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Practice
Eric Mazur
E
Physics 11b
Spring 2007
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of the electrons, or the force exerted on one of the electrons by the entire nucleus?
Section 26.7
14. Consider some arrangement of charge carriers, labeled 1, 2, and 3. What is wrong with this
expression for the force on charge carrier 1?
!"
! F1 = k qr22q1 rˆ21 +k qr32q1 ˆr31 + k qr32q2 ˆr32
21
31
32
2
Developing a feel
Calculate or estimate the following quantities:
1. The minimum electric force on a tiny bit of paper lifted by attraction to a charged comb. (C, O)
2. The magnitude of electric force between a proton and an electron in an atom. (R, S)
3. The magnitude of electric force between the two ions in a salt molecule. (D, V, S)
4. The total number of protons in a large cola. (E, N, H, W)
5. The total number of electrons in the Earth. (Y, T, M, J).
6. The magnitude of electric force between you and a friend, standing 10 meters apart, if each of you
had 1% more electrons than protons. (E, H, K, W, A)
7. The percentage of Earth’s electrons which, if transferred to the Moon, would create an electrical
attraction equal to the gravitational attraction of these bodies. (G, L, Y, U, results of number 5
above)
8. The net charge on each pith ball in Figure 26.11(b). (X, Q, I)
9. The maximum charge you can put equally on two pith balls that you are holding without feeling
their force of repulsion. (B, F, P)
Hints:
A. What is the magnitude of charge carried by 1% of a person’s electrons?
B. What minimum weight (i.e., force) can you detect with your fingers?
C. To lift, what force must be overcome?
D. What is the separation of the ions in the salt molecule?
E. What is the main chemical ingredient?
F. How can you minimize the force between the pith balls you are holding.
G. If you equate the gravitational and electrical force expressions, what cancels?
H. What is the inertia of a molecule of water?
I. What value of electric force is needed?
J. How many electrons for each proton in the Earth?
K. What is the inertia of a person?
L. What is the inertia of the Moon?
M. What is the inertia of a proton?
N. What is the inertia of a large cola?
O. What is the inertia of a small bit of paper?
P. What maximium distance can you create between your hands.
3
Q. What is the inertia of a pith ball?
R. What is a typical separation of proton and electron in an atom?
S. What is the magnitude of charge carried by an electron or proton?
T. What fraction of this inertia is due to protons?
U. What is the ratio of gravitational to electrical constants in the force expressions?
V. What charge does each ion carry?
W. How many protons are in one molecule of water?
X. What is the angle from the vertical to the line of each string?
Y. What is the inertia of the Earth?
Key: A. ~4x107 C. B. about 1/10 the weight of a 0.025-kg penny, ~2x10-2 N. C. gravitational
force. D. ~1.5x10-10 m. E. water. F. By getting them as far apart as you can. G. Earth-Moon
separation cancels. H. Two H plus one O is ~3x10-26 kg. I. for equilibrium, F ~4x10-4 N. J. One.
K. ~70 kg. L. ~7x1022 kg. M. ~2x10-27 kg. N. ~1 kg. O. ~1x10-5 kg. P. about 2 m. Q. ~1x10-4 kg.
R. ~5x10-11 m. S. 1.6x10-19 C, the elementary charge. T. ~half, because atoms have about equal
numbers of protons and neutrons which have nearly identical masses. U. G/k ~ 7x10-21 with both
constants in SI units. V. Na+ and Cl-, each with an elementary charge. W. ten. X. ~20°. Y. ~6x1024
kg.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any particular
section. Typically, an example that is worked out in detail is followed immediately by an example
whose solution you should work out by following the guidelines provided.
WP 26.1: Mutual Attraction. We often feel a physical attraction to other people. What if this were
literally true? Suppose you and your partner stand 1 m apart and feel an attractive electrostatic force
of 10 N between you. Estimate the fraction of excess charge to total charge of a given type in your
bodies.
!Focus problem
We begin with a simplified sketch of the basic elements the physical situation.
–
+
The force of attraction is determined by Coulomb's law. We need to find the amount of charge on
each body that produces 10 N of force between them and we will have to make some simplifying
assumptions about its location. Once the necessary charge is determined, we will need to know what
fraction it is of the total charges in the body. To find the fraction of excess charge we will need to
estimate the total number of charged particles of a given type in the body. These charges are protons
and electrons and reside in molecules. Multiply the number of those charges by the magnitude of
their elementary charge will give the total charge of that type in the human body. This is obviously
going require many estimates.
!Plan approach
We know the force of attraction and the separation between you and your partner. Since the force is
attractive, you must have opposite charges. It is easiest to assume that you both have about the same
mass with equal amounts of excess charge (though of differing signs). The human body is extended
and it would be very difficult to calculate the net force if the charge were uniformly distributed
throughout the bodies. To simply things, we will assume that all of it can be effectively
concentrated at the body's center, with center-to-center distance of 1 m. The human body is mostly
water, H2O. Dividing the mass of a "standard" human body by the mass of a water molecule will
give us the number of water molecules. Multiplying that number by the number of protons or
electrons in each molecule will allow us to find the total charge in the human body. Dividing the
needed excess charge to produce the force by the total charge gives us the desired fraction of excess
charge.
!Execute plan
5
We use Coulomb’s law to solve for the magnitude of charge q to produce the requisite force:
| q (−q ) |
F =k
r2
q=
Fr 2
k
The first step indicating an attractive force on the left and charge magnitudes of opposite sign on the
right. Putting in F = 10 N and r = 1 m, k = 9x109 Nm2/C2, the amount of excess charge required is
q!"x10-4 C.
There are several ways to proceed to estimate the number of water molecules in a human body. You
might know from chemistry that water has an atomic mass of 18 g/mole and that one mole is 6x1023
particles, which makes the inertia of a single water molecule 18x10-3 kg/(6x1023)= 310-26 kg.
Alternatively, you can use the fact that water consists of two hydrogen atoms (1 proton each) and
one oxygen atom (8 protons and 8 neutrons) for a total of 18 nucleons (protons plus neutrons) per
molecule. The masses of a proton and neutron are nearly the same (1.7x10-27 kg), so the mass of a
water molecule is 18 (1.7x10-27 kg) = 3x10-26 kg. (Electrons have a mass 2000 times smaller than
that of a proton, so we can ignore their contribution.) A reasonable value for you mass might be 60
kg, so the number of protons and neutrons in your body is 60 kg/3x10-26 kg = 2x1027 water
molecules in your body. Because neutrons have zero charge, each molecule has 10 protons (and 10
electrons), so the total charge Q of a given type (positive or negative) in the body is
Q!(2x1027 molec)×(10 elementary charges/molec) × (1.6x10-19C/elementary charge) ! 3x109 C.
From this, we can finally estimate fraction of excess charge
q 13 ×10−4 C
≈
≈ 10−14
Q 3 × 109 C
This is approximately one excess electron (or proton) for every 10 trillion (1013) water molecules!
!Evaluate answer
It is very hard to estimate whether this answer is reasonable or not because most people do not have
a good practical feeling for the magnitudes of electrostatic force. In reality, such an attractive
electrostatic forces are not observed between humans. The actual fraction of excess charge is
significantly lower, essentially zero. However you can observe some consequences of excess bodily
charge when your rub your feet on the carpet, transferring charge to your body. You might get a
trillion or so excess elementary charges in this case, for an excess charge fraction of ~10-16. This is
what gives you a shock when you touch a doorknob as the excess charge leaves your body.
6
WP 26.2: Electroscope. An electroscope is a device for measuring electric charge. In one simple
configuration shown in Figure 26-0, two small identical spheres are each attached to a very fine
non-conducting thread affixed to the ceiling. Initially, both spheres are in contact until an equal
amount of charge is put on each. You might imagine that the two are touched with a third charged
object, distributing the charge uniformly on both spheres faster than they have time to separate.
Treat the two spheres as particles each with inertia of 0.017 kg. The thread is 120 mm long, and the
spheres come to rest at a separation of 9.3 cm. What is the electric charge on each sphere?
d
Figure 26-0
!Questions and suggestions
1. In what sense in the problem similar to WP 26.1? In what sense different?
2. Why do the push apart but then come to an equilibrium configuration? What forces produce this?
What types are they?
3. Draw the free body diagram for each particle. Identify and list any assumptions that you make.
4. Draw the free-body diagram. What quantity will you need to know to resolve the forces?
5. Separate the forces into x and y components based on the coordinate system you choose.
6. Work through the algebra to solve for the desired electric charge q. Avoid solving for any
intermediate quantities unless you have to – you will only make more work for yourself.
7. To evaluate your answer, think about how q should depend upon on the separation d? Does your
algebraic answer reflect this expectation?
WP 26.3: Levitation. One possible way of levitating an object might be to use the forces
associated with charged objects. For example, you have two charged particles that are fixed on a
vertical pole 0.5 m apart. The lower one has a fixed charge of -3.0 µC. The upper one has a charge
qA that can be adjusted. A 30-mg particle with a charge of +8.0 µC can move freely on the pole
below the other two. You wish to levitate (i.e., float) this particle at a distance of 1.0 m below the
lower fixed charge. What should the adjustable charge qA be to achieve this feat?
!Focus problem
This sounds a bit complicated, so we have to make a careful sketch of the situation.
7
qA
0.5 m
-3.0 !C
1.0 m
+8.0 !C
In order for the particle to float we need to balance the electrostatic and gravitational forces acting
on the middle charge so the vector sum of these forces is zero. We have to assume here that we’re
near Earth, so that g is a constant. The gravitational attraction between the two objects is much
smaller than the electrostatic force (something which you should quickly verify this) so that we need
only consider the gravitational pull downward of the earth on the middle particle, the repulsive force
upward due to the lower charge, and the needed electrostatic force acting on it due to the top
particle.
!Plan approach
We want to the sum of forces to be zero. A free-body diagram is always helpful in force problems.
Because all the action is along a straight vertical line, we’ll choose the reference axis to be an
downward pointing y-axis. To keep the particles straight in the mathematics, we will label them A,
B and C.
qA
-3.0 !C
A
B
0
0.5 m
y
C
1.0 m
+8.0 !C
0.03 kg
"F eBC
e
"F geC "F AC
!
