗r a − ⃗r b 

|⃗r a − ⃗r b | 3 + q aq c 

4πɛ 0 

−0.6î − 0.1ĵ 

⃗r a − ⃗r c 

|⃗r a − ⃗r c | 3 

−0.5î + 0.2ĵ 

= 0.072N − 0.108N 

| − 0.6î − 0.1ĵ| 

3 

| − 0.5î + 0.2ĵ| 3 

−0.6î − 0.1ĵ 

−0.5î + 0.2ĵ 

= 0.072N − 0.108N 

(0.6 2 + 0.1 2 ) 

3/2 

(0.5 2 + 0.2 2 ) 3/2 

= 0.32N(−0.6î − 0.1ĵ) − 0.69N(−0.5î + 0.2ĵ) 

= (0.15î − 0.17ĵ)N 

The net force is to the right and down at about a 45 ◦ . 

If we reworked the previous example but with the charge q a doubled, 

then we would find that the force on the charge q a would also be 

doubled, ⃗ Fa = 2(0.15î − 0.17ĵ)N. In general the force on a charge is 

proportional to the charge. For example, in the previous example 

⃗F a = q a (37.5 î − 42.5 ĵ) N C . 

So we find that the ratio of the force and charge is a constant. 

⃗F a 

q a 

= (37.5 î − 42.5 ĵ) N C .


This observation leads to the definition of the electric field, which occurs 

in the next section. 

When computing the net electric force it is sometimes easier to 

deal with the directions “by hand”. If you can tell the direction of the 

force simply by looking at the configuration, then there is no reason to 

use the vector form of Coulmb’s law. In such a situation just use the 

magnitude form F = qaq b 1 

4πɛ o r 

. The following example will demonstrate 

2 

the “by hand” method. 

Example 

There are three equal charges q at each vertex of an equilateral triangle 

with sides of length L. What is the force on each charge? 

Let us find the force on the charge at the top. We know that the forces 

are repulsive so that we can draw the free body diagram for the charge 

on the top. 

F 1 

F 1 

F 2 

F net 

F 2 

By inspection we see that the two forces are both at a 30 ◦ angle from 

the vertical, and that the horizontal components of the two forces are 

equal and opposite. So that when we add the two forces together, to 

get the net force, the horizontal components will cancel. The net force 

will be just the sum of the vertical components. 

⃗F net = F 1y ĵ + F 2y ĵ = F 1 cos 30 ◦ ĵ + F 2 cos 30 ◦ ĵ 

= q2 1 

4πɛ 0 L 2 cos 30◦ ĵ + 

q2 1 

4πɛ 0 L 2 cos 30◦ ĵ 

= 2 q2 1 

4πɛ 0 L 2 cos 30◦ ĵ 

By the symmetry of the configuration the force on the other two charges 

will be the same magnitude and also pointing directly away from the 

center of the group. 

⊲ Problem 1.1 

There are two +q charges and two −q charges on the corners of a square 

of size a as shown.

1.3 Electric Field 11 

+ - 

Compute the net force on each charge? 


You have three charges as shown. 

- 

+ 

a 

y (m) 

0.2 

0.1 

0.0 

- 0.1 

- 0.2 

- 0.3 

1.0 µC 

3.0 µC 

0.2 0.4 0.6 0.8 1.0 

-2.0 µC 

x (m) 

What is the net force on each charge? 

§ 1.3 Electric Field 

Definition: Electric Field 

If a particle with charge q a is placed at a point in space and F a is 

the net electric force on this particle due to all other charges, then 

the electric field at that point in space is the ratio of the force on 

the particle and the charge of the particle. 

⃗E = ⃗ F a 

q a 

You can think of the electric field as the force per charge in the 

same way that pressure is the force per area. It is important to understand 

the following points about the electric field: 

• The electric field does not depend on the test charge q a in any way. 

• The electric field represents the effect of all the other charges. 

• The electric field is different at each point in space. 

We will now work a specific example in order to demonstrate these 

three properties of the electric field. It is important to follow the details 

of the computation in this example very closely since there is much to 

learn from it. Consider the configuration of two charges shown below.


0.4 

0.2 

y (m) 

r a 

r b 

2.0 nC 

r c 

-3.0 nC 

0.0 

0.0 

0.2 0.4 0.6 0.8 1.0 

We calculated this configuration in a previous example where the test 

charge q a was at the point ⃗r a = (0.2î + 0.4ĵ)m and we found that the 

force at this point was 

x (m) 

⃗F a = q a (37.5 î − 42.5 ĵ) N C 

so that the electric field at this point is 

F 

⃗E(0.2m, 0.4m) = ⃗ a 

= (37.5 î − 42.5 ĵ) N q a 

C 

We can find the electric field at other points ⃗r a as well. 

⃗E(⃗r a ) = ⃗ F a 

q a 

= ⃗ F ab 

q a 

+ ⃗ F ac 

q a 

= q b ⃗r a − ⃗r b 

4πɛ 0 |⃗r a − ⃗r b | 3 + ⃗r a − ⃗r c 

4πɛ 0 |⃗r a − ⃗r c | 3 

We will stop at this point to notice that the charge q a has already 

dropped out of the equation. So that we can write the electric field at 

an arbitrary location ⃗r as 

⃗E(⃗r) = 

q b ⃗r − ⃗r b 

4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c 

4πɛ 0 |⃗r − ⃗r c | 3 

This observation will be used after this example is finished. Let us now 

find the electric field at various other points. Let ⃗r = xî + yĵ so that 

the electric field at ⃗r is 

⃗E(x, y) = 

q b ⃗r − ⃗r b 

4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c 

4πɛ 0 |⃗r − ⃗r c | 3 

(x − 0.8)î + (y − 0.5)ĵ 

= 18 N C 

[(x − 0.8) 2 + (y − 0.5) 2 ] − 27 (x − 0.7)î + (y − 0.2)ĵ 

3/2 N 

C 

[(x − 0.7) 2 + (y − 0.2) 2 ] 3/2 

Evaluating this at a few locations we find the following. 

q c 

⃗E(0, 0.4) = (22î − 17ĵ) N C 

⃗E(0, 0.3) = (28î − 14ĵ) N C 

⃗E(0, 0.1) = (33î − 2ĵ) N C 

⃗E(0, 0.0) = (32î + 3ĵ) N C 

⃗E(0, 0.2) = (32î − 9ĵ) N C 

⃗E(0, −0.1) = (28î + 8ĵ) N C 

The fields at these six points are graphed as the six bold arrows on 

the left side of the following plot. The field is graphed at many other


points a well. 

+2 

-3 

This graph gives some idea of how the field changes as you move around 

in the space. The graph is a little confusing in the region where the 

arrows cross each other. 

This happens wherever the field is strong. 

Because of this there is another kind of graph that is usually used to 

map out the electric field. Take a look at the following graph, which 

has both styles of maps together.


+2 

-3 

The second style has curved field lines that follow the direction of the 

electric field. Check to see that the arrows are parallel to the field lines 

at the base of the arrow. This type of graph clearly shows the direction 

of the field. You can get an idea of the magnitude of the field by how 

close the field lines are to each other: the field is strong where the field 

lines are close together. Here is a graph showing just the field lines.


Look back now to the middle of the example, where we observed 

that in the calculation of the electric field the test charge q a had 

dropped out of the equation. We observed, at that time, that the 

electric field at any location ⃗r in space could be written out as 

⃗E(⃗r) = 

q b ⃗r − ⃗r b 

4πɛ 0 |⃗r − ⃗r b | 3 + q c ⃗r − ⃗r c 

4πɛ 0 |⃗r − ⃗r c | 3 

⃗r−⃗r s 

|⃗r−⃗r s| 3 

We see the contribution of each source charge is of the form 

qs 

4πɛ 0 

and that the net field is the sum of the contributions of each source 

charge. This is so because the net force is the vector sum of the individual 

force. Following this idea we would arrive at these two theorems. 

Theorem: Electric Field due to a Point Charge 

The electric field at the location ⃗r due to a point charge q s at the 

location ⃗r s is given by the following expression. 

⃗E(⃗r) = 

q s ⃗r − ⃗r s 

4πɛ 0 |⃗r − ⃗r s | 3 

The magnitude of the electric field due to a point charge at a 

distance r from the point charge is 

E(r) = 

q 1 

4πɛ 0 r 2 

Theorem: Superposition Theorem 

The electric field due to a collection of point charges is the sum of 

the fields of the individual charges. 

⃗E(⃗r) = ∑ q n ⃗r − ⃗r n 

4πɛ 

n 0 |⃗r − ⃗r n | 3 


A charge q 1 = +q is placed at the location ⃗r 1 = 0î + aĵ and a second 

charge q 2 = −q is placed at ⃗r 2 = 0î − aĵ. 

(a) Write out the electric field at an arbitrary location on the x-axis: 

⃗r = xî + 0ĵ. 

(b) Sketch a map of the electric field lines, in the first quadrant. Here 

is a table that gives the angle ( from the positive x-axis) of the electric 

field at different locations in the first quadrant.


x/a = 0.0 0.5 1.0 1.5 2.0 2.5 3.0 

y/a = 0.0 -90 -90 -90 -90 -90 -90 -90 

0.5 -90 -54 -48 -52 -57 -61 -65 

1.0 45 -3 -11 -20 -29 -36 -42 

1.5 90 42 19 5 -7 -16 -23 

2.0 90 61 39 23 11 1 -8 

2.5 90 69 51 36 24 13 5 

3.0 90 74 58 45 33 24 15 

§ 1.4 Vector and Scalar Fields 

Definition: Scalar Field 

A scalar field is an entity that has a magnitude at each point in 

space. 

The potential energy is a scalar field, the quantity U = mgy gives 

the amount of potential energy at each point (⃗r = xî+yĵ) in space. 

Definition: Vector Field 

A vector field is an entity that has a magnitude and direction at 

each point in space. 

Gravity is a vector field, the constant g = 9.8 m s 

is the magnitude 

2 

of the field near the surface of the earth, and the direction is toward the 

earth. If you move away from the surface of the earth the gravitational 

field decreases in strength. The gravitational field is similar to the 

electric field in the sense that they both give the force on a body. The 

gravitational field gives the force on a massive body. 

⃗F G = m⃗g 

While, the electric field gives the force on a charged body. 

⃗F E = q ⃗ E 

Definition: Uniform Field 

A uniform vector field is a vector field that has the same magnitude 

and direction at all points in space. A uniform scalar field is a 

scalar field that has the same value at all points in space.

1.5 Continuous Charge Distributions 17 

§ 1.5 Continuous Charge Distributions 

In principle one could compute the electric field in any situation 

by summing the fields of each proton and electron in the system. In 

practice this is not done because there are too many electrons and 

protons. In this section we will see how to compute the electric field 

when there are too many charges to count. 

Suppose that we have a string that has been rubbed on a cat until 

the string has built up a charge Q that is spread uniformly over the 

length of the string. We then take the string and stretch it out in a 

straight line. We wish to calculate the electric field due to the string. 

For convenience assume that the string is stretched out along the x-axis 

from x = −L/2 to x = L/2. 

One way of finding the electric field is to conceptually break the 

string into short little sections, say N of them. Each of the sections 

will be a length dx = L/N, and carry a charge dq = Q/N. We can 

number the sections from 1 to N, and then the n’th section will be 

at the position ⃗r n = x n î (x n = nL/N − L/2). As long as we break 

the string into enough sections so that dx is small, we can treat each 

section as a point charge and then we can compute the electric field as 

a sum of N point charges 

⃗E(⃗r) = ∑ q n ⃗r − ⃗r n 

4πɛ 

n 0 |⃗r − ⃗r n | 3 = ∑ dq ⃗r − ⃗r n 

4πɛ 

n 0 |⃗r − ⃗r n | 3 

Keep in mind when you look at this formula that the vector ⃗r n points 

to the charge q n . With this formula and the aid of a computer it 

is possible to compute the electric field for nearly any distribution of 

charge that you can imagine. 

It is also possible, in some cases, to find a closed form solution, 

without using a computer. If we take the limit as N goes to infinity, 

the sum becomes an integral and the electric field can be written in the 

form 

∫ 

dq ⃗r − ⃗r s 

⃗E(⃗r) = 

4πɛ 0 |⃗r − ⃗r s | 3 

Here again we imagine that we break the object into many small pieces 

of charge dq, the vector ⃗r s points toward dq, ranging over all dq’s and


that we “sum” over all the pieces. 

Let us evaluate this integral for the case of the string that we 

introduced earlier. First we need to relate dq to dx. The entire length 

L of the string has a charge Q so that the charge per length is λ = Q/L. 

If the charge is uniformly spread over the string we expect the charge 

per length to be the same everywhere so that 

dq 

dx = Q = λ −→ dq = λdx 

L 

With this, and the fact that ⃗r s = x s î, we can write 

∫ 

dq ⃗r − ⃗r s 

⃗E(⃗r) = 

4πɛ 0 |⃗r − ⃗r s | 3 

∫ L/2 

λdx s ⃗r − x s î 

= 

−L/2 4πɛ 0 |⃗r − x s î| 3 

If we write the field point as ⃗r = xî + yĵ, and then do the integration, 

we find that 

⎡ 

⎤ 

⃗E(x, y) = 

λ ⎣ 

1 

1 

− √ 

⎦ î 

4πɛ 0 

√y 2 + (x − L 2 )2 y 2 + (x + L 2 )2 

⎡ 

⎤ 

+ λ 1 x − 

⎣− 

L 2 

x + L 2 

+ √ 

⎦ ĵ 

4πɛ 0 y 

√y 2 + (x − L 2 )2 y 2 + (x + L 2 )2 


Suppose that you have a circular hoop of radius R with a net charge 

Q spread uniform around the hoop. Compute the electric field at a 

distance z along the axis of the hoop?

1.6 Gauss’s Law 19 

§ 1.6 Gauss’s Law 

We have seen that the electric field due to any charge distribution 

can be computed. From this relationship between the electric field and 

the charge distribution, another one can be derived. 

First we will define a new quantity, the electric flux through a 

surface. To get an idea of what the flux represents first consider the 

following story. You wish to catch some butterflies but you don’t have 

time to chase them around, so you just set up your butterfly net on 

a pole. The butterflies are migrating south for the winter. They are 

flying by your house heading in a southerly direction, so you orient the 

net so that it faces north. The number of butterflies you catch should 

be proportional to both the density of butterflies in the air and the 

area of the mouth of the net. The number of butterflies caught will 

also depend on the orientation of the net relative to the direction the 

butterflies are moving. For example, if the butterflies end up flying to 

the south-west instead of directly south, you will not catch as many 

since the net was not facing the optimal direction. 

The electric flux is similar, it is the amount of electric field that 

“passes through” a surface. There are three things that determine the 

quantity of electric flux: the area of the surface, the magnitude of the 

electric field and the orientation of the field relative to the surface. 

To write this out clearly, we need to have a way to mathematically 

represent the orientation of a surface. We will define a vector area ⃗ A 

as the vector that has a magnitude equal to the area of the surface and 

has a direction that is normal to the surface. A good picture to keep 

in mind is a thumbtack, 

A 

A 

the nail part of the tack is the vector area and the flat part of the 

tack is the surface. This ends up being the best way to use a vector to 

represent a surface. Now we can define the electric flux.


Definition: Electric Flux 

The electric flux (φ e ) through a surface is 

φ e = ⃗ E · ⃗A 

if the field is uniform over the surface. If the field is not uniform 

then one must integrate over the surface, 

∫ 

φ e = ⃗E · dA ⃗ 

We see that the dot product represents the dependence of the flux 

on the relative orientation of the surface and the field. When thinking 

of the flux integral over a surface it can be helpful to imagine gluing 

tacks to the surface with the nail part sticking out, so that you end up 

with a spiky covering of the surface. Each tack represents one small 

surface element dA ⃗ and the integral is the sum of the flux’s through all 

of the small surface element. 

Now we can state the theorem that relates the electric field to the 

charge density. 

Theorem: Gauss’s Law 

The electric flux out of any closed surface is proportional to the 

total charge enclosed within the surface. 

∮ 

⃗E · dA ⃗ = Q in 

ɛ 0 

Note that the integral of the electric field over the surface does 

not depend on the charge density outside the surface in any way. For 

example if there is no charge inside the surface then the integral must 

be zero. 

Example 

We can use Gauss’s law in order to find the electric field strength. 

Here is an example of how this can be done. The result of this example 

is also generally useful.

1.6 Gauss’s Law 21 

Suppose that you have a block of material which has charges inside 

that move freely. We will show later that in side the block the electric 

field is zero and that outside but near the block the field is normal to 

the surface. Also we will see later that there is no net charge inside the 

block, but that on the surface there can be a surface charge density. 

Let us form a Gaussian surface in the shape of a small tin can that is 

half inside and half outside the material, with the can oriented so that 

the ends of the can are parallel to the surface of the material. 

Since the field is zero inside the material we know that the flux is zero 

through the half of the can that is inside the material. We also know 

that the flux is zero through the sides of the can because the field is 

normal to the surface of the material. So we see that the only flux 

through the surface of the can is the flux through the top of the can. 

Assuming that the can is small enough so that the field is uniform over 

the can, then the flux through the top is 

∫ 

⃗E · dA ⃗ = E ⃗ · ⃗A = EA 

top 

where A is the cross sectional area of the can. We have now computed 

the left hand side of Gauss’s law. The right hand side is the charge 

inside the gaussian surface. Since the charge is only on the surface we 

can find the charge inside the can as the area of the surface that is 

inside the can A times the surface charge density σ: so Q in = Aσ. And 

∮ 

∫ 

bottom 

⃗E · ⃗ dA = Q in 

ɛ 0 

∫ 

∫ 

⃗E · dA ⃗ + ⃗E · dA ⃗ + ⃗E · dA ⃗ = Q in 

sides 

top ɛ 0 

−→ 0 + 0 + EA = Aσ −→ E = σ ɛ 0 ɛ 0 

In the previous example, the determination of the flux was greatly


simplified because we picked a the gaussian surface so that the electric 

field was either normal to the surface or in the plane of the surface. 

For example, the field was in the plane of the sides of the can in the 

previous example, thus the flux was zero through the sides. Further, 

in the previous example, the electric field was normal to the part of 

the gaussian surface that did not have a zero flux, in addition the field 

strength was uniform over this part of the surface. Thus the first thing 

to do, when you are using Gauss’s law to find the electric field strength, 

is to choose the gaussian surface so that the field is either parallel or 

normal to all parts of the surface. 


Four closed surfaces are near three charges as shown. 

S 3 

S 4 

-Q 

+Q 

S 2 

-2Q 

S 1 

Find the electric flux through each surface. 


A charge q is placed at one corner of a cube. What is the electric flux 

through each face of the cube. 


Use Gauss’s law to show that the field strength at a distance r from a 

q 1 

point charge is 

4πɛ 0 r 

. 2 


A solid sphere of radius R has a total charge of Q uniformly distributed 

throughout its volume. Calculate the magnitude of the electric field at 

a distance r from the center of the sphere. Be sure to consider the cases 

of r < R and r > R separately. 


A sphere of radius a carries a volume charge density ρ = ρ 0 (r/a) 2 . 

Find the electric field inside and outside the sphere. 


Use Gauss’s law to show that the electric field near a line charge with 

uniform charge density λ is given by E = 

λ 

2πɛ 0r 

where r is the distance 

from the line.

1.7 More Examples 23 


Consider a long cylindrical charge distribution of radius R with a uniform 

charge density ρ. Find the electric field at a distance r from the 

axis for r < R. 


A spherical shell of radius R carries a net charge of Q uniformly distributed 

over it’s surface. 

(a) Find the electric field strength at a point inside the shell. 

(b) Find the electric field strength at a point outside the shell. 

§ 1.7 More Examples 

Example 

Return to the pithballs in the example in section 1: assume the length 

of each string is 10cm. If the masses have the same net electric charge, 

what is the magnitude of the charge on each mass? Can you determine 

the sign of the net charge on each ball? 

Use the following diagram to determine the distances between the 

masses: 

L 

θ/2 

L = 10cm 

d 

d = 2L sin θ 2 = 2(10cm) sin 5◦ = 1.7cm. 

Assume each mass has a charge q, then using Coulomb’s law: 

F E = 1 q 2 

4πɛ 0 d 2 −→ q2 = F Ed 2 

1 

4πɛ 0 

. 

√ 

(0.0034N)(.017m) 

−→ q = 

2 

9 × 10 9 N · m 2 /C 2 = 1.0 × 10−8 C 

The sign of the charge cannot be determined; all that can be said is 

the net charge on the masses have the same sign. 

Example 

Three particles with electric charge are attached to a meter stick, as 

shown. The value of Q is 1 × 10 −6 C (= 1µC). (a) What is the electric 

force on the center charge? (b) To what position could the center charge 

be moved so that the net electric force on it is zero?


0.5m 

+3Q 

+Q +2Q 

(a) The electric forces on the center charge due to the other two charges 

are: 

The net force is 

F 

2Q 

Q 

F 

3Q 

F net = F 3Q − F 2Q 

= 1 

4πɛ 0 

= 

Q(3Q) 

(0.5m) 2 − 1 

4πɛ 0 

Q(2Q) 

(0.5m) 2 

( 

) 

( ) 

9 × 10 9 N·m2 

3 − 2 

(1 × 10 −6 C) 2 

C 2 (0.5m) 2 = +3.6mN 

(b) There are three regions to consider: in between the 2Q and 3Q 

charges, outside of these charges to the left and outside of these charges 

to the right. Here are force diagrams for placing the charge Q outside 

of the two larger charges: 

F 

3Q 

F 

3Q 

F 

2Q 

+Q 

+3Q 

+2Q 

+Q 

F 

2Q 

Since the forces point in the same direction, there is no location outside 

of the two larger charges where the net electric force on the charge Q 

can be zero. For positions inside the two larger charges, the forces will 

point in opposite directions, as seen in part (a), and will cancel at the 

position where the forces have equal magnitudes. Let’s measure the 

location from the+3Q charge: 

F 

2Q 

F 3Q 

+3Q 

+2Q 

The net force on the charge +Q at a position x is 

F net = F 3Q − F 2Q 

= 1 Q(3Q) 

4πɛ 0 x 2 − 1 Q(2Q) 

4πɛ 0 (1 − x) 2 

x


The net force is to be zero: 

1 Q(3Q) 

4πɛ 0 x 2 − 1 Q(2Q) 

4πɛ 0 (1 − x) 2 = 0 

−→ 3(1 − x) 2 − 2x 2 = 0 

Solving the resulting quadratic equation yields two answers: x = 5.45m 

and x = 0.55m. The latter value is the answer, since we already argued 

that the position must lie between the two larger charges. 


Analyze the configuration in the previous example, except replace the +3Q 

charge with a −3Q charge. 

(a) −0.18N (b)4.45m to the right of the +2Q charge. 

Example 

Four charges are located at the corners of a square as shown in the 

diagram below. A fifth charge is located at the center of the square. 

For the charge values indicated compute the net electric force on the 

charge at the center of the square. 

25cm 

+4μC 

+1μC 

+4μC 

Draw a force diagram: 

+4μC 

-1μC 

+2μC 

+1μC 

F +2 

45 o 

y 

x 

F -1 

F +1 

45 o 

F +4 

-1μC 

+2μC 

The distances between each corner charge and the center charge are 

the same: 

√ (a ) 2 ( a 

) √ 

2 2 

d = + = 

2 2 2 a,


where a = 25cm. To find the net force, add all the forces using vector 

addition: 

⃗F net = îF x + ĵF y , 

where 

F x = (F +4 − F +2 − F +1 − F −1 ) sin(45 ◦ ) 

F y = (−F +4 + F +2 − F +1 − F −1 ) cos(45 ◦ ) 

Use Coulomb’s law to compute the magnitude of each force: 

F +4 = (9 × 109 )(4 × 10 −6 )(4 × 10 −6 ) 

( √2 ) 2 

= 4.6N 

2 (.25) 

F +1 = F −1 = (9 × 109 )(1 × 10 −6 )(4 × 10 −6 ) 

( √2 

2 (.25) ) 2 

= 1.15N 

F +2 = (9 × 109 )(2 × 10 −6 )(4 × 10 −6 ) 

( √2 

2 (.25) ) 2 

= 2.3N 

Finally, compute the components of the net force: 

F x = (4.6N − 2.3N − 1.15N − 1.15N) sin(45 ◦ ) = 0 

F y = (−4.6N + 2.3N − 1.15N − 1.15N) cos(45 ◦ ) = −3.25N 

So, the net force points straight down with a magnitude of 3.25N. 

Example 

Consider the configuration of three charges along a line as shown below. 

Assuming the only forces acting are mutual electric forces between the 

charges, show that the configuration is in equilibrium, meaning that 

the net force on each charge is zero. 

a 

a 

+4q -q 

+4q 

Here is a force diagram for all the charges: 

+4q -q 

+4q 

The net force on the center charge must be zero, since the two forces 

are in opposite directions and caused by equal charges that are the 

same distance away. One down, two to go. Use the force law to write


out the net force on the leftmost charge: 

F L = 1 (4q)(q) 

4πɛ 0 a 2 − 1 (4q)(4q) 

4πɛ 0 (2a) 2 

= 1 (4q)(q) 

4πɛ 0 a 2 − 1 (4q)(q) 

4πɛ 0 a 2 

= 0 

So the net force on the leftmost charge is zero. Since the rightmost 

charge is the mirror image of the leftmost charge, it also has no net 

force acting on it. Thus, all three charges have a net force of zero acting 

on them for the configuration given, and the system is in equilibrium. 

Is the equilibrium stable? 

By stability we mean the following. If we move one of the charges 

a little bit, does the configuration return to the original equilibrium 

configuration, or does the configuration fall apart? Let’s move one of 

the charges a little bit and see what happens. Move the left most charge 

a small distance ɛ to the left. If the configuration is stable, then the 

net force on it must push it to the right. Check: 

a+ε 

a 

The net force is now: 

+4q -q 

+4q 

ε 

F L ′ = 1 (4q)(q) 

4πɛ 0 (a + ɛ) 2 − 1 (4q)(4q) 

4πɛ 0 (2a + ɛ) 2 

If this force points to the right, it will push the charge back from 

where it came and the equilibrium can be stable. If the force is to the 

left, the charge will be pushed away from the equilibrium position and 

the equilibrium can not be stable. With a little algebraic muscle the 

formula above for F L ′ can be rewritten: 

( 

) 

F L ′ = 

q2 1 

πɛ 0 a 2 (1 + ɛ − 1 

a )2 (1 + ɛ 

2a )2 

It is easy to see that this is a negative number since the denominator 

in the subtracted term is smallest, no matter how small ɛ is, as long 

as it is larger than zero. Thus, the force on the leftmost charge points 

to the left and the charge is pushed away from the original equilibrium 

position. The equilibrium is unstable. 

⊙ Do This Now 1.5


Determine what the other two charges in the previous example do when the 

leftmost charge is moved to the left. 

They accelerate towards each other 


An electron accelerates east due to an electric field. What direction does the 

electric field point? 

west 

Example 

A dust particle with mass 2µg has a net electric charge 3µC. The 

piece of dust is in a region of uniform electric field and is observed to 

accelerate at a rate of 180 m s 2 . What is the magnitude of the electric 

field in that region of space? 

The net force on the dust particle can be computed using Newton’s 

second law: 

F net = (2 × 10 −9 kg)(180 m s 2 ) = 3.6 × 10 −7 N. 

Using the definition of electric field and assuming no other forces act 

on the dust particle: 

E = F q = 3.6 × 10−7 N 

3 × 10 −6 C = 0.12 N C . 

Example 

If a proton is placed in the same electric field from the previous example, 

what is the resulting acceleration? 

Work backwards: 

F = qE = (1.6 × 10 −19 C)(0.12N/C) = 1.9 × 10 −20 N 

a = F m = 

1.9 × 10−20 N 

1.67 × 10 −27 kg = 1.15 × 107 m s 2 

Example 

Three charges lie along a line as shown below. What is the electric 

field at a position that is a horizontal distance a to the right of the −q 

charge? 

-4q +2q -q 

2a 

a 

Here is a diagram indicating the electric field due to each charge at the 

point:


-4q +2q -q 

E -1 

E -4 

E +2 

2a 

a 

a 

In the diagram, the electric field due to the −4q charge has been labeled 

E −4 , etc. The electric field at the point indicated is just the 

superposition of the individual electric fields: 

E = E +2 − E −1 − E −4 = 1 (2q) 

4πɛ 0 (2a) 2 − 1 q 

4πɛ 0 a 2 − 1 (4q) 

4πɛ 0 (4a) 2 

−→ E = − 3 q 

16πɛ 0 a 2 . 

Example 

Using the configuration of charges in the previous example, determine 

the electric field a vertical distance a above the +2q charge. 

First, let’s add a coordinate system, since this problem will involve 

vectors in two dimensions: 

y 

E +2 

(0, a) 

E -4 

E -1 

-4q θ 1 

+2q θ 2 -q 

x 

(-2a, 0) (0, 0) (a, 0) 

The angles θ 1 and θ 2 can be computed using some geometry: 

cos θ 1 = √ 2 sin θ 1 = √ 1 

5 5 

cos θ 2 = √ 1 sin θ 2 = √ 1 

2 2 

To compute the electric field at the point (0, a), we add the individual 

electric fields as vectors:


E +2 

θ 1 

θ 2 

E -4 

E -1 

⃗E = î(−E −4 cos θ 1 + E −1 cos θ 2 ) 

+ ĵ(−E −4 sin θ 1 − E −1 sin θ 2 + E +2 ) 

Use Coulomb’s law to compute the magnitudes of the electric field 

contribution from each charge: 

E −4 = 1 (4q) 

4πɛ 0 ( √ 5a) = 1 4q 

2 4πɛ 0 5a 2 

E +2 = 1 (2q) 

4πɛ 0 a 2 = 1 2q 

4πɛ 0 a 2 

Finally, 

( 

⃗E = î − 1 

+ ĵ 

E −1 = 1 

4πɛ 0 

4πɛ 0 

( 

− 1 

4πɛ 0 

(q) 

( √ 2a) = 1 

2 4πɛ 0 

q 2 √5 

5a 2 + 1 

4πɛ 0 

4q 1 √5 

5a 2 − 1 

4πɛ 0 

q 

2a 2 

) 

q 1 √2 

2a 2 

q 1 √2 

2a 2 + 1 ) 

2q 

4πɛ 0 a 2 

−→ E ⃗ = 1 q 

(−î(0.36) + ĵ(1.29)) 

4πɛ 0 a2 Flux example: 

Example 

A uniform electric field passes through a 1cm×1cm square in the xy 

plane. The electric field is ⃗ E = (5 N C )î + (3 N C )ˆk. What is the electric 

flux through the square? 

z 

y 

x 

Since the electric field is constant: ∫ 

φ E = ⃗E · dA ⃗ = E ⃗ · ⃗A.


The area vector points in the ˆk direction, so 

( 

) 

φ E = (5 N C )î + (3 N C )ˆk · (0.01m) 2ˆk = 0.03 

N·m 2 

C 

Gauss’s Law Examples: 

Example 

A 1.5 × 10 −9 C point charge is located at the center of a cylinder. The 

electric flux through the sides of the cylinder is known to be 100 N·m2 

C . 

What is the electric flux through one of the endcaps? 

By Gauss’s law: 

∫ 

sides 

−→ 

⃗E · ⃗A + 

100 N·m2 

C 

∫ 

caps 

∫ 

∮ 

caps 

+ ∫ 

caps 

⃗E · ⃗A = Q in 

⃗E · ⃗A = 

⃗E · ⃗A = 169 N·m2 

C 

ɛ 0 

⃗E · ⃗A = 169 N·m2 

C 

1.5 × 10−9 C 

8.85 × 10 −12 C 2 

N·m 2 

N·m2 

− 100 

C 

= 69 N·m2 

C 

This is the flux through both caps. Since the charge is located symmetrically 

with respect to the two endcaps, the flux through one cap is 

just half this: 34.5 N·m2 

C . 

Example 

A large, thin rectangular plate has a uniform charge density σ distributed 

over its area: 

σ 

Assuming that the plate is so large that the electric field above and 

below the plate is uniform, compute the magnitude of the electric field.


With the assumptions above, symmetry tells us that the electric field 

will point in a direction perpendicular to the surface; for a positive 

charge it will point away from the plate: 

E 

A 

σ 

E 

In the figure a Gaussian surface is indicated. Take it to be a small 

cylinder with cross-sectional area A. The electric flux in non-zero only 

through the endcaps: 

∮ 

∫ 

∫ 

∫ 

⃗E · dA ⃗ = E · dA ⃗ + E · dA ⃗ + E · dA 

⃗ 

upper cap 

sides 

= EA + 0 + EA = 2EA 

Using Gauss’s law: 

2EA = σA 

ɛ 0 

−→ E = 

σ 

2ɛ 0 

. 

lower cap 

§ 1.8 Homework 


y 

7.0 µC 

x 

0.50 m 

0.50 m 

2.0 µC 

0.50 m 

−4.0 µC 

Three point charges are located at the corners of an equilateral triangle 

as shown. Calculate the net electric force on the 7.0µC charge.

1.8 Homework 33 


Richard Feynman said that if two people were at arm’s length from 

each other and each person had 1% more electrons than protons, the 

force of repulsion between them would be about equal to the “weight” 

of the entire Earth. Was Feynman exaggerating? 


A point charge q is at the position (x 0 , y 0 ). Find the electric field at 

an arbitrary point (x, y) due to the charge q. 


Two 2.0 µC charges are located on the x axis at x = 1.0m and at 

x = −1.0m. 

(a) Determine the electric field on the y axis at y = 0.5.m. 

(b) Calculate the electric force on a −3.0µC charged place at this point. 


A uniformly charged insulating rod of length 14 cm is bent into the 

shape of a semicircle. If the rod has a total charge of −7.5µC, find 

the magnitude and direction of the electric field at the center of the 

semicircle. 


An electron moves at a speed of 3 × 10 6 m s 

into a uniform electric field 

of magnitude 1000 N/C. The field is parallel to the electron’s velocity 

and acts to decelerate the electron. How far does the electron travel 

before it is brought to rest? 


A proton moves at 4.5 × 10 5 m s 

in the horizontal direction. It enters a 

uniform electric field of 9.6 × 10 3 N/C directed downward. Ignore any 

gravitational effects and find the time it takes the proton to travel 5.0 

cm horizontally, its vertical displacement after it has traveled 5.0 cm 

horizontally, and the horizontal and vertical components of its velocity 

after it has traveled 5.0 cm horizontally. 


y 

x 

θ 

E


A charged ball of mass 1.00 g is suspended on a light string in the 

presence of a uniform electric field as shown. When ⃗ E = (3.00î + 

5.00ĵ) × 10 5 N/C, the ball is in equilibrium at θ = 37.0 ◦ . 

(a) Find the charge on the ball. 

(b) Find the tension in the string. 


A triangular box is in a horizontal electric field of magnitude E = 

7.8 × 10 4 N/C as shown 

30 cm 

10 cm 60° 

E 

(a) Compute the electric flux through each face of the box. 

(b) Compute the net electric flux through the entire surface of the box. 


The nose cone of a rocket is in a uniform electric field of magnitude E 0 

as shown. 

E 

E 

It is parabolic in cross section, of length d and of radius r. Compute 

the electric flux through the paraboloidal surface. 


A charge Q is at the center of a cube of side L. 

(a) Find the flux through each face of the cube. 

(b) Find the flux through the entire surface of the cube. 

(c) Would these answers change if the charge was not at the center? 


A charge Q is located just above the center of the flat face of a solid 

hemisphere of radius R. 

(a) What is the electric flux through the curved surface. 

(b) What is the electric flux through the flat surface. 


Find the electric field at the origin if there is a charged rod that goes 

from x = a to x = b assuming that the charge density of the rod is 

λ = cx n . 

d

1.9 Summary 35 

§ 1.9 Summary 

Definitions 

Electric Field: 

Electric Flux: 

⃗E = ⃗ F a 

q a 

dφ e = E ⃗ · dA ⃗ 

∫ 

φ e = ⃗E · dA ⃗ 

Facts 

Coulomb’s Law 

⃗F ab = 1 ˆr ab 

q a q b 

4πɛ 0 rab 

2 = 1 ⃗r ab 

q a q b 

4πɛ 0 rab 

3 

Theorems 

Electric Field of a Point Charge: 

⃗E(⃗r) = 1 q ⃗r − ⃗r s 

4πɛ 0 |⃗r − ⃗r s | 3 

where ⃗r points to the field point and ⃗r s points to the charge q. 

Electric Field of a Distribution of Charge: 

⃗E(⃗r) = 1 ∫ 

dq ⃗r − ⃗r s 

4πɛ 0 |⃗r − ⃗r s | 3 

where ⃗r points to the field point and ⃗r s points to the charge element 

dq. 

Gauss’ Law: ∮ 

⃗E · dA ⃗ = Q in 

ɛ 0 

Superposition Theorem: If a charge distribution ρ 1 produces an 

electric field ⃗ E 1 and the charge distribution ρ 2 produces an electric 

field ⃗ E 2 then if both charge distributions are present the electric field 

is ⃗ E = ⃗ E 1 + ⃗ E 2 .

36 Electric Potential 2.9

2.1 Electric Potential 37 

2 Electric Potential 

§ 2.1 Electric Potential 

The Coulomb force is a conservative force and thus we can consider 

the potential energy of a particle in an electric field that is created by 

a static charge distribution. Suppose that you have a charged particle 

in an electric field, and that you move the particle from point A to 

point B. As the particle is moved, the electric field will do work on the 

particle. 

W A→B = 

= 

= q 

∫ B 

A 

∫ B 

A 

∫ B 

A 

⃗F E · ⃗dr 

q ⃗ E · ⃗dr 

⃗E · ⃗dr 

Thus when the particle is moved from A to B the change in the electric 

potential energy is 

∫ B 

∆U = −W A→B = −q ⃗E · ⃗dr 

A 

and the change in potential energy per charge is 

∫ B 

∆U 

= − ⃗E · 

q 

⃗dr. 

A 

Notice that the potential energy per charge does not depend on the 

test charge q, it only depends on the electric field. 

Definition: Electric Potential 

The electric potential, V , is the electric potential energy per charge. 

The electric potential is related to the electric field: 

∫ B 

V B − V A = ∆V = ∆U = − 

q 

A 

For a uniform electric field this simplifies to 

∆V = ∆U = −E q 

⃗ · ∆⃗r 

⃗E · ⃗dr

38 Electric Potential 2.1 

It is important to notice that the definition of electric potential 

only tells us how to find the difference (V B − V A ) between the electric 

potential at two different points, A and B . The value of the electric 

potential itself is not defined. This is not a problem because only the 

difference has physical significance. You may recall that this is true 

for potential energy also. The potential energy at ten meters above 

the surface of the earth is not defined, but the difference between the 

potential energy at ten meters and three meters is defined, ∆U = 

mg∆y. One can choose some point in space that is defined to be 

the zero of the electric potential. This point is sometimes called the 

ground. Then in reference to ground, all points have an absolute electric 

potential. This is like choosing the surface of the earth as the zero of 

gravitational potential energy. 

Often the electric potential is simply referred to as the potential. 

The units of electric potential is evidently the units of energy divided 

by the units of charge. This unit occurs frequently and has it’s own 

name, the Volt. 

1Volt = 

1Joule 

1Coulomb 

This unit is abbreviated as V. Be warned that this is a bad coincidence 

since the symbol used for the electric potential is V , which is very 

similar to V. This is not a serious problem, but if you mix up the 

symbols, you can end up doing silly things with the algebra. 

Example 

Suppose that there is a uniform electric field ⃗ E = (0.2 N C 

)î. The electric 

potential difference between the two points ⃗r A = (1.0m)î+(2.0m)ĵ and 

⃗r B = (5.0m)î + (−3.0m)ĵ is computed as follows. 

V B − V A = ∆V = − ⃗ E · ∆⃗r 

= −(0.2 N C 

)î · [⃗r B − ⃗r A ] 

= −(0.2 N C 

)î · [(4.0m)î + (−5.0m)ĵ] 

= −(0.2 N C 

)(4.0m) + 0 = −0.8V 

This example demonstrates that the electric potential decreases as 

one moves in the direction of the electric field (positive x in this case). 

This is a general rule: the electric field points from regions of high 

electric potential to regions of low electric potential.

2.1 Electric Potential 39 

Example 

Suppose that an electron moved from a region where the electric potential 

is 150 volts to a region where the electric potential is 100 volts. 

There is only the electric field acting on the electron. What would be 

the change in the kinetic energy of the electron? We know that the 

potential energy and electric potential are related: ∆V = ∆U/q. 

so that 

∆K + ∆U = 0 

∆K = −∆U = −q∆V = −(−e)(100V − 150V) = −8.0 × 10 −18 J 

In this example we see that negatively charged particles slow down 

when they go from a region of high electric potential to a region of low 

electric potential. A positively charged particles would speed up. 

Example 

Now let’s try an example where the field is not uniform. 

⃗E = α [ y 2 î + 2xyĵ ] 

If we move from the origin to the point (a, b) what is the change in the 

electric potential. First we need to pick a path from the starting point 

to the ending point. A straight line will do. Let ⃗r(t) = x(t)î + y(t)ĵ = 

atî + btĵ, we see that as the parameter t is varied from 0 to 1, that 

the vector ⃗r(t) points along the path from the origin to the final point 

(a, b): that is ⃗r(t) is the trajectory of a particle following our path. This 

function is called a parameterization of the path. Now we can compute 

the line integral ∫ E ⃗ · dr ⃗ since we can now write 

⃗dr = d⃗r 

dt dt = [ dx 

dt î + dy 

dt ĵ 

] 

dt = [aî + bĵ] dt


so that 

∫ B 

∆V = − 

= − 

= − 

= −α 

= −α 

= −α 

A 

∫ 1 

0 

∫ 1 

0 

∫ 1 

0 

∫ 1 

0 

∫ 1 

0 

= −αab 2 

⃗E · ⃗dr 

⃗E(⃗r(t)) · d⃗r 

dt dt 

α [ y 2 î + 2xyĵ ] · [aî + bĵ] dt 

[ (bt)2î + 2(at)(bt)ĵ ] · [aî + bĵ] dt 

[ 

(bt) 2 a + 2(at)(bt)b ] dt 

3ab 2 t 2 dt 

In general, to compute a line integral, one must first find a parameterization 

of the path. 

The following theorem gives a way of finding the electric field if 

the electric potential field is already know. 

Theorem: Electric Field from the Electric Potential 

[ 

⃗E = −∇V ⃗ ∂V 

= − 

∂x î + ∂V 

∂y ĵ + ∂V ] 

∂z ˆk 

The symbol ⃗ ∇ represents the gradient operator ∂ 

Thus the expression ⃗ ∇f is equal to ∂f 

∂x î + ∂f 

∂y ĵ + ∂f 

∂z ˆk. 

∂xî + ∂ 

∂y ĵ + ∂ ∂z ˆk. 

Example 

The nonuniform electric field, ⃗ E = αy 2 î + 2αxyĵ, that we used in the 

previous example, has the electric potential V = −αxy 2 . Let us check 

the above theorem. 

⃗E = − ⃗ ∇V 

= ∂ 

∂x αxy2 î + ∂ ∂y αxy2 ĵ + ∂ ∂z αxy2ˆk 

= αy 2 î + 2αxy ĵ + 0 ˆk OK

2.2 Equipotentials 41 


You touch the terminals of a 9 Volt battery to your tongue. While it is 

touching your tongue a charge of 0.08 Coulombs passes through your 

tongue from one terminal to the other. How much energy is dissipated 

in your tongue, by the charged particles passing through it? 


Two metal plates are held at an electric potential difference of 1000V. 

An electron is released from the plate that is at a lower electric potential. 

What is the speed of the electron when it reaches the other 

plate? 


The electric field in a region is given by ⃗ E = ayî + axĵ, where a = 

2.0V/m 2 is a constant. What is the electric potential difference between 

the origin and the point x = 1.0m, y = 2.0m? 


The electric potential in a region is given by V = axyz. What is the 

electric field? 

§ 2.2 Equipotentials 

Suppose that we have picked our zero for electric potential. Each 

point in space will now have it’s own value for electric potential. If we 

put a dot at every location that is at an electric potential of, say 4.3V, 

then the collection of all such dots will create a surface. 

Definition: Equipotential 

An equipotential is a surface on which the electric potential is constant. 

Below is a graph of a few equipotentials for the electric field produced 

by three point charges (front left is q = +1, back center is q = +2, 

and front right is q = −3). 

These 3D plots are difficult to generate and read, so often a section 

is cut through the surfaces, and only the cut edge of the surfaces are


shown. The following graph is a section through the equipotentials for 

the same three charges (there are a few more equipotentials drawn in 

this section graph). 

+2 

+1 

-3 

Below is a graph of the same equipotentials with a few electric field 

lines graphed also. 

+2 

+1 

-3 

Notice that while the field lines go from positive charges to negative 

charges, the equipotentials encircle charges. Also notice that wherever 

a field line crosses an equipoential, the field line is perpendicular to 

the equipotential. This must happen, as we will now show. Imagine 

that you move a charge a small distance ⃗ dr along the surface of an 

equipotential. The change in electric potential must be zero since the 

beginning and ending points are both on the same equipotential. But 

we also know that the change in electric potential is given by, 

dV = − ⃗ E · ⃗dr 

So since dV = 0, it must be the case that E ⃗ · ⃗dr = 0, and 

⃗E · ⃗dr = 0 −→ E ⃗ is perpendicular to dr ⃗ 

since ⃗ E ≠ 0 and ⃗ dr ≠ 0.

2.3 Conductors in Equilibrium 43 


(a) Sketch the equipotentials for two like charges. 

(b) Sketch the equipotentials for two opposite charges. 

§ 2.3 Conductors in Equilibrium 

A conductor allows it’s free charges to move through it with some 

amount of ease. So, if there is a force acting on a free charge then that 

charge will move. Thus if there is an electric field inside a conductor 

then the free charges in the conductor will move. In some situations, 

for example if the conductor is electrically insulated, the charges will 

eventually reach an equilibrium configuration in which they no longer 

move around. This makes sense, since if you apply a field to an isolated 

conductor the charges will move around to adjust to the new force but in 

the end they must settle down into the equilibrium configuration, which 

is the state with the lowest energy. But this tells us something about 

the electric field inside a conductor once it has reached equilibrium. 

Since the charges are not moving, there can be no electric field inside 

the conductor. 

Theorem: Conductor in Equilibrium: Field 

The electric field is zero inside a conductor at equilibrium. 

This in turn tells us that there can be no net charge inside the 

volume of a conductor if the conductor is in equilibrium. We can argue 

this as follows. Suppose that there was a net charge in some volume 

inside a conductor. By Gauss’s Law we know that there must be a 

net electric flux through the surface of this volume, but the flux is the 

integral of the electric field over the surface, so the electric field cannot 

be zero if there is a region with a net charge. But by the previous 

theorem we know that if there is an electric field inside a conductor then 

the conductor is not in equilibrium. Thus there can be a net charge 

inside a conductor only if the conductor is not at equilibrium. 

Theorem: Conductor in Equilibrium: Charge 

There is no net charge in any volume inside a conductor in equilibrium. 

If a charge is placed on a conductor it will reside on the 

surface of the conductor. 

We can arrive at one more very important result by using the fact 

that there is no field in a conductor in equilibrium. Since the field is 

zero, this means that ∆V = − ∫ ⃗ E · ⃗ dr = 0 for any path inside the 

conductor. This tells us that a conductor in equilibrium is all at the 

same electric potential.


Theorem: Conductor in Equilibrium: Potential 

All points in a conductor at equilibrium have the same electric 

potential. 

Theorem: Conductor in Equilibrium: Surface Field 

The field at the outside surface of a conductor in equilibrium is 

normal to the surface of the conductor. The magnitude of the 

field is E = σ ɛ 0 

where σ is the surface charge density. 

We can see that the previous theorem must be true by considering 

that the surface of the conductor is an equipotential, and noting that 

the electric field is normal to any equipotential surface. 


There is a solid conductor with a cavity within it. Floating within this 

cavity there is a second conductor. This has been drawn below with a 

quarter of the outer conductor removed so that you can see the inner 

conductor. 

A total charge Q a is place on the inner conductor and a charge Q b is 

placed on the outer conductor. 

(a) What is the amount of charge on the inside surface of the outer 

conductor? 

(b) What is the amount of charge on the outer surface of the outer 

conductor? 


A conducting spherical shell having an inner radius of 4.0 cm and outer 

radius of 5.0 cm carries a net charge of +10µC. If a +2.0µC point charge 

is placed a the center of this shell, determine the surface charge density 

on the inner and outer surfaces.

2.4 Capacitors 45 

§ 2.4 Capacitors 

Suppose that we place two conductors near each other. Now suppose 

that we remove a quantity of charge Q from one conductor and 

place it on the other. One conductor will end up with a charge +Q and 

the other will end up with a charge −Q. In addition an electric field 

will be created between the conductors. 

Since there is an electric field between the conductors, there will also 

be an electric potential difference between the two conductors. 

∆V = V + − V − = − 

∫ ⃗r+ 

⃗r − 

⃗ E · ⃗ dr 

Because the electric field strength is proportional to the charge Q, the 

electric potential difference will also be proportional to Q. 

∆V ∝ Q 

Example 

Here is a specific example of this general result. Place two conducting 

plates parallel to each other as shown, and charge the top plate to a 

net charge Q and the other plate to a net charge −Q 

In a previous example it was shown that the electric field strength near 

a plate with uniform charge density σ is E = σ/2ɛ 0 . Between the 

plates the fields of the two plates are in the same direction (toward 

the negatively charged plate) so that the strength of the net field is 

twice the field strength of each plate alone. So the net field between 

the plates is E = σ/ɛ 0 . If the area of the plate is A then the charge


density is σ = Q/A so that we find, 

E = 

Q 

Aɛ 0 

As long as the separation between the plates d is much less than the 

width of the plates the electric field will be effectively uniform between 

the plates, so that the potential difference between the plates is 

∆V = V + − V − = − ⃗ E · ∆⃗r 

Since ∆⃗r points from the negative plate to the positive plate and E ⃗ 

points the opposite direction, the dot product of these two vectors is 

negative the product of the magnitudes; E ⃗ · ∆⃗r = −Ed. 

∆V = −(−Ed) = 

Q d = 

d Q 

Aɛ 0 Aɛ 0 

We see then that the electric potential difference is proportional to the 

charge on the plates. 

Since the electric potential difference is proportional to the charge, 

the ratio is a constant. 

Q 

∆V 

Definition: Capacitance 

The capacitance, C, of two conductors is the ratio of the charge on 

the conductors to the electric potential difference. 

C = 

Q or Q = C ∆V 

∆V 

The unit of capacitance is a Farad and is equal to a Coulomb per 

Volt: F = C V . 


A 6.0µF capacitor is connected to a 1.5 Volt battery so that a electric 

potential difference of 1.5 volts is maintained between the two conductors 

of the capacitor. What is the charge on each of the conductors? 


Show that the capacitance of two parallel plates is C = ɛ 0 A/d, where 

d is the distance between the plates and A is the area of the plates. 


Show that the capacitance of two concentric spherical conducting shells 

4πɛ 

is 0 

1/a−1/b 

, where a and b are the radii of the shells. 


Coaxial cables are commonly used to carry high frequency signals. Your 

cable for your cable-TV is a coaxial cable. A coaxial cable consists of

2.5 Energy in an Electric Field 47 

a wire (radius a) surrounded by a cylindrical conducting shield (radius 

b) Show that the capacitance for a length L of coaxial cable is 2πɛ0L 

ln(b/a) . 


Consider two long, parallel, and oppositely charged wires of radius a 

with their centers separated by a distance b. Assuming the charge 

is distributed uniformly on the surface of each wire, show that the 

capacitance per unit length of this pair of wires is 

C 

l = πɛ o 

ln( b a − 1) 

§ 2.5 Energy in an Electric Field 

Let us consider how much work must be done to charge a capacitor. 

Suppose that we have already moved an amount of charge q from the 

negative plate to the positive plate of the capacitor, so that the electric 

potential difference between the plates is ∆V = q/C. In order to move 

a little bit more charge dq to the positive plate, we need to do an 

amount of work 

dW = dU = dq ∆V = dq q C 

The total amount of work to bring the capacitor from a charge of q = 0 

to a charge of q = Q is 

U = 

∫ Q 

0 

dq q C = Q2 

2C 

= 

(C∆V )2 

2C 

= 1 2C(∆V )2 

Theorem: Energy Stored in a Capacitor 

A capacitor charged to an electric potential difference of V C stores 

an amount of energy 

U = 1 2 CV 2 C 

This energy is stored in the electric field that has been created 

between the plates. The energy density (energy per volume) is 

u = energy 1 

1 ɛ 

volume = 2C(∆V )2 

0A 

2 d 

= 

(Ed)2 = 1 

V 

A d 

2 ɛ 0E 2 

In the intermediate step in the above equation, the properties of a parallel 

plate capacitor was used, but the final result is true in general. 

Theorem: Energy Density of an Electric Field 

The electric field contains an amount of energy per volume u. 

u = 1 2 ɛ 0E 2



A 120µF capacitor is charged to an electric potential difference of 100V. 

(a) How much energy is stored in the capacitor? 

(b) If the field strength in a capacitor becomes to great then the charge 

will jump across the gap between the plates. Assume that this break 

down occurs when the field strength reaches 3 × 10 6 V m. What is the 

minimum volume that is required for a 120µF capacitor to be able to 

hold an electric potential difference of 100V? 

§ 2.6 Electric Potential of a Point Charge 

We wish to find the electric potential for a point charge. The 

electric field strength around a point charge is 

E = 

q 1 

4πɛ 0 r 2 

where r is the distance from the point charge. The electric field is 

either pointed away or toward the charge depending on if the charge is 

positive or negative. Let us start with the positive charge. 

V b − V a = − 

∫ rb 

r a 

⃗ E · ⃗ dr 

If we let r b > r a then dr ⃗ is pointed outward. This is the same direction 

as E ⃗ so that E ⃗ · ⃗dr = E dr. Thus 

∫ rb 

V b − V a = − E dr = − q ∫ rb 

1 

r a 

4πɛ 0 r a 

r 2 dr = − q [− 1 ] rb 

4πɛ 0 r 

r a 

= q 1 

− 

q 1 

4πɛ 0 r b 4πɛ 0 r a 

Thus we see that the electric potential at a distance r from a point 

charge is 

V (r) = 

q 1 

4πɛ 0 r + constant 

For simplicity one normally chooses for the constant to be zero. Note 

that this choice for the constant implies that the zero of the electric 

potential is at the position r = ∞. 

Theorem: Electric Potential of a Point Charge 

V (r) = 

q 1 

4πɛ 0 r 


Show that V (r) = 

q 1 

4πɛ 0 r 

is correct for a negative charge as well.



An electron is released from a distance of 2.0 cm from a proton. How 

fast will the electron be going when it is 0.5 cm from the proton? 

(Assume that the proton does not move.) 


Example 

An electric field is given by E ⃗ = (3 

C·m 

)x 2 î). What is the electric 

2 

potential difference between the points (1m, 0, 0) and (3m, 0, 0)? What 

is the minimum work needed to move a +6µC charge between the two 

points, starting from (1m, 0, 0)? 

Since the electric potential is path independent (it comes from a conservative 

force), we can integrate along any path we want; choose the 

x-axis: 

∫ 

∫3m 

( ) 

∆V = − ⃗E · (îdx) = − 3 N 

x 2 dx 

C·m 2 

( ) [ 

−→ ∆V = − 3 N x 3 ] 3 

= −26 N·m 

C·m 2 3 

= −26V 

1 

C 

N 

1m 

The work done by an external agent will be 

W = −W E = q∆V = (6 × 10 −6 )(−26V) = −1.56 × 10 −4 J 

Example 

The figure below is a schematic representation of an electron “gun.” 

A potential difference is maintained between the left and right plates, 

with the right plate having a higher potential. By heating the left 

plate, an electron is “boiled” off the plate. The electron, starting from 

rest, then moves toward the right plate, directed toward a small hole in 

the plate so that it “shoots” out of the gun. If the potential difference 

between the plates is 9V, how fast will the electron be moving when it 

reaches the right plate? 

- + 

s = ? 

ΔV


The change in potential energy of the electron will be: 

Use conservation of energy: 

∆U = q∆V = (−e)∆V. 

∆U + ∆K = −e∆V + ( 1 2 m ev 2 − 0) = 0 

√ √ 

2e∆V 2(1.6 × 10 

−→ v = = 

−19 C(+9V) 

m e 9.11 × 10 −31 kg 

= 1.78 × 10 6 m s 

Example 

Near the surface of the Earth, there is always a background electric 

field that averages 100N/C and points down. Assuming the Earth to be 

a perfect conductor, how much electric charge is stored on the Earth’s 

surface? 

Since the electric field points down, we know the Earth’s charge is negative. 

So, lets find the magnitude. Use the previous result to determine 

the surface charge density: 

E = σ ɛ 0 

−→ σ = Eɛ 0 . 

Multiply this by the Earth’s surface area to get the total charge: 

Q E = σ(4πR 2 E) = Eɛ 0 (4πR 2 E) = 453, 000C 

Example 

Two equal and opposite charges are located a distance a apart. Find 

the electric field and the electric potential at a) the point directly between 

the two charges and b) the point a distance a/2 directly above 

the point inbetween the two charges. 

+q -q 

The electric fields due to each charge point in the same direction at the 

center point c: 

a 

E + 

c 

+q -q 

E -


The resulting electric field at c is 

E c = E + + E − 

= 1 q 

4πɛ 0 ( a + 1 

2 )2 4πɛ 0 

q 

( a 2 )2 

= 2q 

πɛ 0 a 2 

To find the potential we will have to come up with a rule for what to 

do with more than a single point charge. As just illustrated, for the 

electric field due to a system of charges, the resulting field at any point 

is just the superposition of the electric fields due to all charges at that 

point: 

⃗E = ⃗ E 1 + ⃗ E 2 + ⃗ E 3 + · · · 

This followed because the electric field came from a net force. Likewise, 

because the electric potential is defined using the electric field, at any 

point in space we will just add up the potentials due to all the individual 

charges to find the resulting potential: 

For our problem: 

V = V 1 + V 2 + V 3 + · · · 

V c = V + + V − 

= 1 q 

4πɛ 0 ( a 2 ) + 1 (−q) 

4πɛ 0 ( a 2 ) 

= 0 

Now repeat the analysis for the second point, labeled d: 

E + 

d 

45 o 

45 o 

y 

a/2 

E - 

x 

+q 

a/2 

a/2 -q 

The electric field is 

⃗E d = î(E + + E − ) cos 45 ◦ + ĵ(E + − E − ) sin 45 ◦ 

Compute the magnitudes of the individual fields: 

E + = E − = 1 q 

4πɛ 0 

= 

q 

2πɛ 0 a 2 

( √ 2 

2 a)2


And, 

The potential is: 

⃗E d = î(2E + ) + ĵ(0) 

√ √ 

q 2 2q 

E d = 2 

2πɛ 0 a 2 2 = 2πɛ 0 a 2 

V d = V + + V − = 1 

4πɛ 0 

( 

q 

( √ 2 

2 

) 

(−q) 

+ 

a) ( √ 2 

2 a) 

= 0 

The potential is zero at both points! The electric field is not zero, and 

it points in the x direction at both points. You should be able to argue 

for any point lying on the line defined by the points c and d that the 

electric field always points in the x direction and the potential is always 

zero. 