The gravitational force FeCg on levitating particle C is downward. Because the charges on B and C
!e
upward. However, we don’t
are opposite in sign, B and C attract, making the electrostatic force FBC
!e
know a priori whether to make the electrostatic force FAC caused by the adjustable charge attractive
or repulsive (through the sign of qA) because we need to know the strength of the other two forces.
We will draw the force vector in the direction of our chosen y-axis, but never implicitly assume that
its component along that axis is positive. It may physically point in the opposite direction,
depending on the calculated sign of qA.
!Execute plan
For stable levitation (a=0), we must have
8
!F
y
e
e
g
e
e
= FeCg , y +FBC
, y + FBC , y = FeC + ( − FBC ) + FAC , y = 0
e
We do not know the charge qA, so that we have to be careful not to give FAC
has a known direction
and let the mathematics determine the sign of the component. We can do this reliably by using the
mathematical expression for the electrostatic force (Equation 26.10):
e
FAC
,y = k
q A qC
( rˆAC ) y
2
rAC
Remembering that rˆAC is the unit vector pointing from A to C, we see ( rˆAC ) y = +1 in this coordinate
e
system. (We could have done something similar for FBC
, y , but we chose to use sign and magnitude
of the component since we know the vector direction of the attractive force on particle C.) Thus
" | q q |# " q q #
! Fy = + mC g + $−k rB2 C % + $k rA2 C % = 0
BC
&
' & AC '
Solving for the desired quantity qA and then substituting numerical values, we find
2
" | qB qC |
#
rAC
qA =
$ k 2 − mC g %
kqC & rBC
'
=
−6
"
#
(1.5 m) 2
C )(8 ×10−6 C ) |
9
2 | ( −3 × 10
×
⋅
− (0.03 kg )(9.8 m / s 2 ) %
(9
10
N
m
/
C
)
$
9
2
2
−6
(9 × 10 N ⋅ m / C )(8 ×10 C ) &
(1.0 m)
'
= −2.4 × 10−6 C = −2.4 µ C
This result shows us that we must put a negative charge on the adjustable particle A to produce a
force that is actually aimed upward (attractive) to help particle B balance the force of gravity.
!Evaluate answer
e
Let us first check to see we would expect the charge qA to be negative, which would make FAC
attractive (i.e., aimed upward). If we make rough estimates of our known forces, we see FeCg !3x10-1
e
!2x10-1 N. Thus the earth gravitationally pulls C down more than B electrostatically
N, while FBC
pulls it up, so A needs to supply an upward force to get C into balance. Our expression for qA
correctly predicts the sign for the charge qA. (Is this system in stable or unstable equilibrium? Does
it depend on the sign of qA?) As a further check of your understanding of the problem, re-derive the
expression for qA if the charges on B and C have the same sign (i.e. both positive or both negative).
Note: If we had not been careful with our decisions about signs, we may well have made an error in
the sign between the two terms (electrostatic and gravitational) in the expression for qA. It is very
important to keep track of the directions and signs of the vectors. Our procedure automatically does
this correctly if applied conscientiously.
9
WP 26.4: Electron Orbit. In the classical model of the hydrogen atom, a single electron orbits the
single proton of the hydrogen’s nucleus at a radius of 0.053 nm. (a) How fast is the electron
moving? (b) How long does it take to complete one orbit?
!Questions and suggestions
1. Draw a sketch with an electron going around it in a circle and a proton fixed at the center of the
orbit. Does this make any assumptions?
2. What force(s) act on the electron? Draw a free-body diagram for the electron.
3. Do see any analogy between this problem and issues discussed in Chapter 14?
4. What is the relationship between the electron's speed and the period of the orbit?
5. Does your answer have the right units?
WP 26.5: Charge Square. Four charged particles are arranged in a square as shown in Figure 26-0,
with q = 3.9×10-4 C and a = 6.9 mm. What is the net force on the particle at the upper right corner
due to the other three?
q
q
a
q
–2q
a
Figure 26-0
!Focus problem
This is an application of Coulomb’s law: we know the charge magnitude and we can find the
separations between the particles.
!Plan approach
We can significantly simplify the problem with a proper choice of axes to exploit the symmetry in
the problem. Let me redraw the figure with a free-body diagram included for are subject particle D,
remembering to label the four different objects (A, B, C and D) to keep track of their various
contributions to the mathematics.
10
C q
!e
FBD
B
–2q
!e
FAD
a
q
A
y
D
45#
!e
FCD
x
0
q
Notice that the y-components of the force cancel because the charged particles are located
symmetrically about the x-axis. This means we only need to calculate the x-component of the force.
We get these x-components from the vector form of Coulomb’s law.
!Execute plan
Because #Fy = 0, the net force on qur is totally due to the components of the forces along the x-axis:
!F
D,x
e
e
e
e
e
e
= FAD
, x +FBD , x + FCd , x = FAD cos 45° + ( − FBD ) + FCD cos 45°
q2
(2q )q
( qq ) 2
= 2 F cos 45° + (− F ) = 2 * k 2 +
−k
=k 2
2
a
, a - 2
2a
e
AD
e
BD
(
)
(
)
2 −1
where we remember that F represents the magnitude of the electrostatic force between particles 1
and 2. Substituting in numerical values yields
e
12
!
F = 11.9 kN in the + x direction
!Evaluate answer
Because two of the forces are repulsive and one is attractive, we expect to see terms of different
signs in the algebraic answer. Reassuringly, we do see that a difference of two terms are to be
computed: the first term due to the two repulsive charges on the y-axis and the second term due to
the attractive charge located on the x-axis. The computed force does seem large. However, the
particles have a sizable charge (remember, a Coulomb is a lot of charge), and they are close to each
other.
We exploited the symmetry of the problem to simplify the work. Symmetry is a very powerful
tool in approaching physics problems, and you should take advantage of it whenever possible.
WP 26.6: Charge Pyramid. The molecule C$F5 has approximately square pyramidal structure:
five fluorine atoms surround a single chlorine atom as shown in Figure 26-0. Assuming a bond
length of a = 0.13 nm, what would be the force on the top fluorine atom if one electron were
simultaneously pulled every atom in the molecule?
11
a
a
Figure 26-0
!Questions and suggestions
1. What causes the resulting force on the top fluorine atom?
2. Look for any symmetry that you can exploit to make the problem simpler. Remember that this is
a three-dimensional situation.
3. How many electrons are in C$ to be pulled away? In F? What does that tell you about the
charged particles that remain?
4. Which equations allow you to express the force in terms of known quantities?
5. Work through the algebra to solve for the force in terms of these variables. Substitute values you
know to get a numerical result.
6. Does your algebraic expression and numerical result completely answer the question posed? Are
the units and direction correct?
WP 26.7: Atom Smashing. To study the fundamental constituents of matter, physicists “smash
atoms” by shooting high-energy subatomic particles at each other. A proton is fired with an initial
velocity of 1×107 m/s at an alpha particle (a helium atomic nucleus with two protons and two
neutrons). Use %K=W to find how close does the proton get to the alpha particle?
!Focus problem
We make an initial sketch of the action and given quantities.
v
d
We know that the electrostatic force is repulsive and causes the proton to slow down by doing
negative work on it. We want to find out at what distance d the proton's speed becomes zero. We
assume that the alpha particle remains effectively fixed because of its significantly higher mass than
the incoming proton. We also assume that the alpha particle acts like a point particle, rather than an
extended charge distribution with different distances to the colliding proton. Finally we assume that
the incoming proton starts very far away (effectively infinitely far) from the alpha particle.
!Plan approach
1. The distance of closest approach, d is the only thing we need to find. But somehow, we need to
relate the deceleration of the incoming particle to the electrostatic force.
12
2. The Coulomb force is not constant as the proton approaches the alpha particle. But we can use
the work-energy theorem to relate the incoming kinetic energy to the work done by the electrostatic
force. We can apply the work-energy theorem to the proton
f !
!
∆K = W = . Fext • ds
i
by noting that the external force acting on it is just the electrostatic force caused by the alpha
particle. We know the change in kinetic energy because the final speed is zero. The total work will
depend upon d, for which we can solve.
!Execute plan
To find the work done by the electrostatic force, we substitute in the explicit expression for the
electrostatic force and then do the integration (being very attentive to signs):
d !
d
qq
!
W = . Fαep • ds = . k α 2 p rˆα p • drˆα p
∞
∞
rα p
d
=. k
qα q p
∞
rα2p
= −4k
qα q p
d
" 1 #
(1 1)
drα p = 4e2 $− % = −kqα q p * − +
,d ∞$& rα p %' ∞
d
We now invoke the work-energy theorem to relate the initial velocity to the work done by the
electrostatic force.
∆K = 12 m p v 2f − 12 m p vi2 = W
0 − 12 m p vi2 = −4k
qα q p
d
This yields an expression for the distance of closest approach:
d = 2k
qα q p
(+2e)(+ e)
e2
2
4
=
k
=
k
m p vi2
m p vi2
m p vi2
Putting in the numbers gives us the distance of closest approach:
d = 5.52×10-15 m
!Evaluate answer
Our algebraic expression for the closest approach distance is at least plausible. Firstly, it is a
positive number, which is reassuring since we want to determine a distance (a magnitude).
Secondly, the greater the proton’s initial speed, the closer it should get to the alpha particle before
turning around. We see that is exactly what is predicted with the square of the speed in the
denominator. The computed closest approach distance brings the proton pretty close to the alpha
particle – about the size of an atomic nucleus. But that is just what is needed if we want to “disturb”
the nucleus.
13
WP 26.8: Shooting the gap. Two objects of equal positive charge +Q are fixed a distance 2s apart.
A particle with negative charge –Q and mass M is held on the perpendicular bisector of the line
joining the two positive charges at a distance d from the midpoint. The negatively charged particle
is released from rest. How fast will it be going when it is closest to the positive charges? How far
will it shoot past the gap?
!Questions and suggestions
1. We must sketch the physical situation, showing all relevant lines and distances.
+Q
-Q
d
2s
+Q
2. What forces act on the particle? What path will those forces cause the particle to take?
3. What is the net force acting on the particle at an arbitrary point in its motion?
4. Is the force constant as the particle moves? If not, how can you specify how it varies as a function
of distance x from the midpoint?
5. What is the relationship between speed and work?
6. Now write an expression for the net work done on the particle from the initial state to its final
state at the midpoint between the positive charges. Take the dot product inside the integral.
7. Your integral may difficult, but note that the differential x dx = ½d(x2).
8. Can you see any simplification by making a change of variables, say using z for x2? Do you have
to change the limits of integration if you make the substitution?
9. Evaluate the integral and determine the speed at the midpoint between the positive charges.
10. Does your algebraic answer give the right units for speed? Does your expression for speed
behave as you would expect with respect to changes in value of Q and M?
14
Questions & problems
Section 26.1-26.2
1.(I) Identify which of the following phenomena are due to the electrostatic interaction: a)
dissolution of salt in water, b) surface tension in water, c) the elliptical orbit of comets, d) the
binding of protons in the nucleus, e) “traction” between tires and pavement.
2.(I) How would results from electrostatic experiments with tape strips described in section 26.2
differ from those actually observed if there were three types of electrical charges in the universe?
3.(II) If there were a natural excess of negative charge over positive charge in ordinary matter, how
might you expect the results of electrostatic experiments with tape strips to change?
Section 26.3
4.(I) Draw a free-body diagram of a balloon clinging to the side of your head (shown in Figure 260) after you’ve rubbed the two together.
Figure 26-0
5.(I) Gasoline trucks used to have light chains that hung from the chassis to drag on the ground.
What do you think the purpose of those chains was? (Hint: nowadays, tires for these trucks have
metal flakes embedded in the rubber.)
6.(II) Some survivors of lightning strikes have reported feeling their hair stand on end shortly before
the strike. Where did this charge come from?
7.(I) Technicians who repair electronics know that a spark can damage electronic chips. They will
therefore wear a strap around their wrist, which is connected by a wire to the metal leg of the
worktable which is electrically "grounded" to the earth. How does this help prevent damage?
Section 26.4-26.5
8.(II) Suppose the electrical force did not decrease with distance but was independent of distance.
What impact would that have on the interaction of a charged object with a neutral object?
9.(I) If electrons move easily in conductors, why aren’t pieces of metal usually negatively charged
on the bottom and positively charged on top, due to gravitational settling of the electrons to the
bottom?
15
10.(I) Metals are often good heat conductors as well as good electrical conductors. In fact, usually
the better the metal is as a heat conductor, the better it is as an electrical conductor. Give a physical
explanation for this.
11.(I) Physicists now believe that protons and neutrons are each made of three quarks. “Up” quarks
carry a charge of +&e and “down” quarks carry a charge of – "e. How could you assemble a proton
from up and down quarks to account for its charge? How could you assemble a neutron?
12.(II) Suppose a steel ball has been negatively charged. How would you expect the excess
electrons to be distributed in the metal?
13.(I) You hold a positively charged insulating rod near the end of a neutral aluminum rod that is
lying on an insulating table. Near the other end of the metal rod is a steel ball as shown in Figure 260. Does the ball roll toward the metal rod, away from it, or stay where it is, if the ball is (a) neutral,
(b) positively charged?
+
+
Figure 26-0
Section 26.6
14.(III) By using information from the packaging of a roll of transparent tape, estimate what
fraction of atoms get an electron displaced in order to produce the effect seen in Figure 26.2. (You
can assume, for this estimate, that the material of the tape is mostly carbon and hydrogen atoms in
the ratio 1 to 2.)
15.(II) A small, charged pith ball rolls smoothly on a table toward an oppositely charged metal ball
that is held in place on the table. Which of the curves shown in Figure 26-0 most realistically reflects
the speed v of the pith ball as a function of the distance r between the balls?
A
v
B
C
Figure 26-0
r
16.(I) (a) How far apart do two objects with 1 C of charge have to be separated before the force on
each object is 1 N? (b) How much charge should be placed on each of two objects separated by 1 m
so that the force on each is 1 N?
17.(II) Assume (not quite correctly, as we shall see in a later chapter) that the electron in a hydrogen
atom orbits the proton in a circle of radius 53 pm. If the force responsible for their attraction were
16
gravitational, what would be the orbital period of the electron in seconds? What is the orbital period,
using the electrostatic attraction instead?
18.(II) You give two crunched-up aluminum foil balls identical charges q. You then stick them on
wooden stakes at an arm’s length distance apart. The force on either ball has magnitude F. You have
a third aluminum-foil ball on a wooden stake and with charge –2q, and you touch it first to one foil
ball, then to the other, and then take it away. (a) What is the magnitude of the force on either of the
original balls (in terms of F) after you’ve done this. (b) Has the direction of the force changed or
stayed the same?
19.(I) In Figure 26-0, the charge on the left-hand ball is +2.1 µC, and the charge on the right-hand
ball is –6.3 µC, and they are separated by 0.11 m, center to center. a) Draw and label the forces on
each of the charged balls. b) Now redraw the forces to the same scale if the charge on the left-hand
ball is changed to –6.3 µC.
+2.1 'C
-6.3 'C
Figure 26-0
20.(I) You want to give two objects positive charges such that one has 50% more charge. If you
have a large supply of identical steel marbles and a rubber rod and a rabbit fur, what procedure can
you use to get these charges?
21.(I) Sphere A carries 6 nC of charge. It is placed 100 mm from a sphere B that carries 3 nC of
charge. Assume the spheres are tiny compared to the separation. (a) Draw a figure showing the
charges and the electric force vectors on each. (b) What is the magnitude of the electric force on A?
!
(c) Draw and label rAB on your diagram.
22.(I) You have a 3-g copper penny. (a) About how many electrons are in it? (b) If somehow you
could isolate just the electrons, how much charge would you have? (c) Estimate the force it would
require to bring one more electron within 1 nm of the rest of the electrons by assuming they are just
one small charged object. (d) Based on the size of this force, do you think it would be possible to
isolate this negative charge?
23.(I) Two objects have the same magnitude of charge, and the magnitude of force exerted by one
on the other is 0.1 N when they are 30 mm apart. (a) What is the magnitude of charge on the two
objects, assuming they are much smaller than 30 mm in diameter? (b) If the magnitude of the force
increases when the charged objects are released, are the objects of the same charge or opposite
charge?
24.(II) The Earth exerts an electrostatic force on small charges near its surface. The effect can be
modeled by assuming there is a negatively charged object at the center of the Earth with a charge
−6.76×105 C. (a) What is the electric force of the Earth on an electron at its surface? (b) How does
this force compare to the gravitational force of the Earth on the same electron? (c) How much charge
would you have to put on a penny for the electric repulsion of the Earth on the penny to cancel the
attractive force of the Earth's gravity on the penny? (d) How many electrons would you have to add
or remove from the penny to achieve this?
17
25.(II) You have to do work to bring a charged balloon toward a negatively charged sphere. What
is the sign of charge on the balloon?
26.(II) Assume that the positively charged red marbles in checkpoint 26.10 have a charge that is
twice the magnitude of that of the negatively charged blue marbles. The red marbles have a charge
of +1 µC. (a) What is the magnitude of the force between the marbles if they are 10 cm apart? (b) If
the marbles are released, what will happen? (c) If the marbles share charge when they come in
contact, what is the force between the marbles when they are again moved 10 cm apart, assuming no
charge is transferred between the two-marble system and their surroundings?
27.(I) A small particle carrying a -4.0 µC charge is located at the origin of an xy coordinate system.
What is electric force on a +1.0 µC charge located at (a) x = 10 m? (b) y = -6 m?
28.(II) A small particle carrying a 6.0 µC charge is located at the origin of an xy coordinate system.
Another small particle carrying a 4.0 µC charge is located along the y-axis at y=3 m. (a) What is the
magnitude and direction of the electric force on the particle carrying the 4.0 µC charge? (b) What is
the magnitude and direction of the electric force on the particle carrying the 6.0 µC charge?
29.(II) Two small charged particles separated by a distance of 3 m are found to exert an attractive
force of 8.0×10-3 N on each other. If the total charge on the two particles is 6µC, what is the charge
on each particle?
30.(II) Prove that the maximum force between two charged particles with a total charge of Q shared
between the particles is at a maximum when each particle has a charge ½Q.
31.(II) Find the electric force (a vector!) on a 20-nC charged particle located at (2m, 2m) due to a
25-nC charged particle located at the origin in a Cartesian coordinate system. Draw a complete
diagram to illustrate the various quantities in your calculation.
32.(II) The electric force between two identical positively charged ions is observed to be 3.7×10-9 N
when they are 0.5 nm apart. How many electrons are missing from each of the original atoms?
33.(II) Two small particles, each carrying 71 pC of charge, are released from rest on a frictionless,
nonconducting surface. One particle accelerates initially at 7 m/s2. The 0.49-mg particle accelerates
initially at 9 m/s2. (a)What is the inertia of the first particle? (b) What is the separation between the
particles when they are first released?
34.(II) Two identical conducting pith balls are suspended by very light strings from a common
point. One of the pith balls is given a charge q and both are constrained in their motion only by the
tension in the strings, gravity and air resistance. (a) Describe the behavior of the pith balls from the
instant the first one is charged. (b) Find q in terms of m, the inertia of the pith balls, l, the length of
the string, d, the final distance between the pith balls, and any necessary constants.
35.(II) The average orbital distance of electron from the nucleus in singly ionized helium is
2.65×10-2 nm. What is the kinetic energy of the electron?
18
Section 26.7
36.(III) Someone has challenged you in a friendly bet that, given a box of 6 charged balls, three
positive and three negative, you can’t place any three of them in a line (a board with a groove in it
serves in this case) and have them be in equilibrium. You find two positive balls, and by comparing
the force on the third positive ball, you rank them and place them as shown in Figure 26-0 Now the
trick is where (to the left of the pair, between the pair, to the right of the pair) to put the third ball,
and what sign charge on the third ball to choose. After a moment’s thought, you realize there is only
one solution.
smaller q
larger q
Figure 26-0
37.(II) Two negatively charged particles, each of charge –Q, are placed at opposite corners of a
square, while a third positively charged particle with charge Q is placed at a third corner, as shown
in Figure 26-0 (a) Draw the force exerted on a fourth charged particle with charge Q as it is placed at
each of the positions A, B, C indicated. (b) Sketch the trajectory of this particle if released at each
of these three positions.