For the configuration in the previous example, compute the electric potential 

at a point that is distance a above the +q charge. 

q 

8πɛ0 a(2 − 

√ 

2) 

Example 

A charge Q is distributed uniformly along a line L that lies along the 

x-axis. Compute the electric potential a distance x from the left end 

of the charge distribution that is on the x-axis. Take the potential at 

∞ to be zero. 

Locate the origin at the point we want to compute the electric potential 

(labeled P ): 

x 

P 

x' 

dx' 

Consider the very small section of length dx ′ of the line that is at x ′ . 

The amount of charge contained in this small section is 

dq ′ = Q L dx′ . 

Since the section is very small, we can treat it as a point charge and 

the potential it causes at P is 

dV ′ = 1 Q 

L dx′ 

4πɛ 0 x ′ , 

where we have used the potential for a point charge, which takes the 

potential at ∞ to be zero. To find the total electric potential at P we


add up potential due to each small section making up the entire line: 

∫ 

x+L ∫ Q 

V (x) = dV ′ 1 

L 

= 

dx′ 

4πɛ 0 x ′ 

x 

= Q ( ) x + L 

4πɛ 0 L ln x 

Let’s check that this gives the expected result for L → 0, which will 

result in Q being a point charge. We need the limit: 

( ) ( 

x + L 

lim ln = lim ln 1 + L ) 

≈ L 

L→0 x L→0 x x 

So 

lim V (x) = Q 

L→0 4πɛ 0 L · L 

x = Q 

4πɛ 0 x , 

which is indeed the electric potential a distance x from a point charge 

Q. 



Through what potential difference would an electron need to be accelerated 

for it to achieve a speed of 40% of the speed of light, starting 

from rest? 


How much work is required to move one mole of electrons from a region 

where the electric potential is 9V to a region where the electric potential 

is -5V? 


An electron moving along the x axis has an initial speed of 2.7×10 6 m s 

at 

the origin. Its speed is reduced to 1.4 × 10 5 m s 

at the point x = 2.0cm. 

Calculate the potential difference between the origin and this point. 

Which point is at the higher potential? 


In Rutherford’s experiments alpha particles (charge +2e, mass 6.6 × 

10 −27 kg) were fired at a gold nucleus (charge +79e). An alpha particle 

initially very far from the gold nucleus is fired at 2.0 × 10 7 m s 

directly 

toward the center of the nucleus. How close does the alpha particle get 

to this center before turning around? 


Show that the amount of work required to assemble four identical 

charges Q at the corners of a square of side s is 5.41kQ 2 /s.



In a certain region of space the electric potential is V = 5x−3x 2 y+2yz 2 . 

Find the expressions for the x, y, and z components of the electric field 

in this region. What is the magnitude of the field at the point P , which 

has coordinates (1, 0, −2)m? 


A rod of length L lies along the x axis with its left end at the origin 

and has a nonuniform charge density λ = αx. What are the units of 

α? Calculate the electric potential at the point x = −b on the x axis. 


A wire that has a uniform linear charge density λ is bent into the shape 

shown below. Find the electric potential at the center of the circular 

part. 

2R 

R 

2R 


There is a point charge near by. You determine that the electric potential 

at your location (due to the point charge) is 3000V and the electric 

field strength is 500 V m. 

(a) How far away is the charge? 

(b) What is the value of the charge? 


Three charges are placed on the x-axis. Two charges Q at x = d and 

x = −d, and a third charge −2Q at x = 0. 

(a) Show that the electric potential along the x-axis is V = 

2kQd2 

x(x 2 −d 2 ) . 

(b) Show that the electric field along the x-axis is E ⃗ = 2kQd2 (3x 2 −d 2 ) 

x 2 (x 2 −d 2 ) 

î. 2 

(c) In the limit that x ≫ d, show that the field is proportional to 1/x 4 

and that the potential is proportional to 1/x 3 . 


You have drop of conductive fluid with a net charge of Q 0 and a radius 

r 0 . The electric field and electric potential at the surface of this drop 

(due to the charge on the drop) are E 0 and V 0 . Two such drops join 

together to form a larger drop. 

(a) What is the radius of the larger drop? 

(b) What is the surface charge density of the larger drop? 

(c) What is the electric field at the surface of the larger drop? 

(d) What is the electric potential at the surface of the larger drop?



How much work is required to bring a total charge Q to a spherical 

shell of radius R? 


You have two concentric spherical shells of radius a and b. The smaller 

shell has a charge of q 1 = 10nC and the larger shell has a charge of 

q 2 = −15nC. 

a 

b 

(a) Find the electric field at all points in space. 

(b) Find the electric potential at all points in space. 

(c) Sketch the electric potential as a function of the distance from the 

center of the shells, for a = 0.15m and b = 0.30m. 


Two conductors having net charges of +10.0µC and −10.0µC have a 

potential difference of 10.0V. Determine the capacitance of the system 

and the potential difference between the two conductors if the charges 

on each are increased to +100.0µC and −100.0µC. 


Two conductors insulated from each other are charged by transferring 

electrons from one conductor to the other. After 1.6 × 10 12 electrons 

have been transferred, the potential difference between the conductors 

is 14V. What is the capacitance of the system? 


Einstein showed that energy is associated with mass according to the 

famous relationship, E = mc 2 . Estimate the radius of an electron, 

assuming that its charge is distributed uniformly over the surface of a 

sphere of radius R and that the mass-energy of the electron is equal to 

the total energy stored in the resulting nonzero electric field between 

R and infinity.



Definitions 

Electric Potential: 

Capacitance: 

∆V = ∆U 

q 

C = 

Q 

∆V 

Equipotential: an equipotential is a surface over which the electric 

potential is a constant. 

Theorems 

dV = − ⃗ E · ⃗dr 

∆V = − 

∫ rf 

r i 

⃗ E · ⃗ dr 

[ ∂V 

⃗E = − 

∂x î + ∂V 

∂y ĵ + ∂V ] 

∂z ˆk 

≡ − ⃗ ∇V 

Conductors in Equilibrium: 

• Field - the field is zero in the volume of a conductor, and at the 

surface it is normal to the surface. 

• Charge - the net charge in the volume of a conductor is zero. 

• Potential - the potential in the volume of a conductor is constant. 

Electric Potential of a Point Charge: 

V = 1 q 

4πɛ o r 

Electric Potential of a Distributed Charge: 

V (⃗r) = 1 ∫ 

dq 

4πɛ o |⃗r − ⃗r s | 

Energy Density of an Electric Field 

E = 1 2 ɛ oE 2 

Energy Stored on a Capacitor: 

U = 1 2 CV 2 C

3.2 Electric Current 57 

3 Circuits 

§ 3.1 Introduction 

When you turn on a flashlight, a flow of electrons is passed through 

the light bulb. Inside the light bulb is a small filament. As the electrons 

pass through the filament they loose energy and in the process the 

filament heats up. The filament gets so hot that it glows, this is how 

the bulb produces light. 

The electrons are supplied from the negative terminal of the battery 

and flow to the light bulb through a metal wire. After the electrons 

pass through the filament they are carried back to the positive terminal 

of the battery by another wire. 

A system of electrical devices connected with wires, such as the 

flashlight system, is called an electric circuit. Circuits are often represented 

in a wiring diagram. The following circuit diagram represents 

the flashlight system. 

+ 

– 

In this diagram there are three elements and the wires that connect 

them. The element on the left is the battery, with the positive terminal 

being wider and marked with a + sign. The element on the top is the 

switch that turns the flashlight off and on. The element on the right is 

the light bulb. 

This chapter will be an investigation of electrical circuits and the 

common electrical devices from which circuits are built. 

§ 3.2 Electric Current 

A flow of charge, such as the charge flowing through the light bulb 

filament, is called an electric current.

58 Circuits 3.3 

Definition: Current 

The electric current, I, is defined to be the amount of charge that 

flows per time. 

I = dq 

dt 

The unit of current is coulombs per second. This 

combination of units is called the Ampere or Amp, 

abbreviated as just A: (1A = 1C 

1s ). 

Definition: Current Density 

Let A be a small area that is normal 

to the flow of current in a particular 

region. Let I be the current flowing 

through the area A. The current density, 

⃗J, is a vector in the direction of the flow 

of current. The magnitude of the current 

density is equal to the current per area. 

J = I A 


A light bulb draws a current of 1.0 mA from a battery. 

(a) How many electrons pass through the bulb in 160 seconds? 

(b) The filament has a radius of 0.20 mm. What is the current density 

through the filament? 

§ 3.3 Ohm’s Law 

When an electric field is applied to a conductor the free charges in 

the conductor begin to move. If the conductor is isolated the charges 

will distribute themselves so that the field inside the conductor is zero 

and then the charges will cease to move. One the other hand, if the 

conductor is not isolated but part of a circuit, as the filament was in 

the flashlight circuit, then a continual flow of charge can be sustained, 

and the electric field in the conductor will not be zero.

3.3 Ohm’s Law 59 

Some materials, such a copper, allow charge to pass through it 

with very little resistance, and therefore it takes very little electric field 

to sustain a large current. Other materials, such a carbon, resist the 

flow of charge, and it requires more electric field to sustain the same 

current. Carbon is more resistive. This property is quantified in the 

following law. 

Fact: Ohm’s Law: Resistivity 

For many materials the current density and the electric field are 

proportional. 

⃗J = σ ⃗ E 

The constant σ is called the conductivity and depends on the material. 

The inverse of the conductivity is called the resistivity: 

ρ = 1/σ. 

Not all materials follow this relationship: those that do are called 

ohmic materials, those that do not are called non-ohmic. 

Here is a table with the resistivity of a few materials. 

Material Resistivity(×10 −8 Ω · m) 

Silver 1.6 

Copper 1.7 

Aluminum 2.8 

Tungsten 5.5 

Iron 10 

Nichrome 100 

Carbon 3500 

Silcon 64000000000 

Wood 1000000000000000 

Amber 50000000000000000000000 

Imagine a piece of conductive material, with two terminals connected 

to it, and with a current I passing into the material through 

terminal a and draining out through terminal b. Such a circuit element 

is called a Resistor. 

In order to sustain this current there will need to be an electric field 

in the conductor going from terminal a to terminal b. Thus there will


be an electric potential difference ∆V = V a − V b = − ∫ a 

E ⃗ · ⃗dr between 

b 

the two terminals. It can be shown that, regardless of the shape of 

the conductors, this electric potential difference is proportional to the 

current if the material is ohmic. 

Theorem: Ohm’s Law: Resistance 

∆V = IR 

Where R is called the resistance of the element. 

The unit of resistance is the ohm which is one volts per amp. The 

ohm is abbreviated as Ω, so that 1Ω = 1V/1A. 


You have a wire with cross sectional area A and length L. Show that if 

the terminals are placed at the ends of this conductor that the resistance 

of this element is R = ρ L A . 


You have a block of carbon, with sides of length a, 2a, and 3a. If 

terminals are placed on two parallel sides we can make a resistor with 

this block. We have three choices for the placement of the terminals, 

the sides that are a apart, 2a apart or 3a apart. 

(a) Which choice will produce the most resistance. 

(b) Which choice will produce the least resistance. 

§ 3.4 Electric Power 

Suppose that you have a circuit element with two terminals, that 

has a current I running through it and a potential difference ∆V between 

the terminals. In a time dt an amount of charge dq = I dt will 

pass through the element. All of that charge falls through the electric 

potential difference of ∆V so that the charge dq looses an amount of 

potential energy 

dU = dq ∆V = I dt ∆V −→ dU 

dt = I ∆V. 

So we see that the element dissipates a power P = I ∆V . 

Theorem: Electrical Power 

The power dissipated in a circuit element is equal to the product 

of the current through the element and the potential difference 

between the terminals of the element. 

P = I ∆V

3.5 Kirchhoff’s Rules 61 


A 60 Watt light bulb is plugged into the wall receptacle which supplies 

an electric potential of 120 Volts. How much current runs through the 

bulb when you turn it on? 


(a) Show that a resistor with a current I running through it has a power 

of P = I 2 R. 

(b) Show that a resistor with a voltage ∆V across it has a power of 

P = (∆V ) 2 /R. 

§ 3.5 Kirchhoff’s Rules 

There are two theorems that are very useful in analysing a circuit. 

The first theorem stems from the conservation of charge, that is, that 

charge is neither created nor destroyed in a circuit. Consider a junction 

where a number of elements come together. 

Since the current does not build up at the junction the sum of current 

going into the junction must be equal to the sum of the current going 

out of the junction. In the case pictured above: I 1 + I 4 = I 2 + I 3 . 

Theorem: Kirchhoff’s Junction Rule 

The sum of the currents into a junction is equal to the sum of the 

current out of a junction. ∑ 

Iin = ∑ I out 

The second theorem stems from the conservation 

of energy. Since the electric potential is the potential 

energy per charge that is due to the electric field in 

the system, if a charge moves around a loop in a circuit 

and comes back to where it started it must be at 

the same electric potential as when it started. Thus 

if we add the electric potential differences of all the 

elements that are crossed as you go around any loop 

in the circuit the sum must be zero.


Theorem: Kirchhoff’s Loop Rule 

The sum of the electric potential differences around any closed 

loop in a circuit is zero. 

∑ 

∆V = 0 

As an example consider the following circuit with five elements. 

+ – 

+ – 

ΔV 1 

ΔV 2 

+ 

+ 

ΔV 3 

ΔV 4 

– 

+ ΔV 5 – 

There are a number of loop rules that we could write down. First, 

consider the loop starting at the bottom left corner and then taking 

the shortest route going clockwise back to the bottom left corner. You 

cross elements 1 and 3, both time going from high to low potential. 

−∆V 1 − ∆V 3 = 0 

Now the equation cannot be satisfied by positive numbers so one of the 

∆V ’s must be negative. This is alright, the algebra will sort it out in 

the end. 

Now consider starting at the lower right corner and taking the 

shortest clockwise loop. We pass through elements 5, 3, 2, and 4. 

∆V 5 + ∆V 3 − ∆V 2 − ∆V 4 = 0 

Notice which ones are positive and which are negative. 

We could also do the big loop, starting at the lower left and going 

clockwise through elements 1, 2, 4, and 5. 

−∆V 1 − ∆V 2 − ∆V 4 + ∆V 5 = 0 

It is important to notice that this last equation could have been 

arrived at by combining the first two equations, so it has not really given 

us any new information. In general it does not help to write down more 

loop equations than there are “windows” in the circuit. 

§ 3.6 Resistors in Combination 

In a circuit where the same current runs through two resistors, 

those resistors are said to be in series. If we a circuit where the same 

electric potential is across two resistors, those resistors are said to be in 

–

3.6 Resistors in Combination 63 

parallel. The diagram below will help explain why these configurations 

are named in this way. 

Suppose that we have two resistors in series. Since they are in 

series they must carry the same current. 

I 1 = I 2 = I 

We also know that the potential difference across the pair is the sum 

of the potential differences across each individual resistor; 

∆V = ∆V 1 + ∆V 2 

= I 1 R 1 + I 2 R 2 

= IR 1 + IR 2 

= I(R 1 + R 2 ) 

−→ R effective = R 1 + R 2 

We see that the pair of resistors in series still follow Ohm’s law, and 

that the pair act as a single resistor with an effective resistance of 

R effective = R 1 + R 2 . 

Series Resistors 

Parallel Resistors 

+ 

– 

ΔV 

+ 

ΔV 1 

– 

+ 

ΔV 2 

– 

I 1 

I 2 

Now consider two resistors in parallel. Since they are in parallel 

the must have the same electric potential. 

∆V 1 = ∆V 2 = ∆V 

We also know from the junction rule that the net current going into the 

system is equal to the sum of the two currents going into the resistors. 

I = I 1 + I 2 = ∆V 1 

+ ∆V 2 

R 1 R 

[ 2 

1 

= ∆V + 1 ] 

R 1 R 2 

= ∆V 

R 1 

I 

+ ∆V 

R 2 

−→ 

1 

R effective 

= 1 R 1 

+ 1 R 2


Theorem: Effective Resistance 

Series R = R 1 + R 2 

Parallel 

1 

R = 1 + 1 R 1 R 2 

§ 3.7 Capacitors in Combination 

Capacitors in series and parallel can also be treated as a single 

capacitor. The argument is much the same, but with charge rather 

than current. 

Suppose that you have two capacitors in series. Then the charges 

on each must be the same. 

Q 1 = Q 2 = Q 

While the potential drop across the pair is the sum of the potential 

drop across each 

∆V = ∆V 1 + ∆V 2 

+ 

– 

−→ 

Series Capacitors 

= Q 1 

+ Q 2 

= Q + Q C 1 C 2 C 1 C 2 

1 

C = ∆V 

Q = 1 + 1 C 1 C 2 

+ 

Parallel Capacitors 

+Q 

ΔV 1 

-Q – 

+Q 1 +Q 2 

ΔV 

+Q + 

-Q 

ΔV 1 

-Q 2 

2 

-Q 

For capacitors in parallel the potential is the same. 

– 

∆V 1 = ∆V 2 = ∆V 

While the net charge that flows into the system is shared between the 

capacitors. 

Q = Q 1 + Q 2 

= C 1 ∆V 1 + C 2 ∆V 2 = C 1 ∆V + C 2 ∆V 

−→ C = 

Q 

∆V = C 1 + C 2

3.8 Capacitor Circuits 65 

Theorem: Effective Capacitance 

Series 

1 

C = 1 C 1 

+ 1 C 2 

Parallel C = C 1 + C 2 

§ 3.8 Capacitor Circuits 

Consider the following circuit, which is composed of a power supply 

with a fixed electric potential output of V S , a capacitor, a resistor, and 

a double pole switch. 

b 

+ 

a 

V S +Q 

– 

I -Q 

+ 

V C 

– 

– 

To begin with the switch is in position a. In this position all of 

the charge will drain from the capacitor. At the time t = 0 the switch 

is moved to position b. In this position the power supply will begin to 

fill the capacitor with charge. The current I flows into the capacitor. 

Thus the charge on the capacitor increases at the rate 

dQ 

dt = I 

But for a capacitor the charge is proportional to the electric potential 

across the capacitor Q = CV C so that 

I = dQ 

dt = C dV C 

dt 

Kirchhoff’s loop rule gives us the following equation. 

But by Ohm’s Law we have 

V R 

+ 

V S − V C − V R = 0 

V R = RI = RC dV C 

dt 

Putting this into the loop rule equation we find 

V S − V C − RC dV C 

dt 

= 0


This is a differential equation that describes how the voltage on the 

capacitor changes with time, similar to how F = ma is a differential 

equation that describes how a particles position changes with time. In 

this particular case we can find the solution by noting that if we define 

a new variable, f(t) = V C − V S then df 

dt = dV C 

dt 

Plugging this into our differential equation we find 

−f − RC df 

dt = 0 

or 

df 

dt = − 1 

RC f 

since V S is a constant. 

This just says that the derivative of f is simply a constant times f. 

The only function that has this property is the exponential. So let us 

try a function of the form 

f = Ae αt −→ df 

dt = αAeαt = αf 

Comparing this with our differential equation we see that it is the same 

if 

α = − 1 

RC 

Thus 

V C − V S = f = Ae −t/RC 

−→ V C = V S + Ae −t/RC 

In order to determine the constant A we need to use the initial condition 

of the system. At t = 0 we know that there was no charge on the 

capacitor, thus the electric potential on the capacitor was zero at t = 0. 

Putting this into the solution we fine that 

0 = V C (0) = V S + Ae −0/RC = V S + A −→ A = −V S 

So the voltage on the capacitor is 

V C = V S (1 − e −t/RC ) 

The combination RC is called the time constant of the circuit because it 

has units of time and gives the time scale for the change in the voltage: 

in a time RC the difference of the electric potential from it’s final value 

(V S ) decreases by a factor of e −1 ≈ 3/8.


Note that the equation for V C does not depend on R or C separately, 

but only on the combination RC. 


Suppose that the capacitor circuit described above is assembled with 

a 10 Volt power supply, a 5.0kΩ resistor, and a 3.0µF capacitor. How 

long after the switch is put in position b, will it be before the capacitor 

is half charged (5.0V)? 


Assume that the capacitor circuit described above is left with the switch 

in position b for a long time, so that the capacitor is fully charged to 

the voltage V S . The switch is now moved to position a. Show that 

V C = V S e −t/RC . 


Example 

The largest part of the nerve cells in your body is called the “axon.” 

The axon is in the shape of a cylinder and has sections through which 

ions pass, sending electrical signals through the nerve. During a nerve 

impuse, for the cylindrical section of an axon indicated below, 10,000 

sodium ions (Na + ) pass through the surface of the cell membrane in 1 

m/s. What is the current and current density into the cell? 

Na + 

20μm 

Na + 

100μm


Each sodium ion has a charge +1.6 × 10 −19 C, so for 10,000 ions, ∆Q = 

10, 000 × (1.6 × 10 −19 C) = 1.6 × 10 −15 C. Since it takes 1ms: 

I = ∆Q 

∆t = 1.6 × 10−15 C 

= 1 × 10 −12 A = 1pA. 

.001s 

The ions flow through the sides of the cylinder, so the magnitude of 

the current density is 

J = I S = 1.6 × 10 −12 A 

(2π)(10µm)(500µm) = 5.1 × A 10−5 m 2 = 51µA m 2 

Example 

A copper wire, with a diameter of 1mm, has a current of 5A flowing 

through it. What is the electric field in the wire? 

Comment: How can E ≠ 0? Earlier it was stated that E = 0 inside a 

conductor, however this was for the case of electrostatics. In the case 

of a current flowing, since electric charges are not static, the electric 

field will be nonzero. 

For our wire, the current density is 

J = I A = 5A 

π(0.0005m) 2 = 6.4 × A 106 m 2 

From Ohm’s law: 

J = 1 ρ E 

−→ E = ρJ = (1.7 × 10 −8 Ωm)(6.4 × 10 6 A m 2 ) 

= 0.11 Ω·A 

= 0.11 V m 

m 

Example 

A light bulb with a resistance of 3Ω is connected to a 9V battery. In 

one second how many electrons flow through the bulb? 

I 

R = 3Ω 

- 

+ 

9V 

Use Ohm’s law to find the current: 

I = ∆V 

R 

= 9V 

3Ω = 3A


This mean that in one second, 3C of charge flows through the resistor. 

Since each electron (which will flow in a direction opposite the 

conventional positive current) carries 1.6 × 10 −19 C of charge: 

3A 

N = 

1.6 × 10 −19 C × (1s) = 1.9 × 1016 electrons 

Example 

A 20W light bulb is left on in your car’s interior. If the car’s 12V 

battery is fully charged and has a capacity of 200Amp-hours. How 

long will it take to completely discharge the battery (assuming the 

terminal voltage remains 12V throughout the discharge)? 

Use electric power P to compute the current that flows: 

P = I∆V −→ I = P 

∆V = 20W 

12V = 1.67A 

The amount of charge that will pass through the bulb in a time ∆t 

is ∆Q = 1.67A × ∆t. The 200 Amp-hour rating tells us how much 

charge can pass from the positive terminal to the negative terminal of 

the battery before discharging the battery, so 

∆Q200Ahr = (1.67A)∆t 

−→ ∆t = 200Ahr 

1.67A 

How much charge has flowed? 

= 120hr = 5days 

200Ahr = 200A(3600s) = (200 C )(3600s) = 720, 000C 

s 

Example 

Determine the currents in the three resistors of the following circuit: 

12V 

6Ω 

- 

+ 

6V 

+ 

- 

3Ω 

9Ω 

Let’s label the currents. We will do our best to indicate the correct 

directions of the current:


6Ω 

12V 

- 

+ 

6V 

+ 

- 

I 1 

I 2 3Ω I 3 

9Ω 

A 

We now apply the loop rules to the two loops indicated. In both cases, 

start from the point labeled A, moving in the direction indicated, summing 

the potential gains or losses across each circuit element. For batteries 

the potential increases from the negative to the positive terminal. 

For resistors the current flows from high to low potential. Applying the 

loop rules gives two equations, on for each loop: 

−(3Ω)I 2 − (6Ω)I 1 + 6V = 0 

−(3Ω)I 2 + 12V − (9Ω)I 3 = 0 

We also must apply current conservation. Choose the point labeled A: 

I 1 − I 2 + I 3 = 0 

Now we have 3 equations with three unknowns. Solving the linear set 

of equations yields: 

I 1 = 0.364A 

I 2 = 1.27A 

I 3 = 0.91A 

Check that these are correct by putting them back into the loop equations: 

−(3Ω)(1.27A) − (6Ω)(0.364A) + 6 −→ −6.0V + 6.0V = 0 

−(3Ω)(1.27A) + 12 − (9Ω)(0.91A) −→ −12.0V + 12.0V = 0 

Example 

Determine the currents in the resistors of the following circuit:


15Ω 

5Ω 

10Ω 

15V 

+ 

- 

10V 

+ 

- 

15Ω 

Label the currents, including best guesses for the directions: 

I 1 

I 2 

I 3 

15Ω 

5Ω 

10Ω 

+ 

- 

15V 

+ 

- 

10V 

I 1 

A 

15Ω 

B 

Apply Kirchoff’s loop rules. First to the lefthand loop, beginning at 

point A and moving clockwise: 

15V − (15Ω)I 1 − (5Ω)I 2 + 10V − (15Ω)I 1 = 0, 

then to the righthand loop, beginning at point B and moving counterclockwise: 

(10Ω)I 3 − (5Ω)I 2 + 10V = 0. 

Note that for a resistor, the conventional positive current flows from 

high to low potential, which determines the sign to use when moving 

from one side of a resistor to the other. 

Now apply Kirchoff’s junction rule at the point labeled B: 

I 2 + I 3 − I 1 = 0 

Simplifying, we have three equations with three unknowns: 

25 − 30I 1 − 5I 2 = 0 

10 + 10I 3 − 5I 2 = 0 

I 2 + I 3 − I 1 = 0 

To solve, let first eliminate I 3 : 

10 + 10I 3 − 5I 2 = 10 + 10(I 1 − I 2 ) − 5I 2 = 0


−→ 10 + 10I 1 − 15I 2 = 0 

Next multiply the first equation, of the three listed above, by -3: 

−→ −75 + 90I 1 + 15I 2 = 0. 

Add the two modified equations together: 

10 + 10I 1 − 15I 2 − 75 + 90I 1 + 15I 2 = 0 

−→ −65 + 100I 1 = 0 

−→ I 1 = .65A 

Using this value of I ! , we can solve for I 2 and I 3 : 

I 2 = 1.1A 

I 3 = −0.45A 

The negative value obtained for I 3 indicates that the direction for 

this current is opposite the direction that was chosen. So I 3 flows 

up through the 10Ω resistor, not down as indicated in our diagram. 


Repeat the analysis for the circuit in the previous example, using the current 

directions shown below: 

I 1 

I 2 

I 3 

15Ω 

5Ω 

10Ω 

+ 

- 

15V 

+ 

- 

10V 

I 1 

A 

15Ω 

B 

Example 

You should not get any negative-valued currents. 

What is the effective resistance between the points A and B for the 

following circuit. 

R = 30Ω 

4 

A 

R = 15Ω 

2 

R = 10Ω 

1 

R = 4Ω 

3 

B


First combine R 1 and R 2 . Since they are in parallel: 

1 

= 1 + 1 = 1 

R 12 R 1 R 2 10Ω + 1 

15Ω −→ R 12 = 6Ω 

The circuit has been reduced to: 

R = 30Ω 

4 

A 

R = 6Ω 

12 

R = 4Ω 

3 

Now combine R 12 and R 3 , which are in series: 

This leaves the effective circuit: 

R 123 = R 12 + R 3 = 6Ω + 4Ω = 10Ω. 

R = 30Ω 

4 

B 

A 

R = 10Ω 

123 

B 

Finally, combine the last two resistances in parallel: 

1 

= 1 + 1 = 1 

R AB R 123 R 4 10Ω + 1 

30Ω −→ R 12 = 7.5Ω 

Example 

A 12V battery is connected across the points AB in the circuit from the 

previous example. What current flows through the R 3 = 4Ω resistor? 

First find the current that flows from A to B: 

I AB = V AB 

= 12V 

R AB 7.5Ω = 1.6A 

The incoming current splits between the 30Ω resistor and the effective 

resistance R 123 : 

A 

I AB 

I 4 

I 123 

R = 30Ω 

4 

R = 10Ω 

123 

B 

We can easily compute the current through the 30Ω resistor, since the 

potential difference across it is V AB : 

I 4 = V AB 

= 12V 

R 4 30Ω = 0.4A 

Using Kirchoff’s junction rule: 

I 123 = I AB − I 4 = 1.6A − 0.4A = 1.2A.


Because the R 3 = 4Ω resistor in connected in series with the parallel 

combination of R 1 and R 2 , the current through it is I 123 : 

−→ I 3 = 1.2A 



In the Bohr model of the hydrogen atom, an electron in the lowest 

energy state follows a circular path, 5.29 × 10 −11 m from the proton. 

Show that the speed of the electron is 2.19×10 6 m s 

. What is the effective 

current associated with this orbiting electron? 


In a particular cathode ray tube, the measured beam current is 20µA. 

How many electrons strike the tube screen every 1 30 seconds? 


The quantity of charge q (in coulombs) passing through a surface of 

area 2.0cm 2 varies with time as q = 4t 3 + 5t + 6 where t is in seconds. 

What is the instantaneous current through the surface at t = 1.0s ? 

What is the value of the current density? 


An electric current is given by I(t) = 100.0 sin(120πt), where I is in 

amperes and t is in seconds. What is the total charge carried by the 

current from t = 0 to t = 1/240s ? 


Calculate the average drift speed of electrons traveling through a copper 

wire with a cross-sectional area of 1.00mm 2 when carrying a current 

of 1.00A. It is known that about one electron per atom of copper 

contributes to the current. The atomic weight of copper is 63.54 and 

its density is 8.92g/cm 3 . 


Eighteen gauge wire has a diameter of 1.024mm. Calculate the resistance 

of 15.0m of 18-gauge copper wire at 20.0 ◦ C. 


Suppose that a voltage surge produces 140V for a moment. By what 

percentatge will the ouput of a 120V, 100W lightbulb increase, assuming 

its resistance does not change? 


Two cylindrical copper wires have the same mass. Wire A is twice as 

long as wire B. What is the ratio of their resistances?



Batteries are rated in ampere hours (A · hr), where a battery rated at 

1.0A · hr can produce a current of 1.0A for 1.0hr. 

(a) What is the total energy stored in a 12.0V battery rated at 55.0A·hr 

? 

(b) At $0.12 per kilowatt hour, what is the value of the electrical energy 

stored in this battery? 


What is the required resistance of an immersion heater that will increase 

the temperature of a 1.5 kg of water from 10 ◦ C to 50 ◦ C in 10 

min while operating at 110V? 


A battery with an emf of 12V and internal resistance of 0.90Ω is connected 

to a load resistor R. 

(a) If the current in the circuit is 1.4A, what is the value of R? 

(b) What power is dissipated in the internal resistance of the battery? 


(a) Find the equivalent resistance between points a and b in the figure 

below. 

(b) If a potential difference of 34V is applied between points a and b, 

calculate the current in each resistor. 

7 Ω 

a 

4 Ω 

9 Ω 

b 

10 Ω 


For the figure below find the current in the 20Ω resistor and the potential 

difference between points a and b. 

25 V 

10 Ω - + 

a 10 Ω b 

5.0 Ω 20 Ω 

5.0 Ω 


Determine the equivalent capacitance for the capacitor network shown 

below. If the network is connected to a 12V battery, calculate the 

potential difference across each capacitor and the charge on each capacitor.


3.0 µF 6.0 µF 

2.0 µF 


Four capacitors are connected as shown. Find the equivalent capacitance 

between points a and b. Calculate the charge on each capacitor 

in V ab = 15V. 

15 µF 3.0 µF 

20 µF 

6.0 µF 


When two capacitors are connected in parallel, the equivalent capacitance 

is 4.00µF. If the same capacitors are reconnected in series, the 

equivalent capacitance is one-fourth the capacitance of one of the two 

capacitors. Determine the two capacitances. 


For the system of capacitors shown below find the equivalent capacitance 

of the system, the potential across each capacitor, the charge on 

each capacitor, and the total energy stored by the group. 

3 µF 6 µF 

2 µF 4 µF 

90 V 


Calculate the equivalent capacitance between the two points shown 

in the circuit below. Note that this is not a simple series or parallel 

combination.


4 µF 

+ - 

2 µF 

+ 

- 

8 µF 

+ 

- 

+ - 

+ 

- 

4 µF 

2 µF 


If R = 1.0kΩ and E = 250V determine the direction and magnitude of 

the current in the horizontal wire between a and e. 

R c 

b 

2R d 

ε + - 

4R 

3R 

+ 

- 

2ε 

a 


Determine the current in each branch of the figure. 

3.0 Ω 

e 

5.0 Ω 

8.0 Ω 

1.0 Ω 

1.0 Ω 

4 V 

12 V 


A 25-W light bulb is connected in series with a 100-W light bulb and 

a voltage V is placed across the combination. Which bulb is brighter? 

Explain. 


An electric heater is rated at 1500W, a toaster at 750W, and an electric 

grill at 1000W. The three appliances are connected to a common 120V 

circuit. How much current does each draw? Is a 25A circuit sufficient 

in this situation? 


An 8-foot extension cord has two 18-gauge copper wires, each having 

a diamter of 1.024mm. How much power does this cord dissipate when 

carrying a current of 1.0A? How much power does this cord dissipate 

when carrying a current of 10A?



Because aluminum has a greater resistivity than copper, aluminum 

wires heat up more than copper wires when they carry the same currents. 

For this reason aluminum wire of the same gauge (diameter) is 

rated to carry a smaller current. What current in an aluminum wire 

will heat up the wire the same amount as a copper wire of the same 

gauge carrying 20 Amps? 


Consider the following circuit. 

12 kΩ 

9.0 V 

R 2 = 15 kΩ 

10 µF 

3 kΩ 

The switch is closed and the capacitor is allowed to charge up. 

(a) How much charge will there be on the capactor when it is fully 

charged? 

(b) Now the switch is opened and the charge on the capacitor drains 

off. How long will it take for the capacitor to reach 1/5 of the charge 

it had at the moment the switch was opened?



Definitions 

Electric Current: 

Current Density 

I = dq 

dt 

J = I A 

Facts 

Ohm’s Law: For many materials current density is proportional to 

the electric field. 

⃗J = σ ⃗ E 

with σ the conductivity of the material. ρ = 1/σ is called the resistivity. 

∆V = IR 

with R the resistance of the device. 

Theorems 

Electric Power: 

Kirchhoff’s Loop Rule: 

P = I∆V 

∑ 

∆V = 0 

Kierchhoff’s Junction Rule: 

∑ 

Iin = ∑ I out 

Resistors in Parallel: 

Resistors in Series: 

1 

R eff 

= 1 R 1 

+ 1 R 2 

Capacitors in Parallel: 

R eff = R 1 + R 2 

Capacitors in Series: 

C eff = C 1 + C 2 

1 

C eff 

= 1 C 1 

+ 1 C 2

80 Magnetic Fields 4.11

4.2 Magnetic Force on a Current 81 

4 Magnetic Fields 

§ 4.1 Magnetic Field 

We have all played with magnets at some point. Magnets will 

attract some types of metals and can attract and repel other magnets. 

This seems similar to electric forces. Magnetic forces can indeed be 

explained in terms of magnetic fields, but the connection between the 

magnetic field and the magnetic force is significantly different from 

that for electric fields and force. The magnetic field is similar to the 

electric field in that they are both vector fields. We use the symbol B ⃗ to 

represent the magnetic field. If we put a compass needle in a magnetic 

field, the field will turn the needle until it is aligned with the direction 

of the field. At this point we will use the following simple operational 

definition of the magnetic field. The direction of the magnetic field 

is the direction that a compass needle points. The magnitude of the 

magnetic field is related to how strong the force is that aligns the needle 

with the direction of the field. 

A magnetic field does apply a force to charged particles. The 

magnetic force on the particle has the following properties: 

• F ∝ q 

• F ∝ v 

• F ∝ B 

• F ⃗ is perpendicular to both ⃗v and B. ⃗ These can be combined into the 

following succinct formula for the force. 

Fact: Magnetic Force 

A particle with charge q and velocity ⃗v in a magnetic field ⃗ B will 

feel a force 

⃗F = q ⃗v × ⃗ B 

Notice that there is no force if the velocity is parallel to the magnetic 

field. Also notice that the magnetic force is zero on a stationary 

particle.

82 Magnetic Fields 4.2 

We can determine the dimensions of the magnetic 

field from the force equation: 

F = qvB −→ B = F qv 

The unit of magnetic field is the Tesla, abbreviated as 

just T. 

Newton 

Tesla = 

Coulomb · meter/second = Kilogram 

Coulomb · second 

§ 4.2 Magnetic Force on a Current 

If a current carrying wire is placed in a magnetic field it will experience 

a magnetic force. The current in the wire is composed of many 

moving charges. Each of these charges experiences a magnetic force. 

The force on the wire is the sum of the forces on all the moving charges 

in the wire. 

Theorem: Magnetic Force on a Current 

Suppose that you have a wire that is carrying a current I. A small 

section of wire of length ⃗ dl, where the vector points in the direction 

of the current, will experience a force 

⃗ dF = I ⃗ dl × ⃗ B 

The force on a longer section of wire can be found by integrating 

over the length of the section. 

∫ 

⃗F = I dl ⃗ × B ⃗ 

If the magnetic field is uniform over the region containing the wire 

then the following result can be proved. 

Theorem: Force on a Current in a Uniform Field 

Let ⃗ ∆l be a vector that points from the beginning to the end of a 

section of wire carrying a current I. If that section is in a uniform 

magnetic field, the force on that section is 

⃗F = I ⃗ ∆l × ⃗ B 


Consider a semicircular piece of wire or radius R in the first two quadrants 

of the x-y plane. The wire carries of current I in the counterclockwise 

direction.

4.3 Trajectories Under Magnetic Forces 83 

y 

dl 

R 

dθ 

θ 

x 

There is a uniform magnetic field in the y direction, B ⃗ = Bĵ. We 

wish to compute the net force on this section of wire without using the 

theorem F ⃗ = I∆l ⃗ × B. ⃗ 

(a) If we break the semicircle into small sections, they will be small 

sections of arc, as pictured in the diagram above. If we take one of 

these, it will be at a position θ, and will subtend an angle dθ. Show 

that dl ⃗ = (− sin θî + cos θĵ)Rdθ 

(b) With this result you can now compute the integral F ⃗ = ∫ I dl ⃗ × B. ⃗ 

Show that the net force on the semicircle is −I2RBˆk. 

(c) Show that this is the same answer you get when you apply the 

theorem F ⃗ = I∆l ⃗ × B. ⃗ 


Consider the rectangular current loop pictured below. 

w 

I 

B 

B 

The field is uniform and in the direction indicated in the diagram. 

Show that the torque about the axis indicated by the dotted line is 

IAB where A is the area of the loop. 

§ 4.3 Trajectories Under Magnetic Forces 

Because the magnetic force is perpendicular to the velocity of the 

particle, the magnetic force can not do work on the particle. Suppose 

that a particle moves a small distance ⃗ dr. The work done by a force ⃗ F 

as the particle moves is 

dW = ⃗ F · ⃗dr 

But ⃗ dr = ⃗v dt so 

dW = ⃗ F · ⃗v dt 

h


For a magnetic force, ⃗v and ⃗ F are prependicular so that ⃗ F ·⃗v = 0. Thus 

dW = 0 for a magnetic force. 

Theorem: Work by Magnetic Force 

The magnetic force does zero work. Thus the magnetic force can 

change the direction but not the speed of a particle. 

Because the magnetic force changes only the direction of a particle, 

a magnetic field is useful for steering charged particles, once they are 

already moving. 

Consider a particle that is in a region with a uniform magnetic 

field. Suppose that the particles initial velocity is perpendicular to the 

magnetic field. Since the magnetic force is always perpendicular to 

the direction of the magnetic field, the force will have no component 

parallel to the field. So, since the particle started with zero velocity 

parallel to the field it will continue to have zero velocity parallel to the 

field. But this means that the velocity will always be perpendicular to 

the magnetic field, so |⃗v × B| ⃗ = vB. Thus we know that the magnitude 

of the magnetic force is 

F ≡ | ⃗ F | = q|⃗v × ⃗ B| = qvB 

In addition we know by the previous theorem that the speed of the 

particle will be constant, v = v 0 . So that the magnitude of the force 

is constant. In summary, we see that the magnitude of the magnetic 

force on the particle is constant and perpendicular to the velocity, so 

that the acceleration of the particle is constant and perpendicular to 

the velocity. We have run into this situation before: In circular motion 

the acceleration is constant and always perpendicular to the velocity. 

We are lead by this observation to the following theorem. 

Theorem: Circular Trajectories 

If a particle is in a region of uniform magnetic field and entered 

the region with a velocity perpendicular to the magnetic field, then 

the particle will execute circular motion while in the region. 

One can relate the radius and velocity of the motion to the field 

strength and charge by employing Newton’s second law. 

F = ma 

−→ qvB = m v2 

r 


One can use the circular motion of a charge particle to determine it’s 

mass if you already know it’s charge. Suppose that you send a particle

4.4 Lorentz Force 85 

with a charge 1.6 × 10 −19 C into a field 0.11mT with a speed of 3.83 × 

10 5 m s 

. The radius of the resulting motion is 2cm. What is the mass of 

the particle? 


A stream of charged particles 

enter a circular region of uniform 

magnetic field as shown (gray). The 

particles fan-out in the region and 

each particle exits along one of the 

six trajectories A through F . Pick 

the trajectories for particles with the 

following properties: 

(a) Which particles are positively charged? 

(b) If all the particles have the same mass and charge, which have the 

highest speed? 

(c) If all the particles have the same speed and charge, which have the 

highest mass? 

(d) If all the particles have the same mass and speed, which have the 

highest charge? 


A particle is moving in a circular trajectory because of a magnetic field. 

Show that regardless of the velocity of the particle, it will take the same 

amount of time to complete one revolution. The fact that the time is 

a constant eased the development of the cyclotron, an early particle 

accelerator. 

§ 4.4 Lorentz Force 

If there is a magnetic as well as electric field then both forces act 

on a charge particle at the same time. 

Definition: Lorentz Force 

The combined electric and magnetic forces on a charge particle is 

called the Lorentz Force. 

⃗F = q( ⃗ E + ⃗v × ⃗ B) 

A 

F 

E 

B 

D 

C 


There is a region with a uniform magnetic field ⃗ B = Bˆk and electric 

field ⃗ E = Eĵ. Particles with charge q are sent into this region with a 

velocity ⃗v = vî.


(a) Show that if v = E/B that the particles will go straight through 

the region without deflection. 

(b) Show that if v > E/B and q > 0 then the particles will be deflected 

in the negative y direction. 

(c) Show that if v > E/B and q < 0 then the particles will be deflected 

in the positive y direction. 

(d) Show that if v < E/B and q > 0 then the particles will be deflected 

in the positive y direction. 

A device setup with crossed magnetic and electric fields as in the 

previous problem is called a velocity selector. Since it sorts out the 

particles based on their speeds, if you select out just the ones that 

go straight ahead, you have selected particles with a particular speed 

v = E/B. By adjusting the field strengths you can choose what velocity 

you would like to select. 

§ 4.5 Hall Effect 

Let us look with more detail at the flow of current in a wire that 

is in a magnetic field. Suppose that we have a rectangular conductor 

with a current I running through it. 

w 

h 

If we consider the charges in the wire we see that there will be a 

magnetic force on the particles that will tend to deflect them to one 

face of the conductor. 

I 

- 

I 

But this force will be acting on all the free charges, so they will all shift 

upward as they move through the conductor. Keeping in mind that 

the positive charge in the conductor does not move, we see that this 

shift of the (negative) free charges will build up a net negative charge 

on one face of the conductor and net positive charge on the opposing 

face. 

-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+-+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

I 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

I 

- - - - - - - - - - - - - - - - - - - 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+ 

+

4.5 Hall Effect 87 

But this charge separation will create an electric field, that is in the 

opposite direction to the magnetic force, thus the charge will continue 

to build up on the face of the conductor until the electric field builds 

up to the point where the electric force on the charges balances the 

magnetic force on the charges. 

- - - - - - - - - - - - - - - - - - - 

I 

E E E E E 

+ + + + + + + + + + + + + + + + + + + 

This will happen any time that a current carrying wire is in a magnetic 

field: the wire will spontaneously generate an electric field across the 

wire (not end to end but across). This effect, discovered by Edwin 

Hall, is called the Hall Effect. Because there is an electric field across 

the conductor there will also be an electric potential difference (called 

the Hall voltage), ∆V = Ew where w is the width of the conductor. 

But we know that in order for the electric force to cancel the magnetic 

force, we need E = vB. So we see that 

∆V = vBw −→ v = ∆V 

Bw 

Because it is relatively easy to measure the electric potential difference, 

the width of the conductor and the magnetic field strength, this type 

of device can be used to measure the velocity of the charge carriers. 

Knowing the velocity allows one to determine how many charge carriers 

there are per volume. In a time dt a quantity of charge dq = I dt passes 

out the end of the wire, and a number of electrons dn = dq/e = I dt/e 

passes out of the wire. But we can also say that the charges have moved 

a distance dx = v dt in this time, so that a section of charge v dt long 

has passed out of the wire. Thus a volume of charge dV = hw v dt 

has passed out of the wire (where h is the thickness of the wire and hw 

is the cross sectional area of the wire). So the number of conduction 

electrons per volume (η) is 

η = dn 

dV = I dt/e 

hw v dt = I 

ehw v 

The carrier density, η, depends on the type of material, not on the 

geometry of the material or the amount of current flowing through the 

material. 

The Hall effect can be used to make a magnetic field sensor. By 

using the equation v = ∆V 

Bw 

to eliminate the velocity from the equation 

η = 

I 

ehw v 

, and then solving for B we find that 

B = eηh ∆V 

I . 

The e, η and h are all constants for a given device, while ∆V and I are 

I


both easily measured. 

The Hall effect also allows us to determine if the charge carriers 

are positive or negative. In the analysis above we assumed that the 

charge carriers were negatively charged electrons. Consider what would 

happen if the charge carriers were positive instead. The positive charges 

would also be deflected by the magnetic field. 

I 

+ 

I 

Which will build up a field in the opposite direction. 

I 

+ + + + + + + + + + + + + + + + + + + 

E E E E E 

- - - - - - - - - - - - - - - - - - - 

Thus we can determine from the sign of the Hall voltage, if the charge 

carriers are positive or negative. There are some types of semiconductors 

that have positive charge carriers, for example silicon with a little 

bit of aluminum mixed in. Semiconductors with positive charge carriers 

are called p-type semiconductors. Semiconductors with negative 

charge carriers are called n-type semiconductors. 

I 


Example 

An electron is injected horizontally into a parallel plate capacitor with 

a velocity v = 4 × 10 6 m s : A, +Q 

-e 

v 

A, -Q 

The plates are square, with sides 20cm and the charge on the capacitor 

is 7.5µC. Describe what magnetic field is required such that the net 

force on the electron is zero between the capacitor plates. Ignore any 

edge effects. 

The electric field points down, since the top plate on the capacitor is 

positive. However, since the electron has a negative charge, the electric 

force is up. Thus, the force caused by the magnetic field must be down. 

Here is what the force diagram on the electron must look like


-e 

F E 

v 

F B 

= -e(v x B) 

Again, since the electron’s charge is negative, the magnetic force will 

point in the direction opposite the cross-product ⃗v × ⃗ B. So, ⃗ B must 

point into the page. The magnitude of B can be determined since the 

net force is to be zero: 

F B = F E 

−→ evB = eE 

Or, solving for B: 

B = E v 

For a parallel plate capacitor: E = σ/ɛ 0 , so 

B = 

σ Q 

ɛ 0 v = A 

ɛ 0 v 

7.5 × 10 −6 C 

−→ B = 

(0.2m) 2 (8.85 × 10 −12 C 2 

Nm 

)(4 × 10 6 m 

2 s ) = 5.3T 

So, ⃗ B must be perpendicular to the electric field, pointing into the page 

with a magnitude of 5.3T for the electon to have no net force on it. 

Example 

A proton is injected from a region with no magnetic field into a region 

with a strip of uniform magnetic field that is 10cm wide: 

+e 

v = ? 

L = 10cm 

B = 0.3T 

The magnetic field points into the page and has a magnitude of 0.3T. 

What is the maximum velocity the proton can have such that it does 

not make it across the 10cm magnetic-field region and exit through the 

other side? 

Upon entering the magnetic field, the proton will experience a force 

perpendicular the velocity, which will cause it to follow a circular path:


+e 

v = ? 

r 

L = 10cm 

B = 0.3T 

Let’s compute the radius of the path in terms of the velocity. Newton’s 

second law applied to the proton yields 

F B = m v2 

r 

evB = m v2 

r 

−→ r = mv 

eB 

So as v increases so does r. In order that the proton not exit through 

the other side of the magnetic field region we must have r < L; 

mv 

eB < L 

−→ v < eBL 

m 

= 2.9 × 106 m s 

Example 

A straight section of wire that is 0.5m long and carrying a current of 

I = 8A directed along the +x axis. One end of the wire is at the origin 

. The wire is in a magnetic field 

⃗B = ĵ 

(3 T m 2 ) 

x 2 + ˆk 

What is the force on the wire? 

Look at a small section of the wire: 

y 

( ) 

2 T x. 

m 

x 

dx 

x 

z


The force on this small section is: 

dF ⃗ = Id⃗x × B ⃗ 

= (Idx)î × 

( ) 

ĵ 

(3 T m 2 

= (Idx) 

(ˆk(3x 2 ) − ĵ(2x) 

x 2 + ˆk 

( 

) 

2 T m 

) ) 

x 

Integrate over the entire length of the wire to get the total force: 

0.5m ∫ 

) 

⃗F = (Idx) 

(ˆk(3x 2 ) − ĵ(2x) 

0 

= I [ î(x 3 ) − ĵ(x 2 ) ] 0.5m 

0 

= (8A) [ (0.5) 3 î − (0.5) 2 ĵ ] [T · m] 

= (1N)î − (2N)ĵ 

Example 

An L-shaped section of wire, carryies a current I, and is in a constant 

magnetic field, as shown in the figure: 

L 1 

L 2 

I 

B 

In terms of the parameters shown, what is the direction and magnitude 

of the net magnetic force on the wire? 

Use the right hand rule to find the direction of the force on each segment: 

L 1 

L 2 

F 1 

F 2 

I 

B


The net force is 

⃗F = îF 1 − ĵF 2 

= îIL 1 B − ĵIL 2 B 

So the direction is given by 

( ) ( ) 

θ = tan −1 IL2 B 

= tan −1 L2 

, 

IL 1 B 

L 1 

where θ is below the x-axis. The magnitude of the force is 

√ 

F = 

√F1 2 + F 2 2 = IB L 2 1 + L2 2 

Example 

What is the net force on an a × b rectangular loop of wire carrying a 

current I that is in a uniform magnetic field perpendicular to the loop’s 

plane? 

Assume a direction for the magnetic field and use the right hand rule 

to draw the forces on each segment of the loop: 

a 

F 1 

b 

F 2 

F 3 

F 4 

B 

The net force on the loop is: 

⃗F = ⃗ F 1 + ⃗ F 2 + ⃗ F 3 + ⃗ F 4 

= î(F 2 − F 4 ) + ĵ(F 3 − F 1 ) 

= î(IbB − IbB) + ĵ(IaB − IaB) 

= 0 

I 

Example 

Show that the magnetic force due to a uniform magnetic field for any 

closed current loop is zero. 

The force law states 

∮ 

⃗F = I d ⃗ l × B ⃗


Imagine the closed loop to be divided up into many d ⃗ l i , approximated 

by a many-sided polygon: 

B 

I 

So, the integral is just the limit of a sum over all of the small segments: 

∮ 

d ⃗ l × ⃗ B = d ⃗ l 1 × ⃗ B + d ⃗ l 2 × ⃗ B + · · · 

= (d ⃗ l 1 + d ⃗ l 2 + · · ·) × B ⃗ (∮ ) 

= d ⃗ l × B ⃗ 

⃗B is outside the integral because it is constant. The vector sum indicated 

by the integral in the last line must vanish since the vectors form 

a close polygon; graphical vector addition yields zero. Thus, 

∮ 

⃗F = I d ⃗ l × B ⃗ 

(∮ ) 

= I d ⃗ l × B ⃗ 

= 0 



Consider an electron near the equator. In which direction does it tend 

to deflect by the magnetic field of the earth if its velocity is directed 

(a) downward, (b) northward, (c) westward, or (d) southeastward? 


An electron moving along the positive x axis perpendicular to a magnetic 

field experiences a magnetic deflection in the negative y direction. 

What is the direction of the magnetic field? 


An electron in a uniform electric and magnetic field has a velocity of 

1.2×10 4 m s in the positive x direction and an acceleration of 2.0×1012 m s 2


in the positive z direction. If the electric field has a strength of 20N/C 

in the positive z direction, what is the magnetic field in the region? 


A duck flying due north at 15 m s 

passes over Atlanta, where the Earth’s 

magnetic field is 5.0 × 10 −5 T in a direction 60 ◦ below the horizontal 

line running north and south. If the duck has a net positive charge of 

0.040µC, what is the magnetic force acting on it? 


A proton moves with a velocity of v = (2î − 4ĵ + ˆk) m s 

in a region in 

which the magnetic field is B ⃗ = (î + 2ĵ − 3ˆk)T. What is the magnitude 

of the magnetic force this charge experiences? 


Show that the work done by the magnetic force on a charged particle 

moving in a magnetic field is zero for any displacement of the particle. 


Below is shown a cube (40cm on each edge) in a magnetic field with a 

wire carrying a current I = 5.0A over the surface of the cube. If there 

is a magnetic field B ⃗ = 0.020Tĵ. What is the magnitude and direction 

of the magnetic force on each straight segment of the current loop? 

y 

B 

d 

a 

I 

z 

c 

b 

x 


A singly charged positive ion has a mass of 3.20×10 −26 kg. After being 

accelerated through a potential difference of 833V, the ion enters a 

magnetic field of 0.920T along a direction perpendicular to the direction 

of the field. Calculate the radius of the path of the ion in the field. 

Find the velocity of the particle first then use the circular motion 

result. 


A proton (charge +e, mass m p ), a deuteron (charge +e, mass 2m p ), 

and an alpha particle, (charge +2e, mass 4m p ) are accelerated through 

a common potential difference, V . The particles enter a uniform magnetic 

field B, in a direction perpendicular to B. The particles move in


circular orbits. Write the radii of the orbits of the deuteron, r d , and 

the alpha particle, r α , in terms of the radius of the protons orbit r p . 


Indicate the initial direction of the deflection of the charged particles 

as they enter the magnetic fields shown below. 

(a) 

(b) 

+ - 

(c) 

(d) 

+ 


A proton moves through a uniform electric field ⃗ E = 50ĵV/m and a 

uniform magnetic field ⃗ B = (0.20î + 0.30ĵ + 0.40ˆk)T. Determine the 

acceleration of the proton when it has a velocity ⃗v = 200î m s . 

+



Definitions 

Facts 

Force on a moving charge: 

⃗F = q⃗v × ⃗ B 

Force on a piece of wire carrying a current I: 

⃗ dF = I ⃗ dl × ⃗ B 

Theorems

5.1 Sources of Magnetic Field 97 

5 Sources of Magnetic Field 

§ 5.1 Sources of Magnetic Field 

So far we have not considered how magnetic fields are created. As 

you may know it is possible to create an electromagnet by wrapping 

a wire around a nail and running a current through the wire. The 

precise way in which a current produces a magnetic field is captured in 

the following law. 

Fact: Biot-Savart Law 

If you have a wire carrying a current I, a small section of the wire 

⃗dl will produce a magnetic field 

dB ⃗ = µ 0I ⃗dl × ˆr 

4π r 2 

where ⃗r is the vector that points from the location of the element 

⃗dl to the field point. The total field is the sum of the contributions 

of each element of the wire. 

This may look complicated but it represents the simple fact that 

the magnetic field wraps around the wire, and decreases in strength as 

you move away. In the figure below is depicted two magnetic field lines 

that are due to the current element ⃗ dl indicated. 

Note that the field warps around the extension of the current element 

in circles. This figure does not depict how the field strength changes 

with position. The direction the field wraps around the current is the 

same direction the fingers of your right hand will wrap around the wire 

if you grasp the wire with your thumb pointing in the direction of the 

current.

98 Sources of Magnetic Field 5.1 

In order to apply the Biot-Savart law one must first parameterize 

the curve that the wire follows. A parameterization for a curve is a 

vector function ⃗r(t) of one variable t, such that ⃗r(t) points from the 

origin to the curve and sweeps along the curve as the variable t is 

increased. Below is picture such a vector function at the values of t. 

The variable t is not the time, but it is sometimes helpful to think of 

it as the time, so that you can imagine ⃗r(t) is the position of some 

particle that is following the curve. 

As an example consider the following vector function of t. 

⃗r(t) = a cos tî + a sin tĵ 

This is a parameterization of a circle of radius a centered at the origin. 

There are other parameterizations of a circle. For example ⃗r(t) = 

a cos tî − a sin tĵ is also a parameterization of a circle, but as t increases 

the vector sweeps clockwise rather than counterclockwise. 


For the parameterization of a circle ⃗r(t) = a cos tî + a sin tĵ, over what 

range would you need to vary t in order to have the vector sweep over 

the quarter circle in the second quadrant? 

Here are some other parameterizations. 

Straight Line : 

Helix : 

Spiral : 

Vertical Parabola : 

Horizontal Parabola : 

⃗r(t) = atî + btĵ + cˆk 

⃗r(t) = a cos tî + a sin tĵ + btˆk 

⃗r(t) = t cos tî + t sin tĵ 

⃗r(t) = ⃗a + btî + ct 2 ĵ 

⃗r(t) = ⃗a + bt 2 î + ctĵ

5.1 Sources of Magnetic Field 99 


Write a parameterization for a parabola that goes through the three 

points (-1,0), (0,1) and (1,0). 

With the parameterization in hand we can proceed with a computation 

using the Biot-Savart law. First we notice that if we change t 

a little bit dt that the difference between ⃗r(t) and ⃗r(t + dt) is a short 

vector along the curve. 

So we see that the small section along the curve ⃗ dl that appears in the 

Biot-Savart law can be found from our parameterization. Further since 

we can write 

⃗dr = ⃗r(t + dt) − ⃗r(t) = 

⃗r(t + dt) − ⃗r(t) dr 

dt = ⃗ dt 

dt dt 

⃗dl = ⃗ dr 

dt dt 

This was the main point of the parameterization, now we can “add up” 

the contributions from all the section by integrating over the parameter 

t. 

We still need to do another step before we can write out the integral. 