-Q
Q
A
C
B
-Q
Figure 26-0
38.(I) On an x-y grid, a particle with positive charge q is placed at the origin, a charge carrier with –
2q is placed at (1,0), and a charge carrier with +3q is placed at (0,1). What is the angle with respect
to the x-axis of the force exerted on a charge carrier with charge +q placed at (2,0)? (Hint: simple
geometry and ratios should be enough to solve this problem.)
39.(I) Cesium chloride (CsCl) is a crystalline salt that forms in a cubic lattice structure, which you
can imagine as a cube with Cs+ ions at the corners and Cl– ion at the center. The edge of the cube is
412 pm. Suppose that at the edge of a crystal, two cesium atoms have been stripped from adjacent
corners of the cube, as shown in Figure 26-0. What is the net force on the chloride ion in the center,
due to the other atoms in this cubic cell?
Cs+
Cs+
412 pm
Cl-
Figure 26-0
40.(I) A very small sphere carrying a charge of –5 µC is located at (4 m, -2 m, 0). Another very
19
small sphere carrying a charge of 12 µC is located at (1 m, 2 m ,0). A third electron is located at (-1
m, 0, 0). (a) Draw a diagram showing the vectors you need to determine the direction of the forces it
due to the individual charged spheres. (b) Find the net electric force on the third electron.
41.(II) A very small sphere carrying a charge of 5 µC is located at (1m, 3m, 0). Another very small
sphere carrying a charge of –4 µC is located at (2 m, -2 m ,0). A third electron is located at (-3 m, 1
m, 0). (a) Draw a diagram showing the vectors you need to determine the direction of the forces it
due to the individual charged spheres. (b) Find the net electric force on the third electron.
42.(II) Two charged particles are located along the x-axis. One has a charge of +2 nC and is at x=-3
cm, while the other has a charge of the –2 nC and is at x=3 cm. What is the force on a particle with
charge +5 µC that is now put on the y-axis 8 cm from the origin?
43.(II) A particle carrying +6.0 µC of charge is located on the x-axis at x=+3.0 m. An identically
charged particle is located on the x-axis at x=-3.0 m. A particle carrying -4.0 µC of charge is at the
origin. What is the force on the particle at x=-3.0 m?
44.(II) Four very small charged particles are located at the corners of a square, 5 cm on a side. They
all have a magnitude of 3 nC, with positive charges on the lower left corner, and negative charges on
the other corners. You wish to find the force on the particle in the upper right corner. (a) Draw a
diagram showing the charged particles and the unit vectors needed to solve the problem. (b) Find the
force.
45.(II) Two particles carrying the same charge of 6.0 nC are located along the x-axis, at x=-3 cm,
and at x=3 cm. Where along the y-axis is a +2.0 nC charge if it experiences an electric force of
6.9×10-5 N ŷ ?
46.(II) Three charged particles are arranged along a line. The one in the middle has charge 2.0 nC.
The second particle is 1 cm to the left with a charge –4.0 nC. The third charged particle is 3 cm to
the right of the middle particle. If the net electrostatic force on the middle particle is 0, what is the
charge on the particle on the right?
47.(II) Consider the arrangement of charged particles in Figure 26.35. (a) Draw the direction of the
net force on particle 7 if all of the even numbered particles have a charge of -2Q and all of the odd
numbered particles have a charge of +Q. (b) Sketch the trajectory of particle 7 if it is released while
the other particles remain fixed. (c) Describe the subsequent motion of particle 7.
48.(II) Two charged particles are on the y-axis. One particle with charge 4Q is located at the origin
and another particle with charge Q is located at y=0.12 m. A third particle carrying a charge of 2.0
µC can move along the y-axis. It experiences a net force of zero when at y=0.08 m and a net force
of 126.4 N ŷ when at y=0.04 m. What is the value of Q?
49.(II) A small metal sphere is hung from the ceiling by a long light string. It is given a charge. An
identical sphere is suspended nearby with an identical string. The sphere is given a charge of equal
magnitude but opposite sign as the first. Although they are attracted to each other, they are 2 cm
apart when in equilibrium. What happens to the distance between the spheres if a third metal sphere
carrying no net charge is placed between them? Explain your reasoning.
20
50.(II) Conducting spheres are placed at three corners of a square. The two at opposing corners are
oppositely charged, with the negatively charged sphere having twice the magnitude of charge as the
positively charged sphere. The third sphere is neutral. (a) Draw the three spheres, showing the
approximate location of charge on the spheres as in Figure 26.31. (b) Indicate all of the force vectors
on the positive sphere due to electrostatic interactions.
51.(I) If you were able to isolate two identical charged particles of inertia m from all external forces,
what would the charge of the particles have to be in terms of m and any necessary physical constants
if the two particles remained at some fixed separation?
52.(II) Four identical charged particles are constrained along the x-axis. Identify one possible
configuration of the particles that would leave one charge at rest at the origin, if the others were
fixed in place. None of the charges can be at the same location.
53.(II) Four charged particles of magnitude Q are to be located at the corners of a square with sides
of length L. Two of the particles have a negative charge, the other two are positive. (a) Draw an
arrangement of charges that would cause zero net force on a fifth particle of charge +Q placed at the
center of the square. (b) If the fifth particle were instead placed at the midpoint of one of the
square’s sides in this configuration of the charges on the square’s corners, what would be the force
acting on it?
54.(III) A small sphere with charge Q is fixed a few of meters above the moon's surface next to a
spaceship. An astronaut at the ship's hatch holds another charged particle a couple of meters above
it, almost but not quite over the sphere, and drops it. (a) If the dropped particle also has charge Q,
draw a plausible trajectory for the dropped particle. (b) If the dropped particle had a charge of 9Q
instead, how would the trajectory change? (c) If dropped particle had a charge of –Q, what could
the trajectory look like?
55.(I) Charged ping-pong balls A and B repel each other at a distance r1 from each other with a
force of 0.4 N. Charged ping-pong balls A and C repel each other at a distance r2 from each other
with a force of 1.4 N. Now make an arrangement with all three balls. You can move B and C to any
location around A you wish, just as long as the distances r1 and r2 remain constant. What are the
maximum and minimum magnitudes of the force on A that you can produce?
21
Answers to review questions
1. The gravitation interaction between the balloon and the earth makes the balloon fall. After
charging, the electrostatic interaction between your hair and the balloon dominates. (Friction
between your hair and the balloon is the force that opposes the gravitational interaction between
earth and balloon.)
2. The electrostatic interaction is much stronger than the gravitational interaction. There is more
than one kind of electrostatic charge, but there is only one kind of gravitational mass. Related to this
last point, objects can be electrically neutral, but nearly every object has nonzero gravitational mass.
3. The same charged object can interact differently with two different objects, namely repel or
attract. This means that the charge on the other objects must be of different sorts.
4. Without further information, it’s impossible to say which is right. In fact, both might be right.
Positive charge moved from the first to the second, or negative charge from the second to the first,
would have the same effect.
5. Zero. Charge can’t be created or destroyed, only displaced from one place to another (or, as we
shall see in a later chapter, created or destroyed in equal positive and negative portions). Charge is
conserved; that is, in a closed system, the total amount of charge is constant.
6. Your skin is a fair conductor, so that a net charge deposited on your feet can migrate to your
finger.
7. Paper is an insulator; try inserting a slip of paper between two batteries in a flashlight and turning
on the switch. A paper clip is metal and a conductor. Seawater is a conductor, due to the mobile ions
(from salt) dissolved. Tires are rubber and an insulator, just as the rubber insulation on electrical
power cords are. Air is usually an insulator, but like most insulators, a sufficient electrostatic force
can cause a breakdown, or spark, and then charge flow can occur.
8. Breaking wood leaves two pieces of similar material. There’s no reason why the symmetric result
should leave a net positive charge on one piece and a net negative charge on the other. It is usually
in the breaking of bonds in dissimilar materials that charging occurs.
9. There would be no difference at all in the interactions studied so far. Opposite charges would still
be opposite and attract, and like charges would still be like and repel. In fact, the assignment of
positive and negative charges is generally credited to Benjamin Franklin, who made an arbitrary
assignment based on, it turns out, an erroneous guess about charge mobility.
10. a) positively charged or neutral; b) negatively charged only.
11. Suppose A interacts electrostatically with B. The force exerted on B is proportional to the
charge on B, experimentally. Likewise, there is a force on A that must be proportional to the charge
on A. Since the forces in the interaction must be equal in magnitude, they must both be proportional
to both the charges on A and B. The only way to make this happen is for the product of the charges
(qA· qB) to appear in the expression for the force. Note that the combination (qA + qB) does not
reproduce the proportionality behavior; doubling qB does not double the sum.
12. Doubling both charges without changing the distance would increase the force on each by a
factor of four. Doubling the distance would compensate for this factor of four, because the force is
inversely proportional to the square of the distance.
13. The force exerted on either partner in the interaction pair is identical, as is always the case with
any interaction.
14. The last term, though it might seem natural at first, does not involve charge carrier 1 at all, and
therefore has nothing to do with the force acting on that charge carrier. This term should be omitted.
22
Chapter 27
Electric field
Review questions
Answers to these questions can be found at the end of this chapter.
Section 27.1
1. Hold one end of a Slinky™ spring and have a friend hold the other. Then both of you close your
eyes. Jiggle your end and find out how long before your friend can detect what you’ve done. How is
this analogous to the transmission of the electrostatic force by a field?
2. Two charged ping-pong balls, A and B, are held at a small distance from each other. Which ball
is the source of the electric field that B feels?
3. A classmate draws a map like Figure 27-0 to represent the electric field around a balloon that has
been rubbed on the carpet, as indicated by the black central dot. What is wrong with her picture?
Figure 27-0
Section 27.2
4. The units of the gravitational field are those of acceleration. Is that true of the electric field as
well? If not, why not?
5. A charged styrofoam pellet is placed in an electric field due to some other charged object. If the
charge on the pellet were reversed in sign, what would happen to the magnitude and direction of that
external electric field at the pellet’s location?