We need to write out the vector that points from the current 

element ⃗ dl to the field point. So that we do not get the different vectors 

mixed up, let us call the vector that points from the origin to the field 

point ⃗r f , and let us call the parameterization of the current path ⃗r s (t). 

Then the vector that points from the current element to the field point 

can be written as 

⃗r = ⃗r f − ⃗r s (t)


Now we can write out the Biot-Savart integral in terms of the 

parameterization. 

⃗B(r f ) = µ ∫ 

0I ⃗dl × ˆr 

4π r 2 

= µ ∫ 

0I ⃗dl × ⃗r 

4π r 3 

= µ ∫ drs ⃗ 

0I 

dt 

× [⃗r f − ⃗r s (t)] 

4π |⃗r f − ⃗r s (t)| 3 dt 

This gives us a system by which we can find the field at any point due 

to any current that we can parameterize. Of course it is very rare that 

the integral has an analytic solution, but one can use a computer to 

evaluate the integral numerically once you have used this system to 

write out the integral. 

In order to see exactly what all this means let us do an example. 

Example 

Suppose that we have a current of 15 amps going through a section of 

wire that follows the curve below, which has the following parameterization. 

⃗r s (t) = (−t + t 3 )î + t 2 ĵ 

Where the distance is in meters, and t goes from -0.9 to 0.9. 

We want to find the field at the point ⃗r f = 0î + 1ĵ. In preparation for 

plugging into the Biot-Savart law we compute the following quantities. 

⃗r f − ⃗r s (t) = (t − t 3 )î + (1 − t 2 )ĵ = (1 − t 2 )(tî + 1ĵ) 

|⃗r f − ⃗r s (t)| 2 = (1 − t 2 ) 2 (t 2 + 1) 

|⃗r f − ⃗r s (t)| 3 = (1 − t 2 ) 3 (t 2 + 1) 3/2 

and 

⃗ dr s 

dt = (−1 + 3t2 )î + 2tĵ

5.2 Ampere’s Law 101 

and 

⃗ dr s 

dt × [⃗r f − ⃗r s (t)] = [ (−1 + 3t 2 )î + 2tĵ ] × [ (1 − t 2 )(tî + 1ĵ) ] 

= (1 − t 2 ) [ (−1 + 3t 2 )î + 2tĵ ] × [(tî + 1ĵ)] 

= (1 − t 2 ) [ (−1 + 3t 2 ) − 2t 2] ˆk 

= −(1 − t 2 ) 2ˆk 

Now putting this into the parameterized form of the Biot-Savart law 

we find: 

⃗B(⃗r f ) = µ 0I 

4π 

= µ 0I 

4π 

∫ 

⃗ drs 

dt 

∫ 0.9 

= − µ 0I 

4π ˆk 

−0.9 

∫ 0.9 

× [⃗r f − ⃗r s (t)] 

|⃗r f − ⃗r s (t)| 3 dt 

−(1 − t 2 ) 2ˆk 

(1 − t 2 ) 3 (t 2 + 1) 

−0.9 

= − µ 0I 

4π ˆk [1.9367m −1 ] 

= −(2.90mT)ˆk 

3/2 

dt 

1 

dt 

(1 − t 2 )(t 2 + 1) 

3/2 


Consider a straight section of wire along the x-axis that goes from x = a 

to the x = b. The wire carries a current I. What is the magnetic field 

at the position ⃗r f = yĵ. 


A circular section of wire is carrying a current I counterclockwise. The 

arc of the wire subtends an angle of θ. The circle has a radius a. What 

is the magnetic field due to this section of wire at the center of the 

circle? 

! 

§ 5.2 Ampere’s Law 

In addition to the Biot-Savart law, there is another way of stating 

the relationship between a current and the magnetic field it produces.


Fact: Ampere’s Law 

For any closed curve the line integral of the magnetic field around 

the curve is equal to the µ 0 times the net current through the 

surface enclosed by the curve. 

∮ 

∫ 

⃗B · ⃗dl = µ 0 I through = µ 0 

⃗J · dA ⃗ 

Example 

It is not at all apparent that Ampere’s law is equivalent to the Biot- 

Savart law, but it is. One can prove the Biot-Savart law from Ampere’s 

Law and visa versa. While they are mathematically equivalent, they 

are useful in different situations. Ampere’s law is useful for abstract 

reasoning about fields, and for finding the field strength in highly symmetric 

configurations. 

It is important when applying Ampere’s law to keep in mind that 

the amperian loop does not correspond to anything physical. There 

does not need to be anything there in order for the law to work. You 

are free to choose the amperian loop to be any shape you like. Of 

course, as was the case with applying Gauss’s law, if you want to use 

the law to find the field strength you need to pick the loop correctly. 

In order to use Ampere’s Law to find the field strength, you need three 

things from the loop. 

• The loop must pass through the point at which you want to find the 

field strength and be parallel to the field at that point. 

• The loop must be either parallel or perpendicular to the field at all 

points on the loop. 

• In all regions where the loop is parallel to the field, the field must 

have the same strength. 

Ampere’s law can be used to find the field strength a distance r from 

a long straight wire. We will take our loop to be a circle that wraps 

around the wire and has a radius r.

5.2 Ampere’s Law 103 

This satisfies all three of the above requirements. 

parallel to the field at all points on the loop, 

⃗B · ⃗dl = Bdl 

and since B has the same value at all points on the loop, 

∫ ∫ 

Bdl = B dl = B2πr 

But by Ampere’s Law this must also be equal to µ 0 I. Thus 

B2πr = µ 0 I −→ B = µ 0I 

2πr 

Since the loop is 

Example 

In the previous example we found the magnetic field outside of a long 

wire. Now let us find the magnetic field that is inside the wire itself. 

The method is much the same. We choose our loop to be a circle of 

radius r that wraps around the central axis of the wire, which we has 

a radius R. Since we want to find the field inside the wire we will be 

working with r < R. We still find the same results that B = µ0I 

2πr , but 

we must remember the current in this equation is the current that goes 

through our loop. This current is not the full current of the wire, since 

our loop does not enclose the entire wire. So we need to find the current 

through our loop of radius r. Let us assume that the current density 

in the wire is uniform. Then we can say that the total current in the 

wire is I = JπR 2 and the current through the loop is I through = Jπr 2 . 

Combining these two equations to eliminate J we find that 

So that 

B = µ 0I through 

2πr 

I through = r2 

R 2 I. 

= µ 0I r 2 

2πr R 2 = µ 0Ir 

2πR 2 

The following graph shows the magnetic field both inside and outside 

the conductor.


Example 

In this example we will see how to deal with a current density that is 

not uniform. Suppose that we have a long wire with a current density 

that is greater at the center and drops off to zero at the edge: J(r) = 

3I 

πR 

(1 − r/R). We want to find the magnetic field and from Ampere’s 

2 

law we find B = µ0I through 

2πr 

. The difficulty is that since J is not uniform 

we cannot use I = JA to find I through . We need to use dI = JdA to 

find I through = ∫ JdA. This example will show how to do that. 

We need to pick the area elements 

so that the current density is uniform 

over the entire surface of each element. 

Since the current density is a function 

of r only, J is constant in regions that 

are circles around the center. So we will 

use a circular element dA, as pictured in 

gray in the figure. The figure shows a 

cross section of the wire. 

R 

dA 

r+dr 

r 

In the limit that dr is very small the area will be simply the length 

around the circular area element, 2πr times the width of the element 

dr so that dA = 2πr dr. Now we can compute the current inside a loop 

of radius r loop by integrating. 

∫ rloop 

∫ rloop 

3I 

I through = J dA = 

(1 − r/R)2πr dr 

πR2 0 

= I r2 

R 2 (3 − 2 r R 

−→ B = µ 0I through 

2πr 

) 

0 

= µ 0Ir 

2πR 2 (3 − 2 r R 

) 


Consider a long wire where the current density is not uniform but 

instead increases as you approach the center of the wire, so that at a


distance r from the center the current density is J(r) = 

I 

2πRr . Note 

that this is not a very realistic current density but it is easy to work 

with. Find the magnetic field strength both inside and outside of this 

wire. 

§ 5.3 Force Between Parallel Wires 

Now that we know that the strength of the magnetic field produced 

by a long straight current carrying wire we can find the force between 

two parallel wires. 

In the figure above the field produced by current B in the region of 

current A is shown. Using the right hand rule we can see that the 

direction of the force on wire A is toward wire B. We can also see that 

the field and the current are perpendicular so that the magnitude of 

the force on a length L of wire A is 

F = I A LB = I A L µ 0I B 

2πd = µ 0I A I B L 

2πd 

where d is the distance between the wires. 

Theorem: Force Currents 

Two long parallel wires have a force per length between them of 

F 

L = µ 0I A I B 

2πd 

where d is the distance between the currents. 


An electrical cable carries 45 amps in each of two wires that are a 

distance of 4mm apart. The currents are in opposites directions. 

(a) Is the force between the wires attractive or repulsive? 

(b) What is the force per length between the wires? 

§ 5.4 More Examples


Example 

Finite Wire 

A straight wire of length a carries a current I. What is the magnetic 

field caused by the current at a point (x, y) measured from the center 

of the wire? 

What is B here? 

x 

y 

The wire is a straight line, which can be parametrized: 

⃗r s = (at)î, 

where t varies from − 1 2 to + 1 2 

. So we have: 

d⃗r s 

dt = aî 

We want to find the magnetic field at the position 

⃗r f = xî + yĵ, 

Collect the quantities needed for applying the Biot-Savart law: 

⃗r f − ⃗r s = (x − at)î + yĵ 

Using the Biot-Savart law: 

| ⃗r f − ⃗r s | = √ (x − at) 2 + y 2 

d⃗r s 

dt × (⃗r f − ⃗r s ) = (aî) × ((x − at)î + yĵ) 

⃗B(x, y) = µ 0I 

4π 

= µ 0I 

4π 

= ayˆk 

∫ 

1 

∫2 

− 1 2 

= − µ 0Iy 

4π 

d⃗rs 

dt × [⃗r f − ⃗r s ] 

| ⃗r f − ⃗r s | 3 dt 

ayˆk 

dt 

((x − at) 2 + y 2 ) 3 2 

x− 

∫ 

a 2 

x+ a 2 

du 

(u 2 + y 2 ) 3 2 

where I have substituted u = x − at into the last expression. Looking 

ˆk


up the integral: 

[ 

] x+ a 

⃗B(x, y) = µ 2 

0Iy u 

4π y 2√ ˆk 

u 2 + y 2 x− a 2 

⎛ 

⎞ 

= µ 0I x + 

⎝ 

a 2 

x − a 2 

√ 

4πy (x ) 

− √ ⎠ 

+ 

a 2 (x ) 

ˆk 

2 + y 

2 − 

a 2 

2 + y 

2 

The resulting expression gives the magnetic field anywhere in the plane 

we’ve chosen as the xy-plane. What about points that lie out of this 

plane? Let’s rotate our diagram: 

Looking from the side 

Looking down the wire 

B points out 

(x, y) 

B 

(x, y) 

I 

I 

I points out 

Now imagine rotating the xy plane around the wire. Since the distances 

x and y remain the same, only the direction of ˆk will change. The 

following diagram shows how the position of the magnetic field will 

change for different points, that are equidistant from the line: 

I 

The magnetic field “rotates” around the wire. A quick method to 

determine which direction the rotation of the field goes is to use a 

modified right-hand rule: Stick your right thumb in the direction of 

the current and your fingers will point in the direction of the field’s 

rotation. 

Example 

Using Ampere’s Law we found that the magnetic field strength at a


distance r from an infinite wire was µ0I 

2πy 

. Let us take the limit of a → ∞ 

of the results we have for the finite wire, and see if we get the same 

results. 

⎛ 

⎞ 

µ 0 I x + 

B = lim ⎝ 

a 2 

x − a 2 

√ 

a→∞ 4πy (x ) 

− √ ⎠ 

+ 

a 2 

2 + y 

2 (x − 

a 

2 )2 + y 2 

( 

= µ a 

0I 

2 

√ 

4πy ( 

a 

− − a 2 

√ 

2 )2 (− 

a 

2 )2 ) 

= µ 0I 

2πy 

OK 

Example 

A circular wire with radius a carryies a current I. Locate the circle 

in the xy-plane and compute the magnetic field it causes along a line 

through its center. 

Here is the geometry of the loop: 

z 

r f 

r s 

The path along the circle can be parameterized using t = [0, 2π]: 

We also have 

dl 

⃗r s = a cos t î + a sin t ĵ 

dr ⃗ s 

= −a sin t î + a cos t ĵ 

dt 

⃗r f = z ˆk 

⃗r f − ⃗r s = −a cos t î − a sin t ĵ + z ˆk 

√ 

| ⃗r f − ⃗r s | = a 2 cos 2 t + a 2 sin 2 t + z 2 

= √ a 2 + z 2 

Compute the cross product needed for the Biot-Savart law: 

⃗ dr s 

dt × (⃗r f − ⃗r s ) = (−a sin tî + a cos tĵ) × (−a cos tî − a sin tĵ + zˆk) 

= az cos tî + az sin tĵ + (a 2 sin 2 t + a 2 cos 2 t)ˆk 

= az cos tî + az sin tĵ + a 2ˆk


Use the Biot-Savart law: 

⃗B = µ ∫ 

0I 

4π 

= µ 0I 

4π 

∫2π 

0 

⃗ drs 

dt × [⃗r f − ⃗r s ] 

| ⃗r f − ⃗r s | 3 dt 

az cos tî + az sin tĵ + a 2ˆk dt 

(a 2 + z 2 ) 3 2 

Since the integrals of the cosine and sine vanish for the interval [0, 2π], 

the resulting magnetic field is: 

µ 0 Ia 

⃗B 2 ∫2π 

= 

ˆk dt 

4π (a 2 + z 2 ) 3 2 

−→ 

⃗ B = µ 0I 

2 

0 

a 2 

(a 2 + z 2 ) 3 2 

ˆk 

Example 

A semi-circular shaped wire with radius a carryies a current I. What 

is the magnetic field caused at the center of the semi-circle’s diameter? 

B = ? 

a 

Divide the wire into small sections and compute the magentic field 

contribution due to one small section: 

I 

dθ 

dB 

dl 

The direction for all sections is up, by the right hand rule. Use the 

Biot-Savart law for the magnitude: 

d ⃗ B = µ 0 

4π 

= Iµ 0 

4π 

Id ⃗ l × ⃗r ′ 

ˆk 

r ′ 3 

(dl)(a) 

ˆk, 

a 3


since d ⃗ l and the radius vector to the point are perpendicular. The 

length of dl can be approximated by the length of the arc subtended 

by the angle ∆θ, so 

|dB| ⃗ = Iµ 0 (adθ)(a) 

4π a 3 , 

= Iµ 0 dθ 

4π a 

To find the magnetic field due to the entire wire, integrate: 

B = Iµ 0 

4π 

∫ π 

0 

dθ 

a = Iµ 0 

4a 

Example 

Two sections of straight wire carrying electric current lie parallel to 

each other, a distance b apart: 

I 2 

b 

a 

I 1 

The leftmost ends of the wires are aligned. What is the force that 

current I 1 exerts on current I 2 ? Draw a coordinate system: 

y 

dx' 

x' 

I 2 

b 

I 1 

x 

x=- 

a 

2 

The z-axis points out of the page. From a previous example, we know 

that the magnetic field due to I 1 in the xy-plane is 

⎛ 

⎞ 

⃗B 2 (x, y) = µ 0I 1 x + 

⎝ 

a 2 

x − a 2 

√ 

4πy (x ) 

− √ ⎠ ˆk 

+ 

a 2 

2 + y 

2 (x − 

a 

2 )2 + y 2


The magnetic force on the small segment, dx ′ indicated for I 2 is 

dF ⃗ 21 = I 2 (dx ′ î) × B ⃗ 2 (x ′ , b) 

⎛ 

⎞ 

= µ 0I 1 I 2 

x 

(î × ˆk) ⎝ 

′ + a 2 

x ′ − a 2 

√ 

4πb 

(x′ ) 

− √ ⎠ 

+ a 2 (x′ ) 

dx ′ 

2 + b 

2 

− a 2 

2 + b 

2 

⎛ 

⎞ 

= µ 0I 1 I 2 

4πb 

(−ĵ) x 

⎝ 

′ + a 2 

x ′ − a 2 

√ (x′ ) 

− √ ⎠ 

+ a 2 (x′ ) 

dx ′ 

2 + b 

2 

− a 2 

2 + b 

2 

To compute the total force on I 2 , integrate over the length of the wire: 

⎛ 

⎞ 

⃗F 21 = −ĵ µ ∫ 

0I 1 I 0 

2 x 

⎝ 

′ + a 2 

x ′ − a 2 

√ 

4πb (x′ ) 

− √ ⎠ 

+ a 2 (x′ ) 

dx ′ 

2 + b 

2 

− a 2 

2 + b 

2 

−a 

2 

The integrals can be done using simple substitution, and yield: 

⃗F 21 = − µ 0I 1 I 2 

(√ 

a2 + b 

4πb 

2 − b) 

ĵ 

What is the force exerted on I 1 by I 2 ? Newton’s third law gives us the 

answer: 

⃗F 12 = µ 0I 1 I 2 

(√ 

a2 + b 

4πb 

2 − b) 

ĵ 

Notice that the two wires attract each other. What would happen if 

one of the current’s directions was reversed? 

They would repel each other 



Calculate the magnitude of the magnetic field at a point 100 cm from 

a long, thin conductor carrying a current of 1.0 A. 


A wire in which there is a current of 5.00 A is to be formed into a 

circular loop of one turn. If the required value of the magnetic field at 

the center of the loop is 10.0µT, what is the required radius? 


In Neils Bohr’s 1913 model of the hydrogen atom, an electron circles 

the proton at a distance of 5.3 × 10 −11 m with a speed of 2.2 × 10 6 m s . 

Compute the magnetic field strength that this motion produces at the 

location of the proton.



Determine the magnetic field at a point P located a distance x from 

the corner of an infinitely long wire bent at a right angle, as shown. 

The wire carries a steady current I. 

I 

x 

P 


Consider the current-carrying loop shown below, formed of radial lines 

and segments of circles whose centers are at point P . Find the magnitude 

and direction of ⃗ B at P . 

I 

b 


Two long, parallel conductors separated by 10.0 cm carry currents in 

the same direction. If I 1 = 5.00A and I 2 = 8.00A, what is the force 

per unit length exerted on each conductor by the other? 


Imagine a long cylindrical wire of radius R that has a current density 

J(r) = J 0 (1 − r 2 /R 2 ) for r ≤ R and J(r) = 0 for r > R, where r is the 

distance from a point of interest to the central axis running along the 

length of the wire. 

(a) Find the resulting magnetic field inside and outside the wire. 

(b) Plot the magnitude of the magnetic field as a function of r. 

(c) Find the location where the magnetic field strength is a maximum, 

and the value of the maximum field. 


Below is shown a cross-sectional view of a coaxial cable. The center 

conductor is surrounded by a rubber layer, which is surrounded by 

an outer conductor, which is surrounded by another rubber layer. The 

current in the inner conductor is 1.00 A out of the page. The current in 

the outer conductor is 3.00 A into the page. Determine the magnitude 

and direction of the magnetic field at the points indicated at a distance 

1mm and 3mm from the center. 

a 

θ 

P


3.00 A 

1.00 A 

1 mm 1 mm 1 mm 


A long cylindrical conductor of radius R carries a current I. The current 

density J is not uniform over the cross-section of the conductor but is a 

function of the radius J = br, where b is a constant. find the magnetic 

field inside and outside the conductor.



Definitions 

Facts 

Field due to a section of current: 

dB ⃗ = µ 0I ⃗dl × ˆr 

4π r 2 

Theorems 

Ampere’s Law: 

∮ 

⃗B · ⃗dl = µ 0 

∫ 

⃗J · ⃗ dA

6.2 Faraday’s Law 115 

6 Time Varying Fields 

§ 6.1 Introduction 

Let us take a moment to review what we know about the means 

of producing electric and magnetic fields. We have found two means to 

produce fields. 

• Electric fields are produced when there are regions in space with a 

net electric charge. This is encapsulated in Gauss’s law. 

∮ 

⃗E · dA ⃗ = 1 ∫ 

ρ dV 

ɛ 0 

• Magnetic fields are produced when there is a flow of charge. This is 

encapsulated in Ampere’s law. 

∮ 

∫ 

⃗B · ⃗dl = µ 0 

⃗J · dA ⃗ 

So far, we have investigated only steady state fields (fields that don’t 

change in time). A net electric charge and a steady flow of charge are 

the only means to produce steady state electric and magnetic fields. 

But these are not the only means of producing time varying fields. In 

this chapter we will investigate time varying electromagnetic systems. 

§ 6.2 Faraday’s Law 

Let us start by considering what will happen if we have some source 

of time varying magnetic field. It could for example be a wire that is 

carrying a current and the current is changing rapidly in time. Since 

the current is changing in time, the magnetic field that the current 

creates will also change with time. 

If we place a loop of wire into this time varying 

magnetic field, the time varying field will produce 

a current in the wire. This is called an induced 

current. Nikola Tesla was able to produce such 

strong induced currents that he was able to make 

a light bulb glow with the current. In the photo 

to the right Tesla is holding a light bulb which 

has no wires connected to it. The current driving 

the bulb is entirely an induced current, which is 

caused by the time varying magnetic field.

116 Time Varying Fields 6.2 

The induced current is caused by an electric field. This induced 

electric field is not like the other electric fields we have seen. This is a 

new type of electric field. The electric field is created by the changing 

magnetic field. 

The electric field wraps around the changing 

magnetic field somewhat like a static magnetic field 

wraps around a steady current. There are some additional 

subtleties due to the fact that the induced 

electric field is caushed by the change in the magnetic 

field. The relationship between the induced electric 

field and the changing magnetic field is stated in the 

following law of physics. 

Fact: Faraday’s Law 

The line integral of the electric field around any closed curve is 

equal to negative the rate of change of the magnetic flux through 

any surface bounded by the curve. 

∮ 

⃗E · ⃗dl = − d ∫ 

dt 

⃗B · ⃗ dA 

Notice that this is very similar to Ampere’s Law: 

∮ 

∫ 

⃗B · ⃗dl = µ 0 

⃗J · dA ⃗ 

Ampere’s Law and Faraday’s Law both relate the line integral around 

closed curve to the flux of a vector field through the area enclosed by 

the curve. The difference is that Faraday’s Law has the time derivative 

of the flux. 

Another important and potentially confusing thing to notice, is 

that the line integral of the electric field ( ∫ ⃗ E · ⃗ dl) has so far been 

referred to as the electric potential difference between the end points 

of the curve. The confusing thing is that since it is a closed curve the 

electric potential difference must be zero, ∆V = 0, while Faraday’s Law 

tells us that the line integral of the induced electric field is not zero, but 

equal to the rate of change of the magnetic flux. So we learn from this 

that the force caused by the induced electric field is not a conservative 

force field. We also see that there is no electric potential associated with 

this electric field, ( ⃗ E ≠ − ⃗ ∇V ). To avoid confusion we will refer to the 

line integral of the induced electric field by its historical name, the 

electromotive force, or EMF. In equations the EMF is represented by 

the symbol E, so that E = ∮ ⃗ E · ⃗ dl. Note that the unit of EMF is the 

volt.

6.2 Faraday’s Law 117 

Definition: Magnetic Flux 

The integral of the magnetic field over the area of the loop is called 

the magnetic flux. 

∫ 

φ m = ⃗B · dA ⃗ 

Note that this is of the same form as the definition of the electric 

flux (φ e = ∫ E ⃗ · dA), ⃗ in addition you may have noticed that current 

is the flux of current density, (I = ∫ J ⃗ · dA). ⃗ With the definition of 

magnetic flux and EMF we can rewrite Faraday’s Law in a simpler 

looking form. 

Theorem: Faraday’s Law (alternate form) 

The induced EMF in a loop is equal to the negative rate of change 

of the magnetic flux through the loop. 

E = − dφ m 

dt 

Example 

Suppose that we are in a region where the magnetic field is uniform 

and increasing with time: 

⃗B = ⃗ B 0 + atî + btĵ + ctˆk. 

We place a loop of wire with an area of A, so that it lies flat in the x-y 

plane. We want to compute the induced EMF in the loop. Since the 

loop is in the x-y plane we know that in computing the magnetic flux 

( ∫ B ⃗ · dA), ⃗ that the area elements will all be pointing in the z direction: 

dA ⃗ = ˆk dA. So ∫ ∫ 

φ m = ⃗B · dA ⃗ = ( B ⃗ 0 + atî + btĵ + ctˆk) · ˆk dA 

∫ 

∫ 

= (B 0z + ct) dA = (B 0z + ct) dA = (B 0z + ct)A 

So we can compute the induced EMF in the loop. 

E = − dφ m 

dt 

= − d dt [(B 0z + ct)A] = −cA 

What can we learn from this example? First we learn that only the 

component of the magnetic field that is normal to the surface of the loop 

contributes to the magnetic flux. Second we see that no component 

of the constant part of the field, ⃗ B 0 , contributes to the induced EMF, 

only the time varying part of the field induces an EMF.


Example 

Consider now a field that is 

⃗B = B 0 cos ωt ˆk 

Again we take the loop in the x-y plane. 

Because the field is uniform we again get the result that the flux is the 

product of the normal component of the field and the area of the loop. 

And thus that 

φ m = B 0 A cos ωt 

E = − dφ m 

dt 

= B 0 Aω sin ωt 

The previous example will be very helpful in understanding the 

relationship between the direction of the magnetic field and the direction 

of the induced electric field. As we said before the induced electric 

field wraps around the magnetic field, but we did not say in which direction 

it wraps itself. The above example will help us understand the 

direction. 

First notice that in the figure above a positive magnetic field is 

one that is in the positive z direction, while a positive electric field is 

one that is counterclockwise. (This choice of positive directions is in 

compliance with another right hand rule, that we will express as Lenz’s 

law in just a moment.) Now examine the graph of the electric and 

magnetic fields. Sometimes the fields have the same sign and sometimes 

they do not. Let us analyze the the graph in four quarters. In the first 

quarter notice that the electric field is counterclockwise, the magnetic 

field is positive, and the magnitude of the magnetic field is decreasing. 

This is represented in symbols in the first row of the table below. The 

other three quarters follow. 

• First Quarter: E and B ↑ and |B| ↘. 

• Second Quarter: E and B ↓ and |B| ↗.

6.3 Maxwell’s Extension of Ampere’s Law 119 

• Third Quarter: E and B ↓ and |B| ↘. 

• Fourth Quarter: E and B ↑ and |B| ↗. 

Here are the same four quarters in diagrams: 

Notice that if the magnitude of the flux is decreasing, the fields have 

the same sign. While if the magnitude of the flux is increasing, the 

fields have opposite sign. This observation is usually stated in terms of 

the current that is caused by the induced electric field if a loop of wire 

is placed in line with the electric field loop. 

Fact: Lenz’s Law 

The induced current creates a magnetic field that opposes the 

change in the magnetic flux. 


Check to see that Lenz’s law is obeyed in all four quarters in the previous 

example. 

§ 6.3 Maxwell’s Extension of Ampere’s Law 

We have seen that a changing magnetic field causes an electric field. 

It ends up that a changing electric field will also cause a magnetic field. 

∮ 

∫ 

⃗B · ⃗dl d 

= µ 0 ɛ 0 

⃗E · dA 

dt 

⃗ 

This magnetic field is added to the magnetic field produced by currents 

so that we arrive at an extension of Ampere’s law that makes it 

applicable to time varying fields. 

Fact: Ampere’s Law - Time Varying 

∮ 

∫ 

∫ 

⃗B · ⃗dl = µ 0 

⃗J · dA ⃗ d 

+ µ 0 ɛ 0 

dt 

= µ 0 I through + µ 0 ɛ 0 

dφ e 

dt 

⃗E · ⃗ dA


dφ 

The quantity ɛ e 0 dt 

is sometimes called the displacement current 

because it acts like a current in Ampere’s Law. 


You have a parallel plate capacitor with circular plates. The capacitor is 

being charged so that the electric field between the plates is increasing 

at a constant rate: E = at, with a = 4.5 × 10 10 V · m −1 s −1 . Use 

the extended version of Ampere’s Law to find the magnetic field at 

a distance r = 4.0cm from the center of the capacitor and half way 

between the plates. Assume that the field is circular about the axis of 

the capacitor. 

§ 6.4 Inductance 

Suppose that you have a loop of wire in a region where there is no 

source of magnetic field. If you run a current through the wire there 

will now be a magnetic field due to the current. This magnetic field will 

create a magnetic flux through the loop. Because the field produced 

is proportional to the current, the flux will also be proportional to the 

current. 

φ m = LI 

The proportionality constant L is called the self inductance of the loop. 

The induced EMF on the loop is the product of the inductance and the 

rate of change of the current. 

E = L dI 

dt 

Circuit elements are manufactured with specific values of inductance, 

and are called inductors. The SI unit of inductance is called the Henry, 

abbreviated as H. Inductors are essentially a coil of wire, sometimes 

wrapped around a core of material designed to increase the inductance. 

The sign of E has been chosen in such a way as to facilitate the application 

if Kirchhoff’s loop rule to inductors. It is the same convention 

that was used for the resistor. 

+ ε – 

+ ΔV – 

I 

For an inductor Lenz’s law can be interpreted as “the induced EMF 

across an inductor is in such a direction as to oppose the change in the 

current through the inductor”. For example, suppose that a current 

is flowing through an inductor, if you now try and reduce the current, 

an EMF will be generated in the inductor that will tend to keep the 

current going. Similarly if you try to increase the current an EMF will 

be generated that will tend to stop the current from increasing. 

I

6.4 Inductance 121 

Example 

Suppose that you have the circuit shown. 

By some means you have established a current 

in the circuit (perhaps it is an induced current 

caused by some external magnetic field) such 

+ ε – 

that at t = 0 the current is I 0 . At t = 0 the 

external cause of the current disappears. What 

will happen to the current? – ΔV + 

I 

Since the inductor opposes the change in the current the current 

will not stop abruptly. Instead it will decrease gradually, the inductor 

will drive the circuit for a while. Note that this implies that an inductor 

can store energy, much like a capacitor can. Let us see how we would 

do the circuit analysis of this system. 

Kirchhoff’s Loop rule give us that 

E + ∆V = 0 

−→ L dI 

dt + IR = 0 

−→ dI 

dt = −R L I 

−→ I(t) = I 0 e − R L t 

Thus we see that the current will decay exponentially. 

I 


The circuit below is assembled with the switch open. At the time t = 0 

the switch is closed. 

+ ε – 

V S 

+ 

– 

I 

– ΔV 

+ 

(a) What is the current as a function of time? 

(b) What is the EMF as a function of time? 

(c) What is the voltage on the resistor as a function of time? 

(d) Check to see if Kirchhoff’s loop rule is satisfied at all times. 

I


(e) Sketch the graphs of all three functions. 

§ 6.5 Alternating Current 

Often the currents in a circuit are sinusoidal. 

I(t) = I 0 cos ωt 

A circuit with such a current is referred to as an alternating current 

circuit, or AC circuit, because the current alternates directions in the 

circuit. In contrast a circuit with constant current is referred to as a 

direct current circuit, or DC circuit. 

§ 6.6 AC Power and RMS Voltage 

The electrical power supplied by the power company is sinusoidal. 

The electric potential between the two sockets of an electrical outlet is 

oscillating. 

V (t) = V 0 cos ωt 

The voltage supplied in the U.S. is said to be 120 volts. If you examine 

the underside of an electrical appliance there will be printed something 

like 

60Hz - 120V - 7.0AMPS 

someplace near the UL listing mark. As you may have guessed the first 

mark indicate that the appliance is designed to operate on AC power 

that has a frequency of 60Hz. You might expect that the amplitude of 

the supplied electric potential, V 0 , would be 120V, and the amplitude 

of the current drawn by the appliance, I 0 , would be 7.0AMPS. This 

is not quite correct. These numbers are not the amplitudes but the 

amplitudes divided by √ 2, so that the actual amplitudes are greater 

by a factor of √ 2 than these values marked on the appliance. Thus V 0 = 

(120V) √ 2 = 170V and I 0 = (7.0A) √ 2=9.9A. In order to understand 

why this is so we need to consider the power dissipated in the appliance. 

Recall that power is the product of the voltage and the current. 

Suppose for simplicity that the appliance that we are using is a toaster. 

This is simple because, as an electrical circuit, a toaster is simply a 

resistor. So our entire system, including the AC source, is diagramed 

as follows.

6.7 AC Circuit Elements 123 

+ 

V S– 

R 

Because the current oscillates in time, the power dissipated in the resistor 

also oscillates in time. 

P (t) = I(t)V (t) 

= I(t)I(t)R 

= [I(t)] 2 R 

= [I 0 cos ωt] 2 R 

= I 2 0 R cos 2 ωt 

Notice that the power oscillates between zero and I 2 0 R so that the 

average power supplied to the toaster is 1 2 I2 0 R. 

P avg = I2 0 R 

= I 0 I 

√ 0 R √2 = I 0 V 

√ √ 0 

= (7.0A)(120V) 

2 2 2 2 

Notice that for a sinusoidal signal the root mean square (RMS) value is 

the amplitude divided by √ 2. Thus we can learn two things from this 

example. First, the values that are printed on the toaster are the RMS 

values. Second, the product of the RMS current and RMS voltage gives 

the average power. So in this example the average power consumed by 

the appliance is P = (7.0A)(120V) = 840W. 


A 60 watt light bulb has an average power output of 60 watts. 

(a) What is the peak power? 

(b) What is the resistance of the bulb? 

(c) Is this resistance the same as the resistance that is required to 

dissipate 60 watts when connected to a 120 volt DC source? 

§ 6.7 AC Circuit Elements 

In many ways an AC circuit can be analyzed using the same techniques 

we used for DC circuits. In fact inductors and capacitors become 

as simple as resistors. In an AC circuit, inductors and capacitors 

follow an adapted Ohm’s Law: the amplitude of the voltage on the


element is proportional to the amplitude of the current through the 

element, V = ZI, just like a resistor. The proportionality constant Z 

is called the impedance. 

Let us see how this works for an inductor. 

Suppose that the current through an inductor is I(t) = I 0 cos ωt. 

Then we know that the EMF (V L ) on the inductor is 

V L = L dI 

dt = −ωLI 0 sin ωt 

So that the amplitude of the voltage oscillation is ωLI 0 . 

V L0 = ωLI 0 = Z L I 0 with Z L ≡ ωL 

So we see that the amplitudes of the voltage and current are proportional. 

The constant Z L = ωL is the equivalent of the resistance for an 

inductor. 

Theorem: Impedance: Inductor 

In an AC circuit the amplitude of the voltage on an inductor is 

proportional to the amplitude of the current flowing through the 

inductor. 

V L0 = Z L I 0 with Z L ≡ ωL 

The impedance of an inductor is Z L = ωL. 

The actual voltage and current are not proportional, since one is 

a sine function and the other is the cosine function. 

V L 

The voltage on an inductor reaches the peak value one quarter of a cycle 

before the current does. For this reason the voltage on an inductor in 

an AC circuit is said to lead the current by a phase of 90 ◦ . 

A capacitor is similar. 

V C = 1 C Q = 1 ∫ 

Idt = 1 

C ωC I 0 sin ωt 

1 

So that the amplitude of the voltage oscillation is 

ωC I 0. 

V C0 = 1 

ωC I 0 = Z C I 0 with Z C ≡ 1 

ωC 

So we see that the amplitudes of the voltage and current are proportional. 

The constant Z C = 1 

ωC 

is the equivalent of the resistance for 

an capacitor.

6.8 Phasor Diagrams 125 

Theorem: Impedance: Capacitor 

In an AC circuit the amplitude of the voltage on a capacitor is 

proportional to the amplitude of the current flowing through the 

capacitor. 

V C0 = Z C I 0 with Z C ≡ 1 

ωC 

The impedance of an capacitor is Z C = 1 

ωC . 

The voltage on a capacitor reaches the peak value one quarter of a cycle 

after the current does. For this reason the voltage on a capacitor in an 

AC circuit is said to follow the current by a phase of 90 ◦ . 

§ 6.8 Phasor Diagrams 

This relationship between the leading and following phases is often 

represented in a phasor diagraph. As you may recall an oscillation can 

be thought of as the horizontal component of a circular motion. So we 

can represent the current or voltage in and AC circuit as the horizontal 

component of a circular motion. 

The current is represented as a vector with constant length I 0 that is 

rotating in the counter clockwise direction. The real physical current 

is the projection of this vector onto the horizontal (Re) axis. 

One thing that is very nice about the phasor representation is that 

it allows us to clearly represent in a diagram the phase relationship between 

different quantities. Let us take the leading phase of the inductor


voltage as an example. The fact that the inductor voltage is one quarter 

of a cycle ahead of the current, means that the inductor voltage 

phasor is 90 ◦ ahead of the current phasor. 

If you imagine the phasors rotating you can see that the projection of 

the voltage, onto the horizontal axis, will reach a peak one quarter of 

a cycle before the projection of the current reaches its peak. 

Here is the phasor diagram for a capacitor. 

At this point the phasor diagram has only allowed us to represent 

what we already know. The phasor diagram is far more useful. For 

example the phasor diagram will alow us to use Kirchhoffs loop rule 

for AC circuits, as we will see in the following example. The idea of a 

phasor is useful in many areas of physics and engineering. We will see 

more examples in the next chapter. 

Example 

Suppose that we have a capacitor and a resistor in series. What is the 

effective impedance of the combination. First imagine that we connect 

the combination to an AC source. 

+ 

+ 

V S– 

C 

R 

+ 

V R 

– 

V C 

–

6.8 Phasor Diagrams 127 

What we want to find is the ratio of the amplitude of the supply voltage 

and the amplitude of the current: V S0 /I 0 . 

Kirchhoff’s loop rule gives us that 

V S (t) − V C (t) − V R (t) = 0 −→ V S (t) = V C (t) + V R (t) 

In the phasor diagram this implies that the sum of the phasors for V C 

and V R must be equal to the phasor of the source voltage V S . Since 

the resistor phasor is parallel to the current phasor we know that the 

phasors for V C must follow the phasor for V R by 90 ◦ . Thus the sum 

(also V S ) forms the hypotenuse of a right triangle. 

We can find the phasor for the sum of the voltage on the resistor and 

capacitor by adding the individual phasors like we add vectors. Also 

notice from the circuit diagram that the sum of the voltages on the 

resistor and capacitor is equal to the voltage on the supply: V S = 

V R + V C . 

Since the lengths of the phasors are the amplitudes of the voltages we 

can use the pythagorian theorem to find that 

V 2 S 0 

= V 2 R 0 

+ V 2 C 0 

= (RI 0 ) 2 + (Z C I 0 ) 2 

−→ V S 2 0 

I0 

2 = R 2 + ZC 

2 √ 

−→ Z eff = V √ 

S 0 

= R 

I 2 + ZC 2 = R 2 + 1 

0 (ωC) 2 

We see that the impedances do not simply add when the components 

are in series. In this case the square of the impedance was the sum of 

the squares of parts. This is not a general rule. We can also find the 

phase angle (φ) between the supply voltage and the current. 

tan φ = −V C 0 

= −Z CI 0 

= −Z C 

V R0 RI 0 R = −1 

ωRC 

There is a very significant difference between the resistance of a 

resistor and the impedance of a capacitor or inductor: the impedance 

depends on the frequency. In the previous example we see that the


effective impedance of the system depends on ω. This can be a useful 

property, since sometimes we wish to treat different frequencies differently. 

For example in an audio system, there are big speakers for 

low frequencies and small speakers for high frequencies, but the signal 

coming from the amplifier to the speakers contains both high and low 

frequencies. Let us see how we can use a capacitor and resistor is series 

to split the signal so that the appropriate signal goes to each speaker. 

Example 

For the circuit in the previous example, let us find the voltage on the 

capacitor and resistor relative to the voltage from the source. 

G R (ω) ≡ V R 0 

= V R 0 

/I 0 R 

= √ 

V S0 V S0 /I 0 R2 + 1/(ωC) = 1 

√ 2 1 + 1/(ωRC) 

2 

G C (ω) ≡ V C 0 

= V C 0 

/I 0 1/ωC 

= √ 

V S0 V S0 /I 0 R2 + 1/(ωC) = 1 

√ 2 (ωRC)2 + 1 

These ratios are called the gain. 

So we see that the voltage on the capacitor is equal the the source 

voltage at low frequencies and drops to zero at high frequencies, while 

the voltage on the resistor does just the reverse. So we see that the 

signal from the source is sorted by this circuit, high frequencies to the 

capacitor and low frequencies to the resistor. Thus we can send the 

voltage from the capacitor to the large speakers and the voltage from 

the resistor to the small speakers. 


Suppose that you build a RL circuit instead of the RC, that is you 

replace the capacitor with an inductor in the previous example. 

(a) Find the gain for the resistor and inductor. 

(b) Graph the gain for the resistor and inductor. 

(c) Which will pass low frequencies better than high frequencies.



Show that this LRC circuit √ has a maximum 

current when ω = 

1 

LC 

. This is 

called the resonance frequency. 

L 

+ 

V S 

– 

R 

C 



Consider the hemispherical closed surface 

of radius R as shown . If the hemisphere 

is in a uniform magnetic field that makes an 

angle θ with the vertical, calculate the magnetic 

flux through the flat surface S 1 . Calculate 

the flux through the hemisphical surface 

S 2 . 

S 1 

S 2 

θ 

B 


A powerful electromagnet has a field of 1.6T and a cross-sectional area 

of 0.20m 2 . If we place a coil having 200 turns and a total resistance 

of 20Ω around the electromagnet and then turn off the power to the 

electromagnet in 20ms, what is the current induced in the coil? 


A rectangular loop of area A is placed in a region where the magnetic 

field is perpendicular to the plane of the loop. The magnitude of the 

field is allowed to vary in time according to B = B o e −t/τ . What is the 

induced emf as a function of time? 


A long, straight wire carries a current 

I = I o sin(ωt+δ) and lies in the plane of a rectangular 

loop of N turns as shown. Determine 

the emf induced in the loop by the magnetic 

field of the wire. 

I 

a 

b 

c 

⊲ Problem 6.11


A magnetic field directed into the page changes 

with time according to B = at 2 + b. The field has 

a circular cross-section of radius R. What are the 

magnitude and direction of the electric field at a 

distance r from the center of the field. 

R 


For the situation in the previous problem if B = (2.0t 3 − 4.0t 2 + 0.80)T 

and R = 2.5cm. Calculate the magnitude and direction of the force 

exerted on an electron located at a distance r = 2R from the center of 

the field at the time t = 2.0s. 


A circular coil enclosing an area of 100cm 2 is made of 200 turns of 

copper wire. Initially a 1.10T uniform magnetic field points perpendicularly 

upward through the plane of the coil. The direction of the 

field then reverses. During the time the field is changing its direction, 

how much charge flows through the coil if the resistance of the coil is 

R = 5.0Ω. 


The rotating loop in an ac generator is a square 10cm on a side. It 

is rotated at 60 Hz in a uniform field of 0.80T. Calculate (a) the flux 

through the loop as a function of time, (b) the emf induced in the loop, 

(c) the current induced in the loop for a loop resistance of 1.0Ω, (d) the 

power dissipated in the loop, and (e) the torque that must be exerted 

to rotate the loop. 


A solenoid has n turns per unit length, radius a and carries a current 

I. 

Long Solenoid 

a 

a 

b 

c 

Large loop of wire 

Small loop of wire 

(a) A large circular loop of radius b > a and N turns encircles the 

solenoid at a point far away from from the ends of the solenoid. Find 

the magnetic flux through the loop.


(b) A small circular loop of N turns and radius c 

inside the solenoid, far from its ends, with its axis parallel to that of 

the solenoid. Find the magnetic flux through this small loop. 


The two circular loops shown below have their planes parallel to each 

other. 

A 

B 

I 

As viewed from A toward B, there is a counter-clockwise current in loop 

A. Give the direction of the induced current in loop B and determine 

whether the loops attract or repel each other if the current in loop A 

is (a) increasing and (b) decreasing. 


A bar magnet moves with constant velocity along the axis of a loop as 

shown in the figure below. 

B 

S 

N 

v 0 

(a) Make a qualitative graph of the flux φ B through the loop as a 

function of time. Indicate the time t 1 when the magnet is halfway 

throught the loop. 

(b) Sketch a graph of the current I in the loop versus time, choosing I 

to be positive when it is counterclockwise as viewed from the left. 


Given the circuit below, assume that the switch S has been closed for 

a long time so that steady state currents exist in the circuit. 

S 

10 Ω 

10 V 

100 Ω 2 Η


Ignore any resistance of the inductor L. (a) Find the battery current, 

the current in the 100 Ω resistor, and the current through the inductor. 

(b) Find the intial voltage across the inductor when switch S is opened. 

(c) Give the current as a function of time measured from the instant 

of opening the switch S. 


A 2.00H inductor carries a steady current of 0.500A. When the switch 

in the circuit is thrown open, the current disappears in 10ms. What is 

the average induced emf in the inductor during this time? 


A coiled telephone cord has 70 turns, a cross sectional diameter of 1.3 

cm, and an unstretched length of 60 cm. Determine an approximate 

value for the self inductance of the unstretched cord. 


A 10.0mH inductor carries a current I = I max sin ωt, with I max = 5.00A 

and ω/2π = 60.0Hz. What is the back emf as a function of time? 


The switch in the circuit below is closed at time t = 0. Find the current 

in the inductor and the current through the switch as functions of time 

if V = 10.0V, R = 4.00Ω and L = 1.00H. 

R 2R 

V 

R 

L 

S 


Show that I = I 0 e −t/τ is a solution of the differential equation 

IR + L dI 

dt = 0 

where τ = L/R and I 0 is the current at t = 0. 


An air-coil solenoid with 68 turns is 8.0 cm long and has a diameter of 

1.2 cm. How much energy is stored in its magnetic field when it carries 

a current of 0.77A. 


The switch in the circuit below is connected to point a for a long time. 

After the switch is thrown to point b find (a) the frequency of oscillation 

in the LC circuit, (b) the maximum charge on the capacitor, (c) the


maximum current in the inductor, and (d) the total energy stored in 

the circuit at time t. 

R a b 

S 

V 

C 

L 


Show that the rms voltage of the pictured “sawtooth” wave is V 0 / √ 3. 

V 

V o 

-V o 


What is the resistance of a light bulb that uses an average power of 75W 

when connected to a 60Hz power source having a maximum voltage of 

170 V? What is the resistance of a 100W bulb? 


An inductor has an AC current of frequency 50Hz passing through 

it. The maximum voltage is 100V and the maximum current is 7.5A. 

What is the inductance of the inductor? The frequency is now changed 

to ω new while the voltage is held constant. The maximum current is 

now 2.5A. What is the angular frequency ω new ? 


A 1.0mF capacitor is connected to a standard wall outlet. Determine 

the current in the capacitor at t = (1/180)s, assuming that at t = 0 

the energy stored on the capacitor is zero. 


For what linear frequencies does a 22.0µF capacitor have a impedance 

below 175Ω? Over this same frequency range , what is the impedance 

of a 44.0µF? 


A sinusoidal voltage v(t) = V max cos ωt is applied to a capacitor. Write 

an expression for the instantaneous charge on the capacitor. What is 

the instantaneous current in the circuit? 


At what frequency does the inductive impedance of a 57µH inductor 

equal the capacitive impedance of a 57µF capacitor. 

t



A series AC circuit contains the following components: R = 150Ω, L = 

250mH, C = 2.00µF, and a generator with V max = 120V operating at 

50.0Hz. Calculate the (a) inductive impedance, (b) capacitive impedance, 

(c) impedance, (d) maximum current, and (e) phase angle. 


An RLC circuit consists of a 150Ω resistor, a 21µF capacitor, and a 

460mH inductor, connected in series with a 120V, 60Hz function generator. 

What is the phase angle between the current and the applied 

voltage? Which reaches its maximum earlier, the current or the voltage? 


Calculate the resonance frequency of a series RLC circuit for which 

C = 8.40µF and L = 120mH. 


An RLC circuit is used in a radio to tune into an FM station broadcasting 

at 99.7MHz. The resistance in the circuit is 12.0Ω, and the 

inductance is 1.40µH. What capacitance should be used?



Definitions 

Facts 

Theorems 

Ampere’s Law: 

∮ 

⃗B · ⃗dl = µ 0 

∫ 

Faraday’s Law: 

∮ 

∫ 

⃗J · dA ⃗ d 

+ µ 0 ɛ 0 

dt 

⃗E · ⃗dl = − d dt 

∫ 

⃗B · ⃗ dA 

⃗E · ⃗ dA 

AC Circuits 

Capacitor: The phase of the voltage across a capacitor is 90 ◦ behind 

the phase of the current through the capacitor. The amplitude of the 

voltage and current are related as follows. 

V C 0 = 1 

ωC I C 0 

Inductor: The phase of the voltage across an inductor is 90 ◦ ahead 

of the phase of the current through the inductor. The amplitude of the 

voltage and current are related as follows. 

V L0 = ωL I L0

136 Wave Optics 7.10

7.1 Maxwell Equations 137 

7 Wave Optics 

§ 7.1 Maxwell Equations 

Let us write down, in one place, all of the fundamental equations 

about the electric and magnetic fields. 

Gauss ′ s Law 

Faraday ′ s Law 

Ampere ′ s Law 

Gauss ′ s Law for B 

∮ 

∮ 

∮ 

∮ 

⃗E · dA ⃗ = 1 ∫ 

ɛ 0 

⃗E · ⃗dl = − d dt 

∫ 

⃗B · ⃗dl = µ 0 

⃗B · ⃗ dA = 0 

ρ dV 

∫ 

⃗B · ⃗ dA 

⃗J · ⃗ dA + µ 0 ɛ 0 

d 

dt 

∫ 

⃗E · ⃗ dA 

These equations as a group are known as the Maxwell Equations. 

The last equation, which we have not discussed before, can be 

understood to say that there is no magnetic equivalent to the electric 

charge. That is, that there are no magnetic charges, places where 

magnetic field lines begin or end, thus that magnetic field lines do not 

end. If you want to think of the magnetic field as a flowing fluid, with 

B the velocity of the fluid, then this law says that the fluid does not 

compress: B flows like water, when it spreads out the velocity decreases, 

when it goes through a narrow region it speeds up, the volume rate of 

flow is a constant. 

Ok, so the new law tells us that there are no magnetic charge, let 

us see what the Maxwell equations look like in a region where there is 

no electric charge (ρ = 0) and no current (J = 0).

138 Wave Optics 7.1 

Gauss ′ s Law 

Faraday ′ s Law 

Ampere ′ s Law 

a new Law 

∮ 

∮ 

∮ 

∮ 

⃗E · dA ⃗ = 0 

⃗E · ⃗dl = − d ∫ 

⃗B · dA 

dt 

⃗ 

∫ 

⃗B · ⃗dl d 

= µ 0 ɛ 0 

⃗E · dA 

dt 

⃗ 

⃗B · ⃗ dA = 0 

You can see that the equations are essentially the same in E and B. 

Investigating these equations led Maxwell to discover that in regions 

where there is no current or charge that there can be traveling waves 

as pictured below. 

x 

y 

B 

E 

E 

B 

z 

time increasing 

E 

B 

E 

B 

E 

B 

E 

B 

B 

E 

E 

B 

B 

E 

E 

B 

What is drawn is the magnetic field and electric field at various posi-

7.2 Describing Oscillations 139 

tions along the z-axis. Notice that the magnitude of the field oscillates, 

as one moves along the z-axis. This pattern of spacial oscillations of 

the field, travels to the right along the axis as time passes. Also notice 

that E and B are perpendicular, and that the direction of travel is 

perpendicular to both E and B. 

Maxwell also found that the equations predict the velocity at which 

these electromagnetic waves travel. 

√ 1 

v EM = 

µ 0 ɛ 0 

From the known values of µ 0 and ɛ 0 , Maxwell computed that the velocity 

of these EM waves is the same as the velocity of light. He 

concluded, rightly, that light and electromagnetic waves are the same 

thing. This was a huge leap in our understanding of light. 

Fact: Electromagnetic Waves and Light 

Light is an electromagnetic wave. 


Show that 1/ √ µ 0 ɛ 0 = c. 

§ 7.2 Describing Oscillations 

We need some terminology to make it easier to talk about sinusoidal 

waves and oscillations. The central idea of a wave is encapsulated 

in the term phase of an oscillation. The word phase is used here in the 

same way as the word is used to describe the cycle of the moon: new 

moon, waxing crescent, first quarter, full moon, etc. 

In a system with a periodic cycle, the phase of the system describes 

what point in the cycle the system is currently occupying. 

Recall from the section on AC circuits that we can describe a 

sinusoidal oscillation as the horizontal component of a rotation. Thus 

the phase of a sinusoidal oscillation can be described by giving the angle 

of rotation, φ. When the phase is φ = 0, the oscillation is at its greatest 

positive extension. When the phase is φ = π/2 the oscillation is at zero 

and with a negative velocity. When the phase is φ = π the oscillation 

is at its greatest negative extension. When the phase is φ = 3π/2 the 

oscillation is zero but with a positive velocity. These and other phases 

of an oscillation are depicted in the figure below.


-A 0 A 

φ = 0 

φ = π/4 

φ = 2π/4 

φ = 3π/4 

φ = 4π/4 

φ = 5π/4 

φ = 6π/4 

φ = 7π/4 

φ = 8π/4 

The actual value of the displacement (x) is the amplitude of the oscillation 

times the cosine of the phase angle. 

Displacement = Amplitude × cos φ 

x = A cos φ

7.3 Describing Waves 141 

For a harmonic oscillator the phase increases at a constant rate: 

dφ 

dt = ω −→ φ = ωt + φ 0 

and so the displacement is 

x = A cos(ωt + φ 0 ) 


An oscillator has an amplitude of 3.2. At this instant the displacement 

of the oscillator is 1.4. What are the two possible phases of the oscillator 

at this instant? 


You have a mass connected to a spring. 

0 x 

You start the mass oscillating in the following ways. What is the initial 

phase φ 0 in each case. 

(a) Stretch the spring to the right, at t = 0 release the mass. 

(b) Compress the spring to the left, at t = 0 release the mass. 

(c) At t = 0 strike the mass so that it begins moving to the left. 

(d) At t = 0 strike the mass so that it begins moving to the right. 

(e) At t = 0 the mass is at the position x 0 = 2.0m and has a velocity 

of v 0 = 3.0 m rad 

s 

(assume that ω = 5.0 

s ). 


It takes a time of T = 0.025s in order for an oscillator to complete one 

cycle. What is the angular frequency (ω) of the oscillator? 

§ 7.3 Describing Waves 

Consider dropping a rock into a pool of still water. Ripples spread 

out from the point at which the rock enters the water. If you examine 

the motion of the water at one fixed point, the surface of the water 

moves up and down as successive waves move past. The height of 

the water at our fixed point is an oscillation. This is true at other 

points a well, at each location the height of the water oscillates. Since 

an oscillation is so simple, the only thing that can differ between one 

point and another is the amplitude of the oscillation and the initial 

phase φ 0 . Let us write the height of the water, at position ⃗r and time 

t, as y(⃗r, t). Since the oscillation at each point is a harmonic oscillation 

we can write: 

y(r, t) = A(⃗r) cos [ωt + φ 0 (⃗r)]


Where the amplitude A(⃗r) and initial phase φ 0 (⃗r) both depend on the 

position ⃗r. 

In the situation where the wave spreads out from a point source 

(such as a rock dropped in the water), the wave travels outward at a 

speed v. Because of this, the oscillation at a distance r from the source 

will be the same as the oscillation of the source, only delayed by the 

time (t d = r/v) it takes for the wave to travel the distance r from 

the source to the observation point. Thus the phase of the oscillation 

at a position r at a time t + t d , will be the same as the phase of the 

oscillation at the source (r = 0) at time t. 

φ(r, t + t d ) = φ(0, t) 

ω(t + t d ) + φ 0 (r) = ωt + φ 0 (0) 

φ 0 (r) = φ 0 (0) − ωt d 

φ 0 (r) = φ 0 (0) − ω r v 

φ 0 (r) = φ 0 (0) − kr with k ≡ ω v 

For convenience we usually pick our zero of time so that φ 0 (0) = 0, 

so that φ 0 (r) = −kr and y(r, t) = A(r) cos(ωt − kr). 

Theorem: Wave due to a Sinusoidal Point Source 

A general wave radiating from a point source can be written down 

mathematically as follows: 

ψ(r, t) = A(r) cos (ωt − kr) 

( 2π 

= A(r) cos 

T t − 2π ) 

λ x 

where k = ω/v and T ≡ 2π ω and λ ≡ 2π k 

are the period and 

wavelength as described below. 

Definition: Wavelength 

If you freeze a sinusoidal wave at one point in time, the distance 

you must travel along the wave in order to go through a complete 

cycle of the oscillation, is called the wavelength. We use the symbol 

λ to represent the wavelength in equations. 

ψ 

λ 

x

7.3 Describing Waves 143 

Definition: Period and Frequency 

If you stand in one place and allow the wave to pass you, the 

time you must wait in order to be back at the same point in the 

oscillation as when you started is called the period. We use the 

symbol T to represent the period in equations. 

ψ 

T 

t 

The frequency is the inverse of the period. 

f = 1 T 

Note that the angular frequency is ω = 2πf. Both frequencies are 

really the same thing, they both measure the rate of the oscillation, it 

is just that one is in the units of radians and the other is in the units 

of cycles. So you can convert between units by using the fact that one 

seconds = radians cycles 

cycle seconds 

cycle is 2π radians: ω = radians 

= 2π f. 

Notice that the realtionship between the angular frequency and the 

wave number can be rewritten in terms of the wavelength and period. 

k = ω 2π/T 

−→ 2π/λ = −→ λ = vT 

v 

v 

So we can also think of the wavelength as the distance the wave travels 

in one period, or writing the equation as T = λ/v, we can think of the 

period as the time it takes to travel one wavelength. 

Let us summarize the relationships between the different parameters 

that describe a wave. 

Time scale: 

ω = 2π 

T = 2πf 

Length scale: 

k = 2π λ 

Relationship between time and length scales: 

k = ω v 

OR λ = vT OR λf = v



For the following wave, graphed at seven different times, determine 

the amplitude, wavelength, period, frequency, angular frequency, wave 

number, and velocity. Write the function y(x, t) that describes the 

wave.

7.4 Electromagnetic Waves 145 

§ 7.4 Electromagnetic Waves 

With our eyes we are able to see electromagnetic (EM) waves that 

have wavelengths between about 400nm to 700nm. We experience light 

with different wavelengths as different colors, as indicated in the following 

table. 

Wavelength Frequency Perceived Color 

740 to 625 nm 405 to 480 THz red 

625 to 590 nm 480 to 510 THz orange 

590 to 565 nm 510 to 530 THz yellow 

565 to 520 nm 530 to 580 THz green 

520 to 500 nm 580 to 600 THz cyan 

500 to 430 nm 600 to 700 THz blue 

430 to 380 nm 700 to 790 THz violet 

Note that the frequency and wavelength of an EM wave are related by 

λf = c where c is the speed of light. 

But this is only a small range of EM spectrum. 