6. Imagine a helicopter hovering over a farmer’s wheat field. The stalks of wheat lie flat as they are
blown, more severely bent nearer the spot directly under the helicopter. Does this more closely
mimic the electric field around a positively charged object or a negatively charged object? With
what would you replace the helicopter to mimic the field around an oppositely charged object?
7. Suppose near the surface of the earth there is an electric field pointing horizontally toward the
east. Does it make sense to add the gravitation and electric fields to determine the behavior of a
proton there?
8. Does it matter if a small amount of charge or a large amount of charge is used on a test object to
1
measure the electric field at a certain place due to a charged, insulating sphere? What if the field
being measured is due to a charged conducting sphere?
9. An electron is initially moving horizontally near the earth’s surface. It enters a region of uniform
electric field and is deflected upward. What can you say about the direction of the electric field in
that area (assuming no other interaction)? What if it were deflected downward instead? (Remember
the old academic adage: Think before you leap!)
Section 27.3
10. Two balloons are charged and taped to a piece of cardboard. A pushpin is put in a third place
on the cardboard. The electric field due to one balloon is measured at the pushpin to be 300 N/C (by
temporarily removing the second balloon), and the electric field due to the second balloon is
measured in a similar way to be 200 N/C at the pushpin. Is it fair to say that the total electric field at
the pushpin is 500 N/C? Why or why not?
11. What would change in Figure 27.12 if both the charged particles were negatively charged
instead? If one were positive and the other negative?
Section 27.4
12. Write down units for electric field in terms of energy (J) and other SI base units.
13. To what extent is the gravitational field uniform inside the room in which you are sitting?
(Neglect the contributions to the gravitational field due to the structure of the building and the
objects in the room.)
14. Is it possible for a dipole initially moving in a straight line to be deflected by a uniform electric
field?
15. An electron is traveling in uniform circular motion due to the effect of an electrostatic field. Is
it possible for the electric field to be uniform?
Section 27.5
16. If the charge on particle 2 in Figure 27.21 were doubled, what aspects of the electric field at
point P would change: the magnitude, the direction, or both?
17. A delicate instrument is two paces away from a highly charged metallic sphere. If you wanted to
lower the electric field at the instrument to 1% of its present value, how many paces would you have
to move the instrument away from its present position?
Section 27.6
18. Which would be larger, the field at a distance r away from a very small object with charge Q, or
the field at a distance r away from a small charged dipole of dipole moment Qd (where r >> d)?
19. Does the magnitude of dipole's electric field remain constant at a given distance r away from it
as its orientation is changed? If not, what is the ratio of the greatest to least magnitudes?
Section 27.7
20. A thin, half-spherical insulating shell of radius R has a charge Q distributed uniformly over it. It
2
is placed so that is entirely in the z!0 half-space with its base resting on the xy plane and centered
about the z axis. Write down the charge on an infinitesimal annular part of the shell of width d!,
where ! is the angle from the z axis to the annular ring.
21. When is it necessary to use an integral to calculate an electric field? Could you have solved the
problem shown in Figure 27.25 by dividing the ring into, say, a hundred small pieces and calculating
a finite sum of their contributions? Could you do the same in the case shown in Figure 27.26?
22. What is the distinguishing radial dependence of the electric field around a small charged object
(extensive in no dimensions), around a long charged wired (extensive in one dimension), and around
a large charged sheet (extensive in two dimensions)?
Section 27.8
23. What are the SI units of polarizability?
24. The electrostatic field due to a permanent dipole falls off as the cube of the distance d away
from the dipole. The force it causes on a small charged particle Q also falls off in the same way.
What is the asymptotic dependence with distance of the force on a permanent dipole caused by the
charged particle?
3
Developing a feel
Calculate or estimate the following quantities:
1. The electric field strength one Angstrom from a proton. (B, P)
2. The electric field strength one meter from a ball of one million electrons. (I)
3. The charge to mass ratio for a sodium (Na+) ion. (G, P)
4. The charge to mass ratio for a particle whose gravitational and electrical interactions with an
identical particle would cancel. (E, R)
5. The volume charge density of a hydrogen nucleus. (F, K, P).
6. The surface charge density on a freshly- pulled strip of adhesive tape. (D, H, M, Q, U)
7. The linear charge density on a rubbed glass rod. (C, H, L, Q, T)
8. The electric field strength 0.1 m from the axis of a rubbed glass rod. (number 7 above, A)
9. The magnitude and direction of the uniform electric field required to “float” a proton near the
surface of the Earth. (N, P, S)
10. The electric dipole moment of a water molecule. (J, O)
Hints:
A. Is this location close or far enough from the rod to use an approximation for E?
B. What is an Angstrom in terms of meters?
C. What is the inertia of the glass rod?
D. What is the inertia of a strip of adhesive tape?
E. If you equate the gravitational and electrical force expressions, what will cancel?
F. What is the radius of a hydrogen nucleus?
G. What is the inertia of a sodium ion?
H. What fraction of this inertia is due to protons?
I. What is the net charge on the ball?
4
J. How can we model the charge distribution?
K. What is the formula for volume of a sphere?
L. What is the total number of electrons in the glass rod?
M. What is the total number of electrons in the strip of tape?
N. What is the gravitational mass of a proton?
O. What is the separation of the centers of charge?
P. What is the charge in Coulombs?
Q. What fraction of electrons will be redistributed during the charging process?
R. What is the ratio of gravitational to electrical constants in the force expressions?
S. If the electric field points upward, in which direction will it exert a force on the proton?
T. What is the length along which the unbalanced charge is distributed?
U. What is the surface area of the strip of tape?
Key: A. No; use Equation 27.23. B. 1x10-10 m. C. ~0.2 kg. D. ~1x10-4 kg. E. The distance between
the particles cancels. F. ~1x10-15 m, the radius of a proton. G. ~4x10-26 kg. H. ~Half, since light
elements contain roughly equal numbers of protons and neutrons. I. -1.6x10-13 C. J. Each hydrogen
essentially gives up half an electron to the oxygen. The dipole consists of one net proton centered on
the H atoms and one extra electron centered on the O. K. V = 4πr3/3. L. ~6x1025 electrons or
protons. M. Same as number of protons, ~3x1022 . N. 1.67x10-27 kg. O. ~5x10-11 m, because of the
bonding angle of 105° between the H atoms. P. 1.6x10-19 C. Q. About one in 1012, according to
Chapter 26. R. G/k ~ 7x10-21 with both constants in SI units. S. Upward. T. ~0.2 m, or two-thirds
the length of the rod. U. ~0.002 m2.
5
Worked and guided problems
These examples involve material from this chapter, but are not associated with any particular
section. Typically, an example that is worked out in detail is followed immediately by an example
whose solution you should work out by following the guidelines provided.
WP 27.1: Charge-Square. Four charged particles are arranged in a square as shown in Figure 270, with q = 3.9x10-4 C and a = 6.9 mm. Find the electric field at the center of the square.
q
-2q
a
-q
2q
a
Figure 27-0
!Focus problem
The path we must take is pretty clear from the diagram and the specification of the desired quantity
“electric field”. We can use superposition of the electric fields caused by the charged particles to
find the total electric field at the center.
!Plan approach
We could simple find the electric field due to each charged particle and then add the components to
get the total field. However, we can simplify the problem a bit by exploiting the symmetry of the
configuration. It does have a lot of symmetry, but it’s not immediately obvious how to use this to our
advantage. One way is to break it into two problems, as shown in the Figure 27-0.
q
-2q
a! 2
!
E1
!
E2
a! 2
-q
2q
!
!
Notice there are now two contributions to electric field, E1 and E2 , that are easier to calculate then
four individual contributions. Each of the two fields are produced by a pair of charges of with equal
magnitudes but opposite signs, with the fields pointing form the positive charge of the pair to the
negative charge. The axes have been aligned in way that facilitates the calculation of components.
6
Because we seek the field at the center of the square, we know that the magnitudes of the two
composite fields are equal (E1=E2) because of the similarities of distance and charge. All we then
have to do is add the two fields to get the total field.
!Execute Plan
For the first half of the problem (the pair with charge ±q), the electric field is given by
!
q
(−q)
q
(−q)
q
(− xˆ ) = 4k 2 xˆ
E1 = k 2 rˆ+ q + k 2 rˆ− q = k
xˆ + k
2
2
r+ q
r− q
a
1
1
2a 2
2a 2
!
Noting that E2 is produced along the y direction by charges that are twice as large, we can infer
(
)
(
)
!
q
E2 = 8k 2 yˆ
a
Combining these together, the total field is
! ! !
q
q
E = E1 + E2 = 4k 2 xˆ + 8k 2 yˆ
a
a
The magnitude of the total electric field is
2
2
q
! 4kq " ! 8kq "
E = Ex2 + E y2 = # 2 $ + # 2 $ = 4 5k 2
a
% a & % a &
Its direction is:
8kq
2
θ = arctan
= arctan a = arctan 2
4kq
Ex
a2
Ey
Putting in numbers gives a field described as
"
E = 6.61×1011 N/C, @ 63.4° above xˆ
!Evaluate result
A large electric field, but then we know from problem 26.5 that the forces between these particles is
large. The direction seems to make sense: it points mostly in the direction of a line connecting the
two ±2q charges.
WP 27.2: Charge Triangle. Three particles form an equilateral triangle with side length d. Two of
the particles have a positive charge +Q while the third particle has a charge –2Q. What is the
electric field at the center of the triangle?
7
!Questions and suggestions
1. Draw a diagram indicating all the particles, charges, and distances.
1. What causes the electric field at the center? Account for all charged particles.
2. Is there any symmetry of which you can take advantage?
3. Will you have to deal with vectors because of the different charge signs and different directions of
the charges? We know that like charges repel and opposite charges attract. Make sure these effects
are in your diagram.
4. What are the relevant distances to the center point?
5. Does the differently signed charges have a difference in your answer? Does your answer behave
in a physically plausible way if the side length or charge magnitudes are changed?
WP 27.3: Torqued dipole. Figure 27-0 depicts a dipole located near a fixed particle A that carries
charge q. What is the torque on the dipole about its midpoint caused by the charged particle in this
configuration? Sketch a graph the torque as the separation s of the dipole increases.