We see that there are a number of familiar items in the spectrum: x- 

rays, microwaves, and radio waves are all EM waves. The range of 

the spectrum just above and just below the visible range are called the 

ultraviolet (UV) and infrared (IR). The prefixes ultra (above) and infra 

(below) refer to the frequency not the wavelength. 


Your microwave oven is filled with EM waves with a frequency of about 

3 GHz. What is the wavelength of the wave? The microwave oven heats 

up the objects in the oven, because the oscillating EM wave causes an 

oscillating electric force on the electric dipoles in the object, which 

causes the dipoles to oscillate. The frequency of a standard microwave


oven is tuned to a natural oscillation frequency of the water molecule. 

Note that a cellular phone uses about the same frequency as an oven, a 

cellular phone is a small microwave emitter, so you are slowly cooking 

your brain when you hold the phone to your head. 


Compute the wavelength of an FM radio station that transmits at a 

frequency of 94.1 MHz. 

§ 7.5 Interference of Waves 

We have claimed that light is an EM wave because it has the same 

speed as an EM wave. We have not looked at the consequences of 

the fact that light is a wave. One of the principle characteristics of 

waves is their ability to interfere with each other when two or more of 

the waves are combined. The following example will demonstrate this 

interference. 

Consider the follow situation. You have a point source of waves, 

and near this source you have a reflecting surface. 

Source 

Detector 

Reflecting Surface 

The wave arrives at the detector by two different paths, a direct path 

and a reflected path. The wave travels a distance r 1 along the direct 

path and a distance r 2 along the reflected path. If the reflected path 

was blocked the oscillation at the detector would be just that due to the 

direct path, ψ 1 (t) = A 1 cos(ωt + φ 1 ) where φ 1 = −kr 1 . Similarly if the 

direct path is blocked the oscillation at the detector would be just that 

due to the reflected path, ψ 2 (t) = A 2 cos(ωt + φ 2 ) where φ 2 = −kr 2 . 

When both paths are open then the resultant wave at the detector will 

be the sum of the waves from each path 

ψ(t) = ψ 1 (t) + ψ 2 (t) 

= A 1 cos(ωt + φ 1 ) + A 2 cos(ωt + φ 2 ) 

Using the phasor diagram for these two oscillations we can find the 

amplitude of their sum.

7.5 Interference of Waves 147 

I 

I 

A 

A 2 

The law of cosines gives the amplitude, A, in terms of the amplitudes 

φ 2 

A 1 

A 1 

A 2 

φ 2 - φ 1 

R 

φ 1 

R 

A 1 and A 2 . 

A 2 = A 2 1 + A 2 2 + 2A 1 A 2 cos(φ 2 − φ 1 ) 

Theorem: Addition of Two Waves 

If two waves (with the same frequency) come together at a detector, 

the amplitude A of the resultant wave at the detector will be as 

follows. 

A 2 = A 2 1 + A 2 2 + 2A 1 A 2 cos(φ 2 − φ 1 ) 

In general the power carried by a wave is proportional to the square 

of the amplitude of the wave, so this result can be rewritten in 

terms of the power. 

P = P 1 + P 2 + 2 √ P 1 

√ 

P 2 cos(φ 2 − φ 1 ) 

Below is graphed the power at a detector where two waves come 

together, The power of one wave is 25 mW and the power of the other 

waves is 4 mW. 

P (mW) 

0 π 2π 3π 4π 

φ 2 - φ 1 

Notice the following points: 

• The power at the detector can be less with both sources on than with 

just one of them on. 

• The maximum power occurs when the phase difference is an even 

multiple of π. The waves as said to be perfectly in-phase.


• The minimum power occurs when the phase difference is an odd 

multiple of π. The waves as said to be directly out-of-phase. 

• When the phase difference is an odd multiple of π/2, the power is 

equal to the sum of the powers of the individual signals. 


Suppose that you are at home talking on your wireless phone, and you 

walk near your refrigerator. Since the refrigerator is made of metal 

it acts as a reflector for the microwave signal going to and from your 

phone. Because of this there are two paths between your phone and 

the transceiver (the base that phone sits in) that it is communicating 

with. 

x 

Phone 

Refrigerator 

Transceiver 

When you are a distance x from the refrigerator the signal that reflects 

from the refrigerator must travel a distance 2x further. Assuming that 

the reflected amplitude is one half of the direct amplitude, graph the 

power of the signal as a function of your distance from the refrigerator. 

Assume that the wavelength of the signal is 10cm. Are there places 

that it would be better to avoid? 

§ 7.6 Interferometer 

It is possible to build a device that measures very small motions 

by using interference. 

The device represented in the diagram below is called an interferometer.

7.7 Interference of Two Sources 149 

Source 

Mirror A 

Half-silvered 

mirror 

Detector 

Mirror B 

A beam of light strikes a half-silvered mirror. The half silvered mirror 

reflects half of the wave toward mirror A and transmits the other half 

toward mirror B. The part that strikes mirror A, is reflected back (to 

the right) toward the half-silvered mirror where it is again split, half 

passing through to the detector on the left and half being reflected back 

toward the source. The part of the beam that strikes mirror B is also 

split so that half reaches the detector. 

So we see that there are two path by which light can reach the 

detector, one path that has been reflected from mirror A, and another 

path that has been reflected from mirror B. By adjusting the location 

of mirror A we can adjust the path length of the light. In this way 

the relative phase of path A and B can be adjusted. Suppose that we 

have adjusted the path length so that the power at the detector is a 

maximum. Then φ A − φ B is an even multiple of π. 

Now suppose that we move mirror A a distance of λ 4 

to the left. 

This will increase the path length of by ∆r A = 2 λ 4 = λ 2 

since the path 

includes the extra bit of length twice, once on the way to the mirror and 

once on the way back from the mirror. But a change in the path length 

of ∆r A will change the phase by −k∆r A = − 2π λ 

λ 2 = −π. Thus φ A −φ B 

is now an odd multiple of π, and the power will be at a minimum. By 

observing the intensity of the light at the detector we can easily tell if 

it has gone from bright to dim (maximum to minimum). Thus we can 

easily detect when the mirror has been moved by λ/4. A common laser 

pointer has a wavelength of something like 650nm so one quarter of this 

is 162nm. So an interferometer can be used as a “ruler” with markings 

about 162nm apart. This is a very small distance, for comparison, the 

finest human hair is about 20,000 nm in diameter.


§ 7.7 Interference of Two Sources 

Suppose that we have two wave sources that are in phase with each 

other. 

Detector 

r A 

r B 

Source A 

Source B 

While the sources are in phase, the waves are not when they reach the 

detector because they must travel a different distance: at the detector 

the phase difference between the waves will be φ A −φ B = k(r B −r A ) = 

k ∆r = 2π ∆r 

λ 

. The minimum and maximum power occur when this 

phase difference is an odd and even multiple of π. That is when 

2π ∆r 

{ max m is even. 

λ = mπ min m is odd 

∆r = m λ { max m is even. 

2 min m is odd 

Theorem: Interference of Two Sources 

If you have two wave sources that are in phase with each other, 

the minimum and maximum of the power occur at points where 

the path difference is and odd (min) or even (max) multiple of half 

of the wavelength of the wave. 

∆r = m λ { max m is even. 



Draw two points that are 3 cm apart on a piece of paper. 

(a) Find all points on the paper that have ∆r = 0. 

(b) Find all points on the paper that have ∆r = 1cm. 

(c) Find all points on the paper that have ∆r = 2cm. 

(d) If λ = 1.0cm what are the locations of the maximum and minimum 

of power? 

(e) If λ = 2.0cm what are the locations of the maximum and minimum 

of power?

7.8 Far Field Approximation 151 

§ 7.8 Far Field Approximation 

In most cases the detector is placed far from the two sources, far 

in the sense that the distance from the sources to the detector r is 

much bigger than the distance between the sources, d. There is a 

useful approximation for the path difference that can be used when the 

detector is far from the source. 

First consider the case when the detector 

is not far, as pictured in the figure to 

Detector 

the right. Notice that if we make of a section 

of arc, with the center at the detector, 

Source 

and going through the closest source, and 

then consider the section of the path from 

the furthest source that is cut off by this 

arc. This cut off section (marked as ∆r in 

the figure) is equal to the path difference. 

Δr 

Also notice that the arc is perpendicular to 

Source 

the path. 

Now imagine moving the detector further from the sources, as 

pictured in the diagram below. The section of arc that is between the 

two paths, subtends a smaller and smaller angle, and thus this section 

of arc becomes closer and closer to a straight line. At the same time 

the grey region becomes a right triangle. 

θ 

Let us examining this right triangle 

more closely. We see from the diagram 

to the right that 

sin θ = ∆r −→ ∆r = d sin θ 

d 

This is a very useful result, that allows 

us to find the path difference from the 

distance between the sources and the angle 

to the detector. 

Combining this with our previous result, ∆r = m λ 2 

, we find the 

d 

θ 

Δr 

θ


angles for the minima and maxima. 

d sin θ m = m λ { max m is even. 



You set up a sheet of aluminum foil to block a microwave transmitter. 

You now punch two holes in the foil, so that the microwaves pass 

through the holes. The holes are 10 cm apart. You now place a microwave 

detector on a track that is parallel to the foil and 3 meters away. 

By sliding the detector along the track you observe that the response 

of the detector is as shown in the diagram. What is the wavelength of 

the microwave? 

Foil with holes 

x 

120cm 

80cm 

Source 

10 cm 

Detector Response 

as a function 

of position 

40cm 

0cm 

300 cm 

§ 7.9 Thin Film Interference 

The swirling colors that you see reflected in a 

soap bubble and in puddles in a parking lot are a 

result of interference. Let us start by considering 

the colors in the puddles. The reason that you 

only see this in a parking lot is because it only 

happens when there is a thin film of oil on the top 

of the water. Consider the diagram to the right, 

Oil t 

of the layers of a puddle. At the bottom we have 

the pavement of the parking lot. Next there is the 

pool of water. On top of the water is a thin layer of 

Water 

Pavement 

oil. The thickness of the oil has been exaggerated 

so that we can see what is happening. 

The light from the sun strikes the surface of the oil and some of it 

is reflected back toward the detector (your eye). The rest of the light

7.9 Thin Film Interference 153 

passes into the oil, were it proceeds until it reaches the water, at which 

point part of the light is reflected back toward the detector. The light 

that is not reflected passes into the water and eventually strikes the 

pavement, where it is mostly absorbed, since the pavement is black. In 

the end then, we have light from the sun reflected into the detector, 

by two paths, one path is reflected from the air-oil interface and the 

other path is reflected from the oil-water interface. If the light is nearly 

normal to the surface the path difference will be twice the thickness of 

the oil: ∆r = 2t. The light will be strongly reflected when the two 

paths are in-phase, that is when ∆r = m λ 2 

with m even. Thus in order 

for the light to be strongly reflected we need, 

2t = m λ 4t 

−→ λ = 

2 

m 

So only wavelengths that “match” the thickness of the oil will be reflected. 

This is why you see swirling colors, what you are seeing is the 

different thicknesses of the oil film, and for each thickness there is a 

particular color that gets reflected. 

There are a two complications to thin films that we need to consider. 

The first complication is that the wavelength of light changes when 

it passes into the oil. This is because the light slows down in the oil. 

Definition: Index of Refraction 

The index of refraction of an optical medium is the ratio of the 

speed of light in a vacuum and the speed of light in the medium. 

n = c v 

Let λ be the wavelength in a vacuum, then λf = c. Let λ ′ be the 

wavelength in the medium, then λ ′ f = v. Taking the ratio of these two 

equations we find 

λf 

λ ′ f = c v = n 

Solving for λ ′ we find the following result. 

Theorem: Wavelength in a Medium 

λ ′ = λ n


The reason that we care the wavelength has changed, is that when 

we use the result that ∆r = m λ 2 

we need to use the wavelength where 

the path difference ∆r occurs. In the case considered above, the extra 

bit of path occurs within the oil, so we need to use the wavelength in 

the oil. 

In order to understand the root cause of the second complication, 

consider a soap bubble again. If you make a soap bubble and watch it 

carefully, you will notice that just before it pops, it appears that the top 

of the bubble has a hole in it. There is not actually a hole, but the top 

of the bubble looks like it has a hole because when the bubble gets very 

thin, no light is reflected from the surface of the bubble. Let us consider 

what our prediction would be for a very thin bubble. From our work 

with the oil film we expect that we will get a maximum reflection when 

2t = mλ/2, and m is even. Recall also that m = 0 gives a maximum, 

so t ≈ 0 should give a maximum for all wavelengths. Thus we expect 

the soap bubble to reflect very well when the film of bubble juice gets 

very thin. The truth is just the opposite. 

The key to this puzzle is to understand that when a wave is reflected 

it can be inverted, that is the direction of the displacement can 

be reversed. But a reversal is the same as a π phase shift. 

Fact: Reflection Phase Shift 

When a wave is reflected at an interface between two media there 

will be a π phase shift in the reflected wave if the index of refraction 

of the incident medium is lower than the index of refraction of the 

reflecting medium. If the incident medium has a higher index, 

there is no phase shift. 


Explain using the reflection phase shift why a thin soap bubble reflects 

no light. 

§ 7.10 Single Slit Diffraction 

If a wave is passed through a rectangular hole that is much narrower 

in one direction than the other we find that we get an interference 

pattern like we do with a two source. Such a hole is usually referred to 

as a slit since it is so narrow in one direction.


Second Min 

First Min 

Source 

θ 

Central Max 

Fact: Single Slit Diffraction 

If light is passed through a single slit of width a, then the angle 

between the central maximum and the first minimum is given by 

a sin θ = λ 



A pair of speakers is configured as shown, with a microphone placed directly 

in front of one of the speakers. Sound waves having a wavelength 

10 cm are coming from the speakers. The speakers are in phase with 

each other. What is the minimum distance d between the speakers so 

that no sound is heard at the microphone labeled P ? 

1 m 

P 

d 


A material having an index of refraction of 1.30 is used to coat a piece 

of glass (n = 1.50). What should be the minimum thickness of this film 

to minimize reflection of 500nm light? 


A soap bubble of index of refraction 1.33 strongly reflects both the red 

and the green components of white light. What film thickness allows 

this to happen? (In air, λ red = 700nm, λ green = 500nm.)



A beam of 560nm light passes through two closely spaced glass plates, as 

shown below. For what minimum nonzero value of the plate separation 

d is the transmitted power a maximum? 

d 


A pair of narrow parallel slits separated by 0.25mm is illuminated by 

green light (λ = 546.2nm). The interference pattern is observed on a 

screen 1.2m away from the plane of the slits. Calculate the distance 

from the central maximum to the first bright region on either side of 

the central maximum and between the first and second dark bands. 


On a day when the speed of sound is 354 m s 

, a 2000Hz sound wave 

impinges on two slits 30.0cm apart. At what angle is the first maximum 

located? If the sound wave is replaced by 3.00cm microwaves, what 

slit separation gives the same angle for the first maximum? If the 

slit separation is 1.00µm, light of what frequency gives the same first 

maximum angle? 


Two waves with amplitudes y 1 = A cos(ωt−kd 1 +δ) and y 2 = A cos(ωt− 

kd 2 ) interfere with each other. Prove that the resultant amplitude, 

y 1 + y 2 = y is 

( k 

y = 2A cos 

2 (d 2 − d 1 ) + 

2) 

δ ( 

cos ωt − k 2 (d 1 + d 2 ) + δ ) 

2 

Choose d 1 = d 2 = 0 and make plots of y 1 (t), y 2 (t), and y(t) for δ = π/8, 

π/4, π/2, and π. 


Two slits are separated by a distance d. Light of wavelength λ comes 

from a far distant source and strikes the slits at an angle θ 1 as in the 

figure below. An interference maximum is formed on a screen that is 

far to the right of the slits. The maximum occurs at an angle θ 2 . Show 

that sin θ 2 − sin θ 1 = mλ/d, where m is an integer.


θ 2 

d 

θ 1 

θ 2 

θ 1 


A material having an index of refraction of 1.30 is used to coat a piece 

of glass (n = 1.50). What should be the minimum thickness of this film 

to minimize reflection of 500nm light? 


A soap bubble of index of refraction 1.33 strongly reflects both the red 

and the green components of white light. What film thickness allows 

this to happen? (In air, λ red = 700nm, λ green = 500nm.) 


Light of wavelength 600 nm is used to illuminate, at near zero angle 

of incidence, two glass plates that are stacked on top of each other. 

At one end the plates are separated slightly because a wire has been 

placed between them on this end. The wire is 0.5 mm in diameter. 

How many bright fringes appear along the total length of the plates, 

when the plates are viewed from the same side as the light source? 


The double slit equation d sin θ = mλ and the equation for a single slit 

a sin θ = mλ are sometimes confused. For each equation, define the 

symbols and explain the equation’s application. 


Light from a He-Ne laser (λ = 632.8nm) is incident on a single slit. 

What is the minimum width for which no diffraction minima are observed? 


A screen is placed 50cm from a single slit, which is illuminated with 

690nm light. If the distance between the first and third minima in the 

diffraction pattern is 3.0mm, what is the width of the slit? 


Light from an argon laser strikes a diffraction grating that has 5310 

lines/cm. The central and first order principal maximum are separated


by 0.488 m on a wall that is 1.72 m from the grating. Determine the 

wavelength of the laser light. 


Paul Revere received information from a person in a church steeple that 

was 1.8 miles away, by the number of lanterns that were displayed: 

”One if by land and, two if by sea.” What would be the minimum 

separation between two lantern for Paul Revere to be able to distinguish 

them? Assume that the diameter of his pupils was 4.00mm and that 

the light had a wavelength of about 580nm. 


Redo all the problems in the book starting from chapter 1, but add one 

to each constant. Add all of the numerical answers and subtract 1 for 

each non numerical answer. What number do you get?



Definitions 

• Index of refraction: n = c v 

Theorems 

• A traveling sinusoidal wave can be represented by 

where ω = 2π 

T and k = 2π λ . 

• For a traveling wave: λf = v. 

y = A cos(ωt − kx) 

• Interference of two waves: If two waves are added together with 

powers P 1 and P 2 then the resultant power is 

P = P 1 + P 2 + 2 √ P 1 P 2 cos(φ 2 − φ 1 ) 

• Maxima and minima of the power occur when ∆φ = φ 2 − φ 1 is an 

even and odd multiple of π, respectively. 

• The phase difference due to a path difference is ∆φ path = 2π ∆r 

λ 

• If a field point is far from two sources a distance d apart then the 

path difference is ∆r = d sin θ where θ is the angle between the path to 

the field point and the normal to the line connecting the two sources. 

• When a wave is reflected at an interface between two media there 

will be a π phase shift in the reflected wave if the index of refraction of 

the incident medium is lower than the that of the reflecting medium. 

• The wave length in a medium is longer than in a vaccum: λ ′ = λ n 

• If light is passed through a single slit of width a, then the angle 

between the central maximum and the first minimum is given by, 

a sin θ = λ

160 Geometric Optics 8.12

8.2 Reflection 161 

8 Geometric Optics 

§ 8.1 Short Wavelength Limit 

In the last section we saw the wave nature of light in the sections 

where we considered interference. In many situations these interference 

effects are small enough to be ignored. For example if light comes in 

through two windows in your house you will not find an unexpected 

dark spots in the room where the light from the two windows combines. 

Fact: Short Wavelength Limit 

When the wavelength of the light is much shorter than the size 

of the objects that the light is traveling through, then the wave 

nature of the light can be ignored. 

For example if you set up a candle and cast shadow of your hand 

on the wall, the shadow is a faithful replica of your hand, there is no 

interference pattern developed because of the interference between the 

light going between different fingers. 

In this short wavelength limit we can think of a light source as sending 

out rays of light that continue in a straight line unless something stops 

them. Because of this use of straight lines, this way of dealing with 

light is called Geometric Optics or Ray Optics. 

§ 8.2 Reflection 

Now consider what happens when a ray of light strikes a mirror.

162 Geometric Optics 8.3 

θ i 

θ r 

Mirror 

Fact: Law of reflection 

The incident and reflected angles are the same. 

θ i = θ r 

§ 8.3 Virtual Image 

Let us draw some of the other rays that come from the source. 

Notice that each ray obeys the law of reflection. This may look 

a little confusing, but if we trace the reflected ray back through the 

mirror, we see that they all converge on the same point. 

Object 

Virtual Image

8.4 Snell’s Law 163 

We see then that from the point of view of an observer of the reflected 

rays, it appears that the object is really behind the mirror, because all 

of the rays appear to come from that one point. Our brain makes the 

simplest conclusions and assumes that the rays come in a straight line, 

not in the bent line that they actually followed. 

This type of image is called a virtual image of the object since the 

light does not actually come from the image. We will see later, ways 

in which it is possible to construct a real image. 

§ 8.4 Snell’s Law 

When a ray of light strikes an interface between two different media, 

some of the light is reflected, and some of the light is transmitted 

through the interface. 

n i 

n t 

θ i 

θ r 

θ t 

Note that both the reflected and transmitted beams are deflected from 

their original direction. The angle at which the transmitted ray leaves 

the interface is given by Snell’s Law. 

Fact: Snell’s Law 

n t sin θ t = n i sin θ i 

We can use Snell’s law to explain a strange phenomenon that you 

may have noticed. If you view a fish tank from the corner, you can see 

objects in the tank through the front glass and the side glass at the 

same time. This is because of the bending of the light when it leaves 

the water. The effect is depicted in the diagram below.


Eye 

Tank 

Front 

Front View 

Both View 

Side 

View 

Side 

The dark grey region is the part of the tank that can be seen from both 

the front and the side. You can actually see around the corner. This is 

really weird. So if you want a wider view from your window, you could 

just fill your room with water. Remember to also install the air tanks. 

§ 8.5 Virtual Image Caused by Refraction 

Consider a shiny coin on the bottom of a swimming pool. 

Air 

Water 

The light comes up from the coin and out into the air. When the light 

comes out it spreads out more. This looks similar to the rays reflected 

from a mirror. This similarity leads one to wonder if there is a virtual 

image of the coin, that is, is there a place where all the rays appear to 

originate. The following diagram traces the rays back into the water.

8.5 Virtual Image Caused by Refraction 165 

Air 

Water 

We see that they do not converge on a single point. The location of 

the image changes as you view it from different angles. When you get 

down low and view it from near the horizon the image is close to the 

surface of the water and also displaced horizontally from the object. 

As you move your view point upward the image moves away from you 

and deeper into the water. When you view from straight overhead the 

image is directly over the object and about 1/5 of the way up toward 

the surface. 

Air 

Water 

Image from horizon 

Image from 45° 

Image from directly above 

Object 

This is very different from the image in a flat mirror. The image in a 

flat mirror is in the same position, from any viewpoint in the room. 


Consider the beam of light shown going through a prism made of glass 

with an index of refraction n.


α 

β 

β 

Show that the net deflection of the beam (2β) is 

§ 8.6 Thin Lens Equation 

Imagine that we take a stack 

of prisms as shown in the diagram. 

Since the prisms farther 

from the middle have a steeper 

angle they bend the light ray 

more sharply. If we pick the angle 

of each prism correctly all 

of the rays will converge on the 

same point. 

2β = 2 arcsin(n sin α 2 ) − α 

Such a device was used in old light houses in order to focus the 

light on the horizon (where there are ships that want to see the light) 

rather than letting it spread out and go into the water or into the sky. 

Now imagine that you make 

the prisms in the middle thicker, 

without changing the angle of 

the faces. We would now end 

up with a smooth curved surface. 

Since the angle of faces was not 

changed the light would still converge 

on the single point. You 

can see that you get a lens. This 

type of lens is called a converging 

lens, since it bends the rays 

toward each other. 

focal length 

focal 

point 

The optical axis of the lens is a line that passes through the center 

of the lens, that is normal to the surface of the lens. A focus is a point 

where rays converge. The focus of rays that are parallel to the optical 

axis is called the principle focus or focal point. The distance between 

the lens and the focal point is called the focal length of the lens. 

Suppose that you set up your lens at sunrise, so that the rays are 

parallel to the optical axis and converge at the focal point. If you come

8.6 Thin Lens Equation 167 

back in an hour the sun will have moved up in the sky, and the rays will 

no longer be parallel to the optical axis. The rays will still converge at 

a point but it will not be the focal point of the lens. The new focus 

will be in the focal plane, a plane parallel to the plane of the lens and 

one focal length away from the lens. 

Focal Plane 

Optical Axis 

Notice that the ray that passes through the center of the lens is not 

deflected. This is a particularly nice ray for doing constructions of 

images, as we will see. In this case it allows us to find the location of 

the focus, since the focus is at the intersection of this straight line and 

the focal plane. Once we have found the focus, using the central ray, 

we can draw the other rays, because they must pass though the focus 

also. 

We can also reverse these diagrams: Light that comes from a point 

in the focal plane and strikes the lens leaves the lens parallel to the ray 

that passes through the center of the lens. 

We have seen what happens to groups of rays that are parallel to 

each other. Let us now see what happens to groups of rays that diverge 

from a point that is not in the focal plane. Suppose that you have light 

radiating from a point. The point is indicated by the fat arrow on 

the left, in the following diagram. We follow the light from this point 

source to it’s focus.


In the diagram we are able to draw the three rays that are bolded, 

because they are, (top ray) parallel to the optical axis, (middle ray) 

through the center, and (bottom ray) through the focal point. These 

three rays are called the principle rays. From the three principle rays 

(we only needed two really) the location of the focus is determined. 

After the location of the focus is determined, the other faint rays are 

drawn so that they go through the focus. 

Notice that the rays of light actually do pass through the focus. If 

we place a piece of paper at that point the paper would be illuminated. 

This type of image is called a real image. 

x o 

x i 

y o 

y i 

f 

f 

The location and size of the image can be computed from the 

location and size of the object. 

Theorem: Thin Lens Equations 

With the dimensions as shown in the figure above, 

1 

x o 

+ 1 x i 

= 1 f 

Theorem: Magnification Equations 

With the dimensions as shown in the figure above, 

= − x i 

y o y i 

where y is negative if below the optical axis. 

x o

8.7 Virtual Image in a Converging Lens 169 


Using geometric arguments, prove the thin lens equations. 


You have a lens with a focal length of 100cm. You place an object at 

150cm from the lens. The object is 3cm tall. 

(a) Construct the image location using the principle rays. 

(b) Find the location of the image using the thin lens equation. 

(c) Find the height of the image from the magnification equation. 

§ 8.7 Virtual Image in a Converging Lens 

You do not get a real image for all positions of the object. When 

the object distance is smaller than the focal length, you get a virtual 

image. You have probably notice this effect before while using a magnifying 

glass: if you look at something far away through a magnifying 

glass, you will se the object upside down. If you move closer to the 

object at some point it flips over and appears right side up. 

Virtual Image 

Real Image 

image 

paper 

lens 

paper 

lens 

image 

In the photograph above you can notice a few things. In the photo on 

the left, the letters on the page are almost in-focus, this means that 

the image is not too far from the page. In contrast, in the photo on the 

right, the letters on the page are very out of focus, this tells us that 

the image is far from the page. This is verified by the ray diagrams. 

Here is a larger diagram, showing the construction of the virtual 

image, that corresponds to the photo on the left.


-x i 

x o 

f f 

Notice that we still use the three principle rays. The topmost ray in 

the figure above strikes the lens as if it came from the focal point. The 

dotted line going back to the focal point is an extension of the actual 

path of the light ray. Notice also that even though the ray constructed 

in this manner does not hit the lens (since the lens was not tall enough) 

we can still use the ray to construct the location of the image. 

The thin lens and magnification equations can still be used in this 

case, but in order to get the algebra to work out correctly you need to 

interpret a negative image distance (x i ) as being on the side of the lens 

where the light originates from. 

Example 

Suppose that in the figure above x 0 = 6cm, y 0 = 2.0cm and f = 9cm. 

Putting this into the thin lens equation we find 

1 

6cm + 1 = 1 

x i 9cm −→ 1 = 1 

x i 9cm − 1 

6cm = − 1 

18cm 

−→ x i = −18cm 

Now we can use the magnification equation to find the size of the image. 

y i 

= − y o 

−→ y i = − x i 

y o = − −18 (2cm) = 6.0cm 

x i x o 

x o 6 


You have a lens with a focal length of 80cm. You place an object at 






For a converging lens with a focal length of f, over what range of object 

distances is the image: 

(a) real?

8.8 Diverging Lenses 171 

(b) virtual? 

(c) upright? 

(d) inverted? 

(e) larger than the object? 

§ 8.8 Diverging Lenses 

Now consider what happens when parallel rays strike a lens that 

is thiner in the middle. The rays diverge from the optical axis as if 

they came from the focal point on the incident side. This type of lens 

is called a Diverging Lens. This is drawn in the figure below. 

-f 

We might also consider what happens to rays that are converging toward 

the focal point on the far side of the lens, these rays exit the lens 

parallel to the optical axis, as indicated in the diagram below. 

-f 

We can, as before, construct the location and size of the image, by 

using three principle rays. The three rays are, (1) a ray that leaves the 

object parallel to the optical axis, (2) a ray going through the center of 

the lens, (3) a ray that leaves the object headed toward the focal point 

on the far side of the lens. This has been done in the figure below.


x o 

-x i 

-f -f 

We can also find the location and size of the image by using the thin 

lens equation and the magnification equation. Once again a negative 

value for image distance x i , indicates that the image is on the same 

side of the lens as the source of the light. We must also use a negative 

value for the focal length of a diverging lens. 

Example 

Suppose that in the figure above x 0 = 20cm, y o = 8cm and f = −12cm. 