+Q
q
s
y
A
-Q
x
d
Figure 27-0
!Focus problem
Thinking about the physical situation depicted the positive charge of the dipole will be repelled by
the fixed particle, while the negative charge will be attracted. This will cause a torque about the axis
connecting the dipole charges. This looks hard – and it is, particularly when we realize that the
forces at the two ends are not even parallel to each other since the electric field of particle A radiates
outward radially. However, careful application of the techniques we have developed will lead us to
the right answer.
!Plan approach
There are many ways to approach this problem. One way is the brute force way: calculate the force
acting at one end of the dipole, determine the torque caused by it about the dipole’s midpoint, do the
same for the other end, add the torques to get their vector sum. However, we can be clever. If we
think of the dipole and the charged particle as a system, then all the interactions are internal. This
means the net torque on the system is zero. If we compute the torque on particle A, then the torque
on the dipole is just the negative of it. Since the text has derived a formula for the force on a
charged particle by a dipole (Equation 27.8), it will be (relatively) easy for us to solve this problem.
8
!Execute Plan
To keep track of the dipole in our mathematics, we will give it the label D. The force due to an
electric field of the dipole on charged particle A is
"
"
FDA = q A ED
From Equation 27.8, the electric field of a dipole is
"
ED = − k
Qs
yˆ
[d + ( 12 s) 2 ]3/ 2
2
The torque on A about the midpoint of the dipole is
"
!
!
!
τ DA = rDA × FDA = (−dxˆ ) × q # − k
%
Qds
[d + 14 s 2 ]3/ 2
2
"
yˆ $
&
By remembering that xˆ × yˆ = zˆ (out of the page), we obtain the torque on the dipole:
"
"
τ AD = −τ DA = −k
qQds
zˆ
[d + 14 s 2 ]3/ 2
2
!Evaluate result
If the particle has a positive charge (q>0), our math says that the torque vector on the dipole is in the
negative z direction, i.e., into the page. This means it should want to start rotating in a clockwise
direction by the right-hand-thumb rule. This is exactly what we would expect on physical ground
because the positive charge q on particle A would cause the top of the dipole to rotate away from it
while it would pull the bottom of the dipole towards it.
The direction of the torque is as we expect, twisting the dipole in the counterclockwise direction.
Notice that the force on the dipole is up, along positive y. Does this make sense? From the equation
it should be clear that as the separation s gets large the torque decreases. Why is this?
9
Graphing the torque as a function of separation, we see that there is actually a maximum torque.
Again, can you think of why this is? Think about what happens to the moment arm as the separation
get small.
WP 27.4: Interacting dipoles. As shown in Figure 27-0, two dipoles separated from each other by
a distance d and each with a charge of q on each particle and a separation of charged particles s.
What is the force of one dipole on the other?
d
s
Figure 27-0
!Questions and suggestions
1. The text has derived the equation for the electric field along a dipole's axis. How can we use this
to determine the forces on the components of the other dipole? You should note that one object in
this example is a dipole, and the other is a system composed of two oppositely charged particle. You
need to identify clearly the dipole and particles you are talking at a given moment so you do not
double count.
2. Can you take advantage of any symmetry in the problem?
3. How can you keep track of the various relevant distances in your diagram? Students often
incorrectly substitute a symbolic quantity given in a problem (i.e., d) for the same symbol in a
derived equation, even though they do not represent the same thing.
4. Does the force in your result point in the correct direction? Think about the relative size of the
terms. What happens to the force as the separation of the dipoles becomes large compared to the
separation of the charged particles making up the dipoles?
WP 27.5: Field of a straight charged rod. A rod of length L has a uniform charge per unit length
" distributed along it. What is the electric field at a distance d from the end of the rod along its axis?
!Focus problem
First, we make a diagram that displays the given information so that we begin thinking about the
problem.
d
L
10
This problem is similar to examples in the text for a continuous. However, in this case we do not
know the total charge on the rod, only the charge per unit length. Nonetheless, we should be able to
use similar techniques to find the electric field.
!Plan approach
We can find the total electric field for a continuous distribution by dividing up the rod into little
segments and "adding" up (integrating) the contributions to the electric field from all these
differential element:
"
"
dq
E = ' dE = ' k 2 rˆP
r
where dq is the little bit of charge on the segment and r is the distance to the point. We have to draw
a diagram to keep track of the contributions from each segment. From symmetry, we know the
electric field will point along the rod's axis.
!Execute Plan
First we draw a diagram that shows how we select a segment and all the elements needed for our
calculation. For convenience, we choose the x axis to be along the rod's axis. If we choose the x
axis to be pointed to the right, rˆP = xˆ .
dx
0
P
x
!
dE
x
d
r
We should label the point in which we are interested as P so that we can refer to it if necessary. In
!
ˆ
is the electric field due to a small segment of the rod dx. We consider a small
this case, dE = xdE
segment of the rod of length dx centered a position x. Remembering that the coordinate x represents
a signed quantity (the position), we see the distance r to point P is just (d – x) in our coordinate
system with the origin at the rod’s right end. The charge dq on the segment is just dq = " dx. Putting
the pieces together, the electric field at point P is:
0
!
E = xˆ ' k
−L
λ dx
(d − x)
2
=k
λ
0
(λ
λ )
1 )
(1
xˆ = k * −
xˆ = k λ * −
xˆ
+
d − x −L
, d d + L +, d d − (− L) -
!Evaluate result
11
We first want to make sure our expression gives the correct direction for the electric field. If the
charge is positive, the electric field at P should point to the right since that is the direction a small
positive test charge would move. This is in the positive x direction for our chosen coordinate
system. For positive ", the factor in front of x̂ is indeed positive, meaning the component of the
electric field in the x direction is positive as expected. If the rod gets very long compared to distance
d, then we can assume L is effectively infinite. As L approaches infinity the first term drops out and
the electric field magnitude becomes E=k"/d, that is, the field approaches a constant value. Notice
that as we approach the end of the rod (d goes to zero) the electric field becomes infinite. At the end
of the rod you are presumably right on top of a charged particle (zero separation) and the electric
field should become infinite because of the 1/r2 dependence of Coulomb’s law. Note also that as L
gets really small compared to d (a very short rod), the electric field becomes zero – which is what we
would expect since there is no charge on a non-existent rod!
WP 27.6: Field of a semi-circular charged rod. A thin semi-circular rod of radius R has a total
charge Q uniformly distributed along its length. What is the electric field at the center of the full
circle?
!Questions and suggestions
1. Draw a sketch with the rod as a half circle with the given quantities placed appropriately.
2. How do you account for the effects of a continuous charge distribution with elements in different
directions from point of interest?
3. How are you going to account for the contributions to the electric field from different portions of
the rod? What is the most natural variable to do the integration in this circular arc problem? How
can an angular measure be related to the small segments of charges on the rod?
4. What simplifications, if any, does the symmetry of the circle allow you to make? Make sure you
take the right component of the vectors at the center.
5. In your result, is the direction of electric field plausible? Does your expression behave in a
physically plausible way if you vary the radius of the semicircle or the magnitude of the charge?
WP 27.7: Dipole corrections. The electric field of a dipole asymptotically falls off as the inverse
cube of distance in any given direction. For a dipole with moment p and separation of charges s,
what is the first level of correction for distances somewhat closer, i.e., not too near, not too far, (a)
along the dipole’s axis, and (b) at a point located perpendicular to the dipole’s axis.
!Focus problem
Figure 27.22 gives a very clear diagram of the situation, so we start by just repeating it and adding
notation for the net field at along the axis (E||) and at a point perpendicular to the axis.
12
The text derives the exact expressions for the electric field along the dipole's axis and perpendicular
to it, and shows how to make an approximation in these expressions to obtain the long-range
variation of the electric field with distance. We just have to make the approximation a little better.
We have to consider that the two corrections may be slightly different because the asymptotic
expressions differ in the two directions. But if they do, at least we can have a good idea of the
corrections.
!Plan approach
Equation 27.12 gives the non-zero component of the electric field for any distance from the midpoint
of the dipole along the axis of the dipole. To get the asymptotic (long-range) behavior of a dipole’s
electric field for far distances along its axis, the distance variable in the exact expression for the
electric field (Equation 27.12) was allowed to become sufficiently large to overwhelm all other
terms. The same was thing was done to get the asymptotic field for large distance from the dipole in
a perpendicular bisecting plane. These asymptotic forms proportional to the inverse distance cubed
can be discerned using the first-order term in a binomial expansion. We probably just have to go to
the next order in the expansion. The relevant expansion approximation is (1+z)n = 1 + nz + n(n–
1)z2/2 + n(n-1)(n-2)z3/6 + order(z4). We hope to get by with just the terms through z2 since the longrange expressions just used 1+nz, but it is often good to go to one higher term in case there is an
exact cancellation.
!Execute Plan
Consider first the net electric field along (parallel to) its axis, given by Equation 27.12:
−2
−2
(!
!
s "
s " )
*# 1 −
( E|| ) y
$ − #1 +
$ +
*,% 2 y &
% 2 y & +Noting that n=-2 in this case and using the expansion of (1+z)n to cubic power with z=s/2y:
Q
=k 2
y
13
−2
2
3
!
s "
s (−2)(−2 − 1) ! s " (−2)(−3)(−4) ! s "
s 3 s 2 1 s3
1
±
1
±
(
−
2)
+
±
=
1
+
#
$
$
#
$
# $
# $
2y
2
6
y 4 y 2 2 y3
% 2y &
% 2y &
% 2y &
Using this in the expression for E|| and letting the distance d=y, we find the first order correction for
intermediate distances is:
(E )
|| y
#k
Q (!
s 3 s 2 1 s3 " !
s 3 s 2 1 s3 ")
Q ! s s3 "
Qs !
s2 "
1
1
2
2
1
k
k
+
+
+
−
−
+
−
=
+
=
+
*#
$ #
$+
#
$
#
$
d 2 ,% d 4 d 2 2 d 3 & % d 4 d 2 2 d 3 & d2 % d d3 &
d 3 % 2d 2 &
This is the similar to the expression at large distances (Equation 27.13), except that we have the
additional term [1+s2/d2]. It is good that we kept terms through z3 = (s/d)3 because the z2 term
cancelled in the difference.