Putting this into the thin lens equation we find 

1 

20cm + 1 1 

= 

x i −12cm −→ 1 = − 1 

x i 12cm − 1 

20cm 

−→ x i = −7.5cm 

Now we can use the magnification equation to find the size of the image. 

y i 

= − y o 

−→ y i = − x i 

y o = − −7.5 (8cm) = 3.0cm 

x i x o 

x o 20 


You have a lens with a focal length of -100cm. You place an object at 






For a diverging lens, show that the object distance must be negative, in 

order for the image distance to be positive. In other words the object 

must be virtual, in order for the image to be real. We will see, in 

the next section, how it is possible for to have a virtual object to be 

negative.

8.10 Multi-Lens Optical Systems 173 

Example 

§ 8.9 Sign Conventions and Coordinates System 

We have used the coordinates 

(x o , y o ) for the object and the coordinates 

(x i , y i ) for the image. The 

direction of positive y is the same 

for both coordinate systems. The 

direction of positive x was opposite. 

Light 

In 

A curved mirror is also a lens. The coordinate 

sign convention, stated as follows, 

covers both types of lenses. 

• The x-axis of the object coordinate system 

is in the opposite direction as the incoming 

light. 

• The x-axis of the image coordinate system 

is in the same direction as the outgoing light. 

y o 

y i 

Light 

x o 

x Out i 

Transmitting Lens 

Light 

In 

Light 

Out 

x o 

x i 

y o 

y i 

Reflecting Lens 

A concave mirror (as shown) is a converging lens, while a convex 

mirror is a diverging lens. The sign convention for focal lengths is the 

same for mirrors. 

§ 8.10 Multi-Lens Optical Systems 

Most optical systems are composed of more than one lens. We do 

not need more theory in order to work with a multi-lens system. The 

light passes from one lens to the next in sequence. In order to do a 

computation with a multi-lens system, we need only use the image of 

one lens as the object of the next lens. 

Suppose that you have a diverging lens, f = −12cm, followed by a 

converging lens, f ′ = 16cm. The lenses are 20 cm apart. You place an 

object that is 9cm tall at a distance of 24 cm from the diverging lens. 

We can construct the final image as follows. 

24 cm 

20 cm 

28 cm 37.3 cm 

Mirror 

Object 1 

8cm 

12cm 

Image 1 

Object 2 

12cm 

16cm 

Or we could compute the final image from the thin lens equation. Given 

f = −12cm and x o = 24cm we can compute 

1 

= 1 x i f − 1 1 

= 

x o −12cm − 1 

24cm = − 3 

24cm −→ x i = −8cm 

16cm 

Image 2


So 

y i = − x i 

y o = − −8 (9cm) = 3cm 

x o 24 

The second lens is 20 cm past the first, so the object distance for the 

second lens is 

x ′ o = 20cm − x i = 20cm − (−8cm) = 28cm. 

From this we can compute the final image location 

1 

x ′ = 1 

i f ′ − 1 x ′ = 1 

o 16cm − 1 

28cm −→ x′ i = 37.33cm 

and 

y i ′ = − x′ i 

x ′ y o = − −37.3 (3cm) = 4cm 

o 28 

Notice that the second lens has formed a real image of the virtual image 

produced by the first lens. 

In the following example, the second lens has a virtual object. 

Example 

Suppose that you have a converging lens, f = 12cm, followed by a 

diverging lens, f ′ = −12cm. The lenses are 18 cm apart. You place an 

object that is 4cm tall at a distance of 24 cm from the converging lens. 

We can construct the final image as follows. 

24 cm 

18 cm 

Image 2 

Image 1 

Object 1 

12cm 

12cm 

12cm 

Object 2 

The top ray is not used to construct the first image, since it is not a 

principle ray of the first lens. The first image is constructed from the 

other two rays, and then the third ray is draw so that it goes to the 

first image and also through the center of the second lens, this makes 

it a principle ray of the second lens. Similarly the center ray is not a 

principle ray of the second lens, it is only used for constructing the first 

image. The bottom ray is a principle ray of both lenses.


We can also compute the final image from the thin lens equation. 

Given f = 12cm and x o = 24cm we can compute 

1 

= 1 x i f − 1 = 1 

x o 12cm − 1 

24cm = 1 

24cm −→ x i = 24cm 

So 

y i = − x i 

y o = − 24 (4cm) = −4cm 

x o 24 

The second lens is 18 cm past the first, so the object distance for the 

second lens is 

x ′ o = 18cm − x i = 18cm − (24cm) = −6cm. 

From this we can compute the final image location 

1 

x ′ = 1 

i f ′ − 1 1 

x ′ = 

o −12cm − 1 

−6cm −→ x′ i = 12cm 

and 

y i ′ = − x′ i 

x ′ y o = − 12 (−4cm) = −8cm 

o −6 

Notice that the diverging lens has formed a real image. Also notice that 

the first lens would have formed a real image, but since the diverging 

lens enters the optical path before the image is formed, the first image 

does not actually get formed. 



You are designing a movie projector. The film is 8mm wide, and you 

wish to project this film onto a screen that is 2.0 meters wide from a 

distance of 10 meters. 

(a) How far from the lens should the film be? 

(b) What should the focal length of the lens be? 


An object is located 20cm to the left of a diverging lens having a focal 

length f = −32cm. Determine the location and magnification of the 

image. Construct a ray diagram for this arrangement. 


An object is placed 40 cm in front of a lens with focal length +10 cm. 

Describe the image (i.e. location, magnification, etc.). 


Two converging lenses, each of focal length 10 cm, are separated by 

35 cm. An object is 20 cm to the left of the first lens. (a) Find the


position of the final image using both a ray diagram and the thin-lens 

equation. (b) Is the image real or virtual? Upright or inverted? (c) 

What is the overall magnification of the image? 


An object is 15 cm in front of a positive lens of focal length 15 cm. A 

second negative lens of focal length -15 cm is 20 cm from the first lens. 

Find the final image and draw a ray diagram. 


A concave mirror has a focal length of 40.0cm. Determine the object 

position for which the resulting image is upright and four times the size 

of the object. 


A concave mirror has a radius of curvature of 60 cm, (f = 30cm). 

Calculate the image position and magnification of an object placed in 

front of the mirror at distances of 90 cm and 20 cm. Draw ray diagrams 

to obtain the image in each case. 


Under what conditions will a concave mirror produce and erect image? 

A virtual image? An image smaller than the object? An image larger 

than the object? Repeat the exercise for a convex mirror. 


A dentist wants a small mirror that will produce an upright image with 

a magnification of 5.5 when the mirror is located 2.1 cm from a tooth. 

(a) What should the radius of curvature of the mirror be? (b) Should 

it be concave or convex?

@ Summary 177 


Facts 

• Law of Reflection: 

θ i = θ r 

• Snell’s Law: 

n i sin θ i = n t sin θ t 

Theorems 

• Thin Lens Equation: 

1 

x o 

+ 1 x i 

= 1 f

178 Hints A 

A 

Hints 

1.1 Do the directions “by hand”. To make it easier to keep track of 

everything write all forces in terms of the quantity F 0 = 

1.2 Use the vector form of Coulomb’s law. 

1.3 The answer is E(x, ⃗ 0) = q 

4πɛ 0 

−2aĵ 

(x 2 +a 2 ) 3/2 . 

q2 

4πɛ 0a 

. 2 

1.4 Use polar coordinates, ⃗r s = R cos θî + R sin θĵ and dq = Q 2π dθ. 

1.5 Use Gauss’s Law. 

1.6 Do not attempt to integrate the flux over the surfaces. First, 

look carefully at the orientation of the electric field through each of the 

three faces that touch the charge. Second, because of the symmetry of 

the configuration the remaining three faces must have the same flux as 

each other. Third, notice that if you placed eight such cubes around 

the charge (forming a larger cube with the charge at the center), that 

all eight smaller cubes would have the same flux. 

1.7 Use a gaussian surface that is a sphere of radius r, centered on 

the charge. 

1.8 Use Gauss’s law. The gaussian surfaces are spheres. If r < R 

then the amount of charge inside the gaussian surface depends on r, 

show that the amount of charge inside is q in = Q r3 

R 3 

1.10 Assume that the field is directed straight out from the line. Let 

the Gaussian surface be a cylinder (like a tin can) with the line charge 

as the axis of the cylinder. Note that the flux through the ends of the 

can is zero because of the orientation relative to the field. 

1.11 The gaussian surface is a cylinder of radius r and length L. 

1.12 Recall that there can be no field inside the body of a conductor 

that is in static equilibrium. 

1.13 The geometry gives you the angle of the forces, but because 

the charges are not of the same size you must still deal with both 

components. Depending on your disposition, you might prefer using 

the vector form of Coulombs law.

A Hints 179 

1.14 Since this is only an estimate, assume that you are composed 

entirely of water. 

1.16 Be sure to add the parts as vectors. 

1.17 Let the charge elements dq be little sections of arc that subtend 

and angle dθ. Notice that the full charge is spread over a half a circle 

so that the charge density (charge per angle) is −7.5µC 

π 

, so that dq = 

−7.5µC 

π 

dθ. 

1.18 Use the work energy theorem. Recall that work is force time 

distance. 

1.19 This is like a projectile motion problem, but this time we have 

a constant electric force rather than an constant gravitational force. 

1.20 Start by making a free body diagram and be sure to include the 

force of gravity and the force of the string. 

1.21 A flux is negative if it is into the volume of the box and positive 

if the flux is out of the volume of the box. 

1.22 There is a very easy way to do this problem, by using the fact 

that there is no charge inside the volume of the nose cone. 

1.25 ⃗r = −xî 

2.1 Look at the definition of electric potential, how is the electric 

potential related to potential energy? 

2.2 Use the work energy theorem. 

2.3 Follow the example in the text for a nonuniform field. 

2.4 Follow the example in the text. 

2.6 Consider a closed surface that is totally within the outer conductor 

and surrounds the inside surface of the outer conductor, as indicated 

by the dotted line in the figure. Using the fact that there is no field 

inside a conductor argue that the electric flux through this surface is 

zero. From this use Gauss’s law to find the charge on the inside surface 

of the outer conductor. Next use the fact that the total charge on the 

outer conductor is equal to the sum of the charge on its two surfaces, 

to find the charge on the outside surface of the outer conductor. 

2.7 Use Gauss’s law. Remember that the electric field is zero inside 

the body of a conductor. 

2.8 Use the definition of capacitance. 

2.9 Use the fact that the electric field strength between the plates is 

both σ/ɛ 0 and ∆V/d, where σ is the charge density on the plates. 

2.10 Assume that the capacitor is charged so that the inside sphere 

has a charge Q and the outside sphere has a charge −Q. Use Gauss’s 

law to get the field strength between the shells.

180 Hints A 

2.11 First suppose that there is a charge Q on the length L of the 

central wire, and a charge −Q on the outer shield. Then use gauss’s 

law to show that the electric field strength between the wire and shield 

is E = 

Q 1 

2πLɛ 0 r 

. Next show that the potential difference between the 

wire and shield is ∆V = 

Q 

2πLɛ 0 

ln(b/a). Finally use the definition of 

capacitance. 

2.12 The field around a line charge has already been found, use this 

to find the electric potential around a single wire. Find the electric 

potential difference as you move from one to the other for each wire 

and then add. Once you have the electric potential difference between 

the wires you can find the capacitance. 

2.13 Since there is a maximum field strength there is also a maximum 

energy density, and the energy density is the energy stored divided by 

the volume of the capacitor. 

2.14 The change in the charge switches the direction of the electric 

field, which alters the dot product. Also q = −|q|. 






2.20 Assemble the particles one at a time. The work to bring the 

first particle in is zero since there is no field to do work against. The 

second particles feels the potential of the first as you bring it in. The 

third particle feels the potential of the first two, and so on. 

2.21 The field is the gradient of the potential. 

2.22 V = ∫ dq 

4πɛ . 0r 

2.23 Find the electric potential for each section first. You have actually 

done a problem like each of the sections before. 

2.24 Write out the field and electric potential of a point charge, then 

do some algebra. 

2.25 The potential is the sum of the three point potentials. The field 

can be found from the derivative of the potential. 

2.26 The answer will be in terms of the radius (r 0 ), field (E 0 ), charge 

density (σ 0 ), and potential (V 0 ) of the original drops. The charge is all 

on the surface of the drop.

A Hints 181 

2.27 Imagine that you build of the charge slowly. Suppose that you 

have already got a charge q accumulated and you want to bring in an 

amount dq more. How much work dW must you do? Once you have 

dW written out you can integrate to get W . 

2.28 Use Gauss’s law to find the field in all three regions. Then 

integrate from infinity inward to find the electric potential. You will 

need to split the integral into three regions because the form of the field 

changes each time you cross the surface of a shell. 

2.29 Look at the definition of capacitance. 

2.30 Look at the definition of capacitance. 

3.1 The amount of charge is equal to the number of particles times 

the charge per particle. 

3.2 You will need to use Ohm’s law, the definition of current density 

and the relationship between the electric field and the electric potential 

∆V = −E ∆x. 

3.3 Use the result that R = ρL/A. 

3.4 Use P = I ∆V . 

3.5 Look up the theorem on electrical power. 

3.6 Use the result of the example before this problem. 

3.7 The current and voltage on the resistor can be negative (and they 

are in this case). 

3.8 Assume that the centripetal acceleration of the electron is caused 

by the Coulomb force of the proton on the electron. Remember the 

definition of electric current. 

3.9 Remember that you know the charge of each electron, and that 

current is the amount of charge per time. 

3.10 Read the definition of current. 

3.11 q = ∫ dq = ∫ dq 

dt dt 

3.12 First compute how many electrons pass a particular point in the 

wire in one second. All of these electrons together fill a volume V of 

the wire. From the number of electrons you can compute the volume of 

electrons. Once you have the volume you can compute the length the 

electrons occupy in the wire, since you know the cross sectional area 

of the wire. But this length is the distance the electrons move in one 

second. 

3.13 Use the result that R = ρl/A. 

3.14 Look up electric power.

182 Hints A 

3.16 Compute the total charge that can flow from the battery. One 

kilowatt hour is a unit of energy 3600kJ. 

3.17 From the specific heat of water you can find the energy required 

and from this you can find the power and thus from the known voltage 

you can find the current and then resistance. 

3.18 The internal resistance acts like it is in series with the external 

resistor and and ideal voltage source. 

3.19 Reduce the system in stages. Look for pairs of resistors in the 

system that are in parallel or in series, combine this pair, then repeat 

until there is only one resistor left. 

3.23 Write out all of the equations and then solve. 

3.25 Connect the circuit to a voltage supply V S and using Kirchhoff’s 

rules for the resultant circuit you can find the charge on each capacitor. 

This will in turn allow you to compute the total charge drawn form the 

power supply and thus the effective capacitance of the system. 

3.26 Use Kirchhoff’s rules 

3.27 Use Kirchhoff’s rules 

3.28 The bulb that draws the most power is brighter. 

3.29 The appliances are connected in parallel to the voltage supply. 

3.30 Use the result of the other problem in which you computed the 

resistance of a 12 gauge copper wire. 

3.31 The amount of heating is proportional to the power dissipated 

by the wire per length. 

3.32 When the capacitor is fully charged, no current flows into it. 

4.1 

(a) Notice that ⃗ dl has length Rdθ and is perpendicular to the radius 

vector. 

4.2 Compute the force on each of the fours section of wire first. Recall 

that torque is ⃗τ = ⃗r × ⃗ F , where ⃗r points from the axis of rotation to 

the point at which the force is applied. 

4.5 For any radius it will take a time 2πm/qB 

4.6 Plug the given values into the Lorentz force formula and do the 

cross product. 

4.7 The magnetic field of the earth points roughly northward. 

4.8 Don’t forget that the electron is negatively charged. 

4.9 This question is asking you to relate acceleration and force so 

start with Newton’s second law.

A Hints 183 

4.12 Consider the vector nature of the definition of work and the 

direction of the velocity relative to the magnetic force. 

4.13 Write out the vectors ⃗ L for each line segment and then do the 

cross products. 

5.1 Try plotting points for various values of t to get a sense of the 

function. 

5.2 Start with ⃗r(t) = ⃗a + btî + ct 2 ĵ and find the value of the constants 

by demanding that the curve goes through the three given points. 

5.3 The parameterization is tî 

5.4 The parameterization is a cos tî + a sin tĵ. 

5.5 Since the current density is not uniform, I in = ∫ JdA. 

5.7 Use the result B = µ 0 I/2πr. 

5.8 Use the Biot-Savart law. 

5.9 Use the method of the previous problem, and also note that the 

current can be found from the electron speed and the radius of the 

orbit. 

5.10 First argue that half of the wire does not produce any field at 

the point of interest. Then use the Biot-Savart law on the other half. 

5.11 Argue that the radial lines do not contribute. Note that the 

field due to the two arcs are in opposite directions to each other. 

5.12 First find the field created by wire 1 at the location of wire 2. 

Then find the force that this field produces on wire 2. 

5.13 Use Ampere’s Law. 


5.15 Since the current density is not uniform, I in = ∫ JdA. 

6.1 The induced current is in the same direction as the electric field. 

Determine the direction of the induced field in the center of the loop, 

from the known electric field direction. Compare the induced field with 

the existing field. 

6.2 Choose the Amperian loop to be a circle with radius r and going 

through the desired field point half way between the plates. Note that 

the current density between the plates is zero. 

6.3 Use Kirchhoff’s loop rule to write an equation in the current and 

the derivative of the current. Try a solution of the form I(t) = a+be −αt . 

6.4 The RMS voltage of the source in the a standard outlet is 120 

volts. 

6.5 Follow the example in the text for an RC circuit.

184 Hints A 

6.6 Use a phasor diagram to draw Kirchhoff’s loop rule, then find the 

frequency that minimizes the effective impedance of the RLC combination. 

6.8 Use Faraday’s law. Assume that the magnetic field decreases at 

a constant rate over the 20 ms that it goes from 1.6 T to zero, that is 

dB/dt = (−1.6T)/(20ms). 

6.10 Since the field strength decreases as you move further from the 

wire you will need to do an integral in order to compute the magnetic 

flux. 


6.13 Consider that Q = ∫ Idt and I = V/R = E/R. 

7.2 Start from x = A cos φ. 

7.3 Take a look at the diagram of the oscillator that shows the phase 

angle at different points in the cycle. For the last part remember that 

v = dx/dt and consider the quantity v/x. 

7.4 Recall that the phase is φ = ωt + φ 0 so that ∆φ = ω∆t. Then 

consider by how much the phase must change in order for the oscillator 

to complete one cycle. 

7.8 First compute the phase difference. Remember that φ 1 = −kr 1 

and φ 2 = −kr 2 . Next use the addition of two waves theorem. 

7.9 The collection of points with the same ∆r will form a continuous 

line, just like the collection of points that are all 5cm from a single 

point makes a circle or radius 5 cm. 

7.10 Treat the two holes as two sources. 

7.11 The phase of each path is now −kr plus the reflection phase 

shift. 

7.12 Find the distance from the lower speaker to the point P as a 

function of the distance d. 

7.15 The path difference occurs in air. 

7.16 Use the far field approximation to find the path difference. 

7.18 Use the trig identities in the appendix. 

7.19 There is a path difference, d sin θ, on both sides of the screen. 

7.22 The air gap between the plates will be wedge shaped, because 

the thickness of the air gap increases as you move toward the wire, 

there will be an increasing phase difference between the two reflected 

beams. This leads to the fringes. 

7.24 As the slit gets narrower the angle of the first minima increases. 

There is a limit to the increase.

A Hints 185 

7.26 The distance from the central maximum to the first order maximum 

is the same as it would be for a double slit with the same distance 

between the two slits as there is between successive lines in the diffraction 

grating. 

8.1 Use geometry to show that 

α 

α/2 

β 

α/2 

Then use Snells law. 

8.2 Each straight line segment that crosses the optical axis, creates 

two similar triangles, one above the axis and one below the axis. Use 

the properties of similar triangles to get an equation for each of these 

pairs. 

8.5 Use the thin lens equation to find under what conditions the image 

distance is negative. 

8.7 Use the thin lens equation. 

8.8 Start with the magnification equation.

186 Index B 

B 

Index 

† Energy Density of an Electric Field, 48 

† Addition of Two Waves, 147 

† Ampere’s Law - Time Varying, 120 

† Ampere’s Law, 102 

† Biot-Savart Law, 97 

† Circular Trajectories, 84 

† Conductor in Equilibrium: Charge, 43 

† Conductor in Equilibrium: Field, 43 

† Conductor in Equilibrium: Potential, 44 

† Conductor in Equilibrium: Surface Field, 44 

† Coulomb’s Law in Vector Form, 8 

† Coulomb’s Law, 8 

† Effective Capacitance, 65 

† Effective Resistance, 64 

† Electric Charge, 4 

† Electric Field due to a Point Charge, 15 

† Electric Field from the Electric Potential, 40 

† Electric Potential of a Point Charge, 48 

† Electrical Power, 61 

† Electromagnetic Waves and Light, 139 

† Energy Stored in a Capacitor, 47 

† Faraday’s Law (alternate form), 117 

† Faraday’s Law, 116 

† Force Currents, 105 

† Force on a Current in a Uniform Field, 82 

† Gauss’s Law, 20 

† Impedance: Capacitor, 125 

† Impedance: Inductor, 124 

† Interference of Two Sources, 150 

† Kirchhoff’s Junction Rule, 61 

† Kirchhoff’s Loop Rule, 62 

† Law of reflection, 162 

† Lenz’s Law, 119 

† Magnetic Force on a Current, 82 

† Magnetic Force, 81 

† Magnification Equations, 169

B Index 187 

† Ohm’s Law: Resistance, 60 

† Ohm’s Law: Resistivity, 59 

† Reflection Phase Shift, 154 

† Short Wavelength Limit, 161 

† Single Slit Diffraction, 155 

† Snell’s Law, 163 

† Superposition Theorem, 15 

† Thin Lens Equations, 168 

† Two Types of Charge, 4 

† Wave due to a Sinusoidal Point Source, 142 

† Wavelength in a Medium, 154 

† Work by Magnetic Force, 84 

alternating current, 122 

capacitance, 46 

conductivity, 59 

conductor, 7 

converging lens, 166 

current density, 58 

current, 58 

cyclotron, 85 

direct current, 122 

displacement current, 120 

Diverging Lens, 171 

electric field, 10 

electric field, 11 

electric flux, 20 

electric potential, 37 

electromagnet, 97 

electromagnetic, 139 

electromotive force, 116 

elements, 57 

EMF, 116 

equipotential, 41 

Farad, 46 

field lines, 14 

focal length, 166 

focal plane, 167 

focal point, 166 

focus, 166 

frequency, 143 

gain, 128 

gradient, 40

188 Index B 

ground, 38 

Hall Effect, 87 

Henry, 120 

in parallel, 62 

in series, 62 

in-phase, 147 

index of refraction, 153 

inductance, 120 

infrared, 145 

Lorentz Force, 85 

magnetic flux, 117 

n-type, 88 

non-ohmic, 59 

ohm, 60 

ohmic, 59 

optical axis, 166 

out-of-phase, 148 

p-type, 88 

parameterization, 39 

parameterize, 98 

period, 143 

phase, 139 

phasor, 125 

principle focus, 166 

principle rays, 168 

real image, 168 

resistance, 60 

resistivity, 59 

Resistor, 59 

resonance, 129 

scalar field, 16 

steady state, 115 

Tesla, 82 

ultraviolet, 145 

uniform, 16 

vector field, 16 

velocity selector, 86 

wavelength, 142 

wiring diagram, 57

B Index 189

190 Useful Mathematics C 

C 

§ C.1 Trigonometry 

Useful Mathematics 

cos(a + b) = cos a cos b − sin a sin b 

sin(a + b) = sin a cos b + cos a sin b 

cos a + cos b = 2 cos b − a 

2 

cos a − cos b = 2 sin b − a 

2 

sin a + sin b = 2 cos b − a 

2 

cos b + a 

2 

sin b + a 

2 

sin b + a 

2 

§ C.2 Fundamental Derivatives and Integrals 

Basic Facts 

d 

dx [xp ] = px p−1 

d 

dx [eax ] = ae ax 

d 

dx [ln(x)] = 1 x 

∫ 

d 

[sin(kx)] = k cos(kx) 

dx 

d 

dx 

[cos(kx)] = −k sin(kx) 

∫ 

∫ 

x p dx = xp+1 

∫ 

p+1 

e ax dx = eax 

∫ 1 

x 

dx = ln(x) 

cos(kx) dx = 

sin(kx) 

k 

sin(kx) dx = − 

cos(kx) 

k 

a 

Divide and Conquer Rules 

Sum Rule: 

f(x) = a g(x) + b h(x) −→ 

d 

Example: 

dx [3x2 + 4x 3 ] = 3 d 

dx [x2 ] + 4 d 

Product Rule: 

Example: 

2x sin(x). 

Chain Rule: 

Example: 

f(x) = g(x) h(x) −→ 

df 

dx = a dg 

dx + bdh dx 

dx [x3 ] = 6x + 12x 2 . 

df 

dx = g(x)dh dx + dg 

dx h(x) 

d 

dx [x2 sin(x)] = x 2 d 

dx [sin(x)] + d 

dx [x2 ] sin(x) = x 2 cos(x) + 

d 

dx [sin(3x2 )] = 

df 

dx = dg dh 

dh dx 

d 

d[3x 2 ] [sin(3x2 )] d 

dx [3x2 ] = cos(3x 2 )6x 

f(x) = g(h(x)) −→

C Power Series Expansions 191 

The Fundamental Theorem of Calculus 

Recall that the distance traveled between time t 1 and time t 2 is 

equal to the area under the v versus t graph between these times. 

v(t) 

Area = ∫ t 2 

t 1 

v(t)dt 

t 

t 1 t 2 

But the distance traveled is also x(t 2 ) − x(t 1 ). So we find that 

∫ t2 

v(t)dt = x(t 2 ) − x(t 1 ) 

t 1 

This works for any pair of functions where one is the derivative of the 

other. So in general 

IF 

f(t) = dg 

dt 

THEN 

∫ t2 

t 1 

f(t)dt = g(t 2 ) − g(t 1 ) 

§ C.3 Power Series Expansions 

(1 + x) N = 

∞ 

1 

1 − x = ∑ 

x n = 1 + x + x 2 + x 3 + x 4 . . . 

sin(x) = 

n=0 

∞∑ 

n=0 

cos(x) = 

e x = 

N∑ 

n=0 

∞∑ 

n=0 

(−1) n 

(2n + 1)! x2n+1 = x − 1 6 x3 + . . . 

∞∑ 

n=0 

(−1) n 

(2n)! x2n = 1 − 1 2 x2 + . . . 

1 

n! xn = 1 + x + 1 2 x2 + 1 6 x3 + . . . 

N! 

N! 

n!(N − n)! xn = 1 + Nx + 

2!(N − 2)! x2 + · · ·

192 Physical Constants and Data D 

D 

Physical Constants and Data 

§ D.1 Index of Refraction 

(at λ=589.3 nm, from wikipedia.org) 

Material Index 

Vacuum 

1 (exactly) 

Helium 1.000036 

Air at STP 1.0002926 

carbon dioxide 1.00045 

water ice 1.31 

liquid water (20C) 1.333 

ethanol 1.36 

glycerine 1.47 

polycarbonate 1.59 

glass (typical) 1.5 to 1.9 

cubic zirconia 2.2 

diamond 2.4 

moissanite 2.7 

gallium phosphide 3.5 

gallium arsenide 3.9 

silicon 4.0 

§ D.2 Approximate Electrical Conductivity 

(from wikipedia.org) 

Material Ω −1 · m −1 

Silver 63.0 × 10 6 

Copper 59.6 × 10 6 

Gold 45.0 × 10 6 

Aluminium 37.8 × 10 6 

Brass 20.0 × 10 6 

Iron 10.0 × 10 6 

Bronze 7.0 × 10 6 

Lead 4.8 × 10 6 

Stainless Steel 1.4 × 10 6 

Seawater 5.0 × 10 0 

Drinking water 5.0 × 10 −3 

Deionized water 5.5 × 10 −6 

Glass 

∼ 10 −12 

Rubber 

∼ 10 −13

D Unit Prefixes 193 

§ D.3 Fundamental Constants 

speed of light c 2.99792458 × 10 8 m s 

(exact) 

Planck constant h 6.6260755(40) × 10 −34 J · s 

hc 1239.8424(93) eV · nm 

1240 eV · nm 

¯h 1.05457266(63) × 10 −34 J · s 

¯hc 197.327053(59) eV · nm 

fundamental charge e 1.60217733(49) × 10 −19 C 

mass of electron m e 9.1093897(54) × 10 −31 kg 

0.51099906(15) MeV/c 2 

mass of proton m p 1.672631(10) × 10 −27 kg 

938.27231(28) MeV/c 2 

Boltzman constant k 1.380658(12) × 10 −23 J/K 

8.617385(73) × 10 −5 eV/K 

Avogadro number N A 6.0221367(36) × 10 23 

permeability of free space µ 0 4π × 10 −7 N/A 2 (exact) 

permittivity of free space ɛ 0 1/(µ 0 c 2 ) (exact) 

8.854187817 × 10 −12 C 2 /N · m 2 

1 

4πɛ 0 

8.99 × 10 9 N · m 2 /C 2 

gravitational constant G 6.67259(85) × 10 −11 m 2 /kg · s 

§ D.4 Unit Conversions 

1 m 3.28 ft 

1.6 km 1 mile 

1 mile 5280 ft 

1 hp 746 W 

1 liter 1 × 10 −3 m 3 

1 gallon 3.79 liters 

1 atm 1.013 × 10 5 Pa 

1 J 0.239 cal 

1 kcal 4186 J 

§ D.5 Unit Prefixes 

f femto 10 −15 

p pico 10 −12 

n nano 10 −9 

µ micro 10 −6 

m milli 10 −3 

k kilo 10 3 

M mega 10 6 

G giga 10 9 

T tera 10 12

Introductory Physics Volume Two

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?