We have to do a similar task for intermediate distances from the dipole in directions perpendicular to
the dipole’s axis. Starting with the exact expression (Equation 27.8) for the net electric field in this
direction
( E⊥ ) y = −k
(x
Qs
2
+ ( 12 s ) 2 )
3/ 2
Qs !
s2 "
= −k 3 # 1 + 14 2 $
x %
x &
−3/ 2
Using distance d=x and doing the same type of approximation, we find that the first level of
correction for this direction is:
( E⊥ ) y # −k
2
)
Qs (
Qs ! 3 s 2 "
3 1 s
+
−
=
−
1
(
)
k
#1 −
$
2 4
d 3 *,
d 2 +d3 % 8 d2 &
We see that the correction to the 1/d3 asymptotic behavior along the dipole’s axis is [1+(s/d)2/2]
while the correction in perpendicular bisecting plane is [1 – 3(s/d)2/8]. It is gratifying to note that in
both cases the correction term is on the order of (s/d)2. We expect that these two cases should
bracket the actual correction for arbitrary directions at intermediate distances.
!Evaluate result
If we let d get very large compared to s in either of our expressions (i.e., we are looking very far
from the dipole) then s/d#0. We see that our expression for (E||)y takes on the correct asymptotic
form obtained in the Principles Volume (Equation 27.13) and our expression for (E⊥)y takes on the
correct asymptotic form of Equation 27.10. The plus sign in front of the (s/d)2 correction for (E||)y is
plausible because, as we approach the dipole with a positive test charge from the positive charge’s
end, the repulsion of the dipole’s positive end will increase more rapidly with closing distance than
the attraction of the negative end.
14
WP 27.8: Advertising imaging. A cathode ray tube, the heart of all television picture tubes, uses
an electric field to steer electrons to the proper place on the television screen. Upon hitting the
screen it lights up a phosphor pixel, creating the image. A sketch of the steering mechanism is
shown in the figure. As shown in Figure 27-0, an electron is fired from an “electron gun” at the left.
Two parallel plates create a uniform filed in the y direction that bends the trajectory of the electron.
By varying the field strength the electron can be directed to impact anywhere vertically from top to
bottom. A separate set of plates controls the horizontal position. For the cathode ray tube depicted,
the plates are 3 cm long, located 20 cm behind the screen. The height of the screen is 30 cm. What
electric field is required between the plates to divert the electron to the bottom of the screen, given
that the initial horizontal velocity of the electron from the gun is 3x105 m/s?
screen
vertical deflection
plates
v
• • •
•
electron gun
L= 3 cm
E
h=30 cm
•
•
•
y
•
0
x
D=20 cm
Figure 27-0
!Questions and suggestions
1. Describe the physical situation and question asked in your own words.
2. What physical interactions control the trajectory of the electron? It looks like the electric field
and possibly gravity are the main operators here. What simplifications and assumptions can or must
you make?
3. What happens to the electron as it traces its path from gun to screen and why?
4. How can you determine the trajectory of the electron? You may remember that you have done
trajectory calculations before with gravity as the dominant force. Is the total force on the electron
constant? Is that decision affected by whether the electron is between the plates or in the region
from the plates to the screen? If not, the trajectory must still be continuous at the boundary between
the regions.
15
Questions & problems
Section 27.1-27.2
1.(I) Two charged pellets, A and B, are held some distance from each other as shown in Figure 270.
A
B
Figure 27-0
The charge on A is then doubled. Which of the following statements is correct?
i) The force on A is doubled because the electric field at the position of A is doubled.
ii) The force on A is doubled because the electric field at the position of B is doubled.
iii) The force on B is doubled because the electric field on the position of B is doubled.
2.(II) Sketch the gravitational field at about fifteen points uniformly distributed along the orbit of a
comet shown in Figure 27-0.
Figure 27-0
3.(III) You are given the topographical map, a section of which is shown in Figure 27-0. Every
curve represents a fixed elevation. You are trying to draw what your guide calls a “gradient map”,
that is, a field of vectors that show the steepness of the hill and the direction of steepest ascent at
representative places. He has marked one for you at point A, but wants you to continue over every
grid intersection on the section.
Insert a topological map here like Fig 23-15 in Giancoli, PSE, 3rd ed.
Figure 27-0
4.(I) Two charged particles, A and B, have equal and opposite charge, Q and –Q, respectively as
shown in Figure 27-0. They are held some distance from each other as shown.
Q
A
B
-Q
Figure 27-0
(a) Draw the electric field at the centers of A and B and the electrostatic forces acting on each pellet.
(b) The charge on B is now doubled. Redraw the electric field at the centers of A and B and the
electric forces acting on each pellet.
5.(II) (a) Plot the value of the radial component of the electric field due to a positively charged
16
pellet as a function of distance from the center of the pellet. (b) Repeat this for a negatively charged
pellet.
Section 27.3
6.(I) The two ping-pong balls shown in Figure 27-0 have the same magnitude of charge but with
opposite signs. Sketch the electric field they produce at the points A, B, C, D and E which are all in
same plane that includes the line joining the centers of the balls. Use a ruler to estimate the
proportions of the distances.
A
B
–
C
+
D
E
Figure 27-0
7.(II) Repeat the previous problem, but with the negative charge doubled, i.e., twice the magnitude
of the positive charge.
8.(I) Sketch the electric field caused by the two particles shown in Figure 27-0 at the points A, B
and C in the plane midway between the two particles and perpendicular to the line connecting them.
+q
A
B
C
+q
Figure 27-0
9.(I) What is the direction of the electric field at the center of the semicircle arc of the uniformly
positively charged plastic thread shown in Figure 27-0?
Figure 27-0
10.(III) The charge on a non-conducting rod increases linearly from end A to end B, and then the
rod is bent in a circle so that ends A and B meet at the top of the circle as in Figure 27-0. What is the
direction of the electric field at the center of this circle? (Hint: Consider contributions from
diametrically opposite segments!)
17
B A
Figure 27-0
11.(II) A bead with charge +Q and another with charge +4Q are separated by distance D. Is there a
point between them where the electric field is zero? If so, where is it? Is there any other point where
the electric field is zero?
12.(II) Charged beads are placed at the corners of a square in the various configurations shown in
Figure 27-0. Red beads have charge +Q, while green bead has charge –Q. Rank the configurations
according to the magnitude of the electric field at the center of the square, least to greatest.
(a)
(b)
(c)
(d)
(e)
Figure 27-0
Section 27.4
13.(II) Assume a positive test charge is put at point P1 in Figure 27.11(c). Is the equilibrium there
stable or unstable? What if the test charge were negative?
14.(II) A box full of charged plastic balls is sitting on a table. The force on a ball at the top,
northeast corner of the box has components 1.2 x 10–3 N to the north, 5.7 x 10–4 N to the east, and
2.2 x 10–4 N vertically upward. The charge on this ball is 120 nC. If this ball were replaced with one
with a charge of –50 nC, what would be the force components on the replacement ball?
15.(II) An electron traveling horizontally east passes between two oppositely charged plates and is
deflected downward. Passing through the same space between the plates, in which direction (if any)
would each of the following be deflected?
a) a proton traveling horizontally east
b) an electron traveling horizontally west
c) a proton traveling horizontally west
d) a proton traveling horizontally north
16.(II) Two dipoles are free to rotate about fixed axes through their centers. They are initially
arranged in line as shown in Figure 27-0. The magnitudes of the charges and masses on the two
dipoles are q, and 3q, and 3m and m, respectively left and right. The red circles indicate positive
charge, and the green circles indicate negative charge. After they are released, what will be the most
likely final orientation of the dipoles if their kinetic energy were dissipated in some way?
18
q, 3m
–q, 3m
–3q, m
3q, m
Figure 27-0
!
17.(II) An electron is launched into a region of constant electric field, E =2×104 N/C ŷ , with an
initial speed of 2.1×107 m/s x̂ . How far has the electron traveled in the y direction by the time it has
moved 4 cm in the x direction?
Section 27.5
18.(I) What is the magnitude of the electric field 20 cm away from a particle with charge 3.0 m·C?
19.(I) You know that if place an electron 0.1 m to the right of a small charged ball, the electron will
accelerate to the right with an initial acceleration of 4.0 m/s2. What is the charge on the ball?
(Neglect gravity in your calculations.)
20.(III) You’ve been given the task of charging up a spherical weather balloon made of aluminized
Mylar™ (a conducting material), and you want to put as much charge on it as possible. An
experienced colleague advises you that the air surrounding the balloon will break down when the
electric field at the surface of the balloon gets to 100,000 N/C, causing a spark that discharges the
balloon. By measuring the balloon’s shadow, you calculate that the diameter of the balloon is 3.5 m.
This gives you enough information to estimate the maximum charge on the balloon.
21.(II) Consider a rectangle with diagonal length 2a in the y-z plane with the origin at the center of
the rectangle. A bead with charge Q is at each corner. Show that for any such rectangle the electric
field along the x-axis is given by
!%
E=k
(x
4Qx
2
+ a2
)
32
ˆx
22.(II) Table salt (NaCl) forms in a cubic crystal structure. A cube in this lattice has sodium (Na+)
and chlorine (Cl–) ions in alternating corners. Na+ ions have a 99 pm radius, and Cl– ions have a 181
pm radius, and in this cubic array the ions are spaced as though they were solid balls of these sizes,
touching their neighbors. For a single cube of this arrangement, what is the electric field and
electrostatic force on a sodium ion in one of the corners, due to the other ions?
23.(II) In an inkjet printer, tiny droplets of ink of approximately uniform inertia m are given a
charge Q in a charging unit. They are fired toward the paper at speed V. They first pass between two
charged plates of length L that create a uniform electric field of magnitude E between them,
perpendicular to the flight path as shown in Figure 27-0. This deflects the droplet from a straight
line and is used to control the formation of characters on the paper. a) Derive an expression for the
deflection distance the drop experiences in passing between the plates. b) For an ink droplet inertia
of 1.5 × 10–10 kg, an electric field strength of 1.2 × 106 N/C, a plate length of 1.0 cm, and a droplet
speed of 20 m/s, calculate the droplet charge necessary to result in a deflection of 1.3 mm. (c) Is it
acceptable to ignore the effects of gravity in this case?
19
L
Figure 27-0
24.(II) (a) What electric field would you have to have to balance the Earth's gravitational force on an
electron? (b) Where would you have to place a proton with respect to the electron to cause this
force?
25.(I) A small sphere carrying 6 nC of charge is placed 10 cm from another small sphere carrying 3
nC of charge. Assume the spheres are tiny compared to the separation. At what point on the line
joining the two charges is the electric field zero?
26.(I) Four very small charged particles are located at the corners of a square, 5 cm on a side. They
all have a magnitude of 3nC, with positive charge on the lower left corner, and negative charges on
the other corners. Find the electric field at the center of the square. Draw all vectors (including
appropriate unit vectors) needed for your solution.
27.(II) Two objects have the same magnitude of charge, and the magnitude of force exerted by one
on the other is 0.1 N when they are 5 cm apart. (a) What is the magnitude of charge on the two
objects, assuming they are much smaller than 5 cm in diameter? (b) What is the magnitude of the
electric field at a point 5 cm directly above one of the charges, if the other charge is located to its
left? (c) Explain whether or not your answer would change if the spheres each had a radius of 1 cm.
28.(II) A small particle carrying a 6.0 µC charge is located at the origin. Another small particle
carrying a 4.0 µC charge is located along the y-axis at y=5 m. (a) What is the magnitude and
direction of the electric field at x=5 m? (b) What is the magnitude and direction of the electric field
at x=-5 m? (c) What is the magnitude and direction of the electric field at y=-5 m?
29.(II) Three particles with equal positive charges are located at the corners of an equilateral
triangle with side length a. (a) What is the magnitude and direction of the electric field at the center
of the triangle? (b) What is the magnitude and direction of the electric field at the midpoint of one of
the sides of the triangle? (c) What is the magnitude and direction of the electric field at a point a
distance a above the apex of the triangle?
30.(II) Replace the negative charge in figure 27.10b with a particle carrying -2q. Assume q=1.0 µC,
and that the scale of the drawing is that the separation between the two charges is 1.5 m. Find the
electric field at points (a) P5, (b) P6, (c) P7 and (d) P8.
31.(II) A small particle carrying a -5.0 µC charge is located at the origin. A small particle carrying a
!
12.0 µC charge is located at (1m, 0.5m). Find the coordinates of the point at which E = 0 .
32.(II) Five small charged particles are placed such that they are equally spaced around a semicircle
of radius 10 cm with a charge placed at each of the endpoints. The semicircle lies in the xy plane
20
such that it is centered on the origin and lies in the half-plane with x<0. If each particle carries a
charge of 1 nC, (a) what is the electric field at the origin? (b) What is the electric field at y=+10 cm?
(c) Where could you put a single -5 nC charge to make the field zero at the origin?
33.(III) A small sphere with charge Q and inertia m is constrained to move inside a narrow vertical
tube as shown in Figure 27-0. Fixed at the bottom of the tube is another small sphere with charge q.
(a) Find the equilibrium height h for the particle carrying charge Q (neglect friction). (b) Show that
if the particle is displaced from this equilibrium position by a small amount and then released, it will
exhibit simple harmonic motion with an angular frequency of (2g/h)1/2.
Q
h
q
Figure 27-0
Section 27.6
34.(II) An electric quadrupole can be constructed by placing charged objects at the corners of a
square, with adjacent corners bearing objects of opposite sign as shown in Figure 27-0. (Note that
there is no net charge on this quadrupole, and no net dipole moment.) How would you expect the
magnitude of the electric field to depend on the distance r from the center of the quadrupole? (Take r
to be substantially larger than the side of the square.)
r
Figure 27-0
35.(II) A water molecule consists of a central oxygen atom and two hydrogen atoms, with the OH
bonds separated by 104.5° as shown in Figure 27-0. Experimentally, water is a notably polar
molecule, with a permanent dipole moment of 6.172 × 10-30 C·m. What is the dipole moment of each
OH bond?
21
Figure 27-0
36.(I) Explain why equation 27.13 holds for Ey for y<0 as well as for y>0 for which it was derived.
37.(I) A water molecule has a dipole moment of 6.17 × 10-30 C·m. The ionic model of chemical
bonding assumes that the electrons from the two hydrogen atoms are completely transferred to the
oxygen atom. If the centers of positive and negative charge are 0.058 nm apart, how accurate is this
assumption?
38.(II) Two plastic bowling balls are rubbed with cloth until they each carry a uniform charge of
magnitude 0.1 nC. The blue one is rubbed with wool cloth and the red one is rubbed with cotton. (a)
If the two balls are held apart by a rigid 60-cm stick stuck through the holes so it runs from the
center of one bowling ball to the center of the other, what is the dipole moment of the arrangement?
39.(II) A small sphere carrying a uniform charge of -3.0 nC experiences a force of magnitude 200
nN in the positive y direction when it is along the perpendicular bisector of a permanent dipole. (a)
If the charges making up the dipole have a magnitude of 10 nC and a separation of 2.0 cm, how far
away from the dipole is the sphere? (b) What is the orientation of the dipole?
40.(III) A dipole is aligned along the y-axis, pointed in the positive y-direction. (a) Show that for
y>>½s and x>>½s, the electric field of a dipole is given by
Ex =
k 3 pxy
( x + y 2 )5 / 2
2
kp(2 y 2 − x 2 )
Ey = 2
( x + y 2 )5 / 2
(b) Show that this general result includes the special results derived in section 27.6.
Section 27.7
41.(II) Two thin plastic rods, each of length L, are joined end to end. One plastic rod is positively
charged with a charge per unit length ", and the other is similarly but negatively charged. What is
the effective dipole moment of the rod?
42.(II) The thin glass rod of length L shown in Figure 27-0 has a linear charge density starts out as
zero at its left end but which increases uniformly from left to right. The total positive charge on the
22
rod is Q. a) What is electric field along the rod’s axis at point P that is distance d from its right end?
b) What is the approximate expression for the electric field at distances is very much larger than L;
that is, far enough away that the rod looks small?
P
L
d
Figure 27-0
43.(III) The thin rod has a linear charge density that starts out at as "o at the left end but decreases
uniformly from left to right in such a way that the total charge on rod comes to zero. What is
electric field along the rod’s axis at point P that is distance d from its right end?
44.(II) As a laboratory assistant for the day, you have a box of small insulating disks, all having
various amounts of fixed charge. The disks are all of the same dimensions and material, and all have
the charge uniformly distributed. You are given a reference disk of the size that has a specified
charge of 1.5 µC. You place the reference disk and an unknown disk a set distance from each other
perpendicular to the line through their centers. You use a small charged pith ball to find the spot
along the common axis between them where the force on the ball is zero. Knowing the distance
between the reference disk and the unknown disk is 3.2 cm, and between the reference disk and the
ball is 8.8 cm, you realize you’ve found a way to measure the charge on the unknown disk.
45.(I) How much of a percentage error in the electric field do you introduce by approximating the
charged disk of Figure 27.26 as an infinite charged sheet with the same surface charge density, for
(a) z = 0.1 R, (b) z = 0.5 R, (c) z = R?
46.(III) Charge is deposited on the end of a long rod with a low conductivity. A charge of 2.2 µC is
deposited in all, but its distribution has the form of a falling exponential: Q(x) = Q0 e −bx , where b is
0.35 cm–1. Calculate the electric field at a distance of 2.0 cm beyond the end of the rod, along its
axis.
47.(II) You wish to determine the electric field along the perpendicular bisector of a 25-cm line of
charge with a total charge of 30 nC. You want to get by with a minimal amount of work, so you
need to know when you can get by with approximating the line charge as a point particle. How
close can you get to the line charge and still get away with this approximation with only a 5% error?
48.(I) A positively charged particle is released from rest along the axis of a fixed ring with an
evenly distributed negative charge. Describe the subsequent motion of the particle, if the ring is
unable to move.
49.(II) You wish to determine the electric field along the axis of a uniformly charged disk of radius 9
cm. You want to get by with a minimal amount of work, so you need to know when you can get by
with approximating the disk as an infinite sheet. How close can you get to the disk and still get
away with this approximation with only a 10% error?
23
50.(II) You wish to determine the electric field along the axis of a thin uniformly charged rod of
length 9 cm. You want to get by with a minimal amount of work, so you need to know when you
can get by with approximating the disk as an infinite sheet. How close can you get to the rod and
still get away with this approximation with only a 10% error?
51.(II) A uniformly charged rod lies along the y-axis from y= -15 cm to y= +15 cm. The total
charge on the rod is 30 nC. Approximate the rod as three point particles to calculate the electric field
along the perpendicular bisector at x=20 cm. (a) Draw a picture showing the location and magnitude
of the three point particles you will use for the calculation. (b) What is the value of the electric field
at x=20 cm using this approximation? (c) How much is this in error? (d) If you were unable to
integrate, what could you do to your approximation to make it better?
52.(I) A uniformly charged rod lies along the z-axis from z= -10 cm to z= +10 cm. The charge per
unit length on the rod is 100 nC/m. What is the vector expression for the electric field at (4 cm, 3
cm, 0)?
53.(II) Three narrow concentric rings of radii 5 cm, 7 cm and 9 cm are centered about the y-axis.
The charge on the inner ring is 1.0 µC, that on the middle ring is -2.0 µC and that on the outer ring is
1.0 µC. (a) What is the electric field at the point (0, 10 cm, 0)? (b) What charge would have to be
placed at (0, -10 cm, 0) to make the field zero for the position in part (a)?
54.(II) Assume that two large oppositely charged parallel plates are separated by 1.0 cm. An
electron is projected directly between the plates and parallel to them with an initial speed of 4.0 ×
106 m/s. The electron hits the top plate a distance of 2 cm from where it was projected. (a) Is the top
plate or the bottom plate negatively charged? (b) What is the surface charge density on the plates if
you know the densities have the same magnitude?
55.(II) A uniformly charged, non-conducting rod is bent to form an arc of a circle of radius R. The
angle subtended by the arc is 2!. If the total charge on the rod is Q, find the electric field due to the
rod at the center of the circular arc.
56.(III) A rod of length $R is composed of three non-conducting segments of equal length. The
middle one is without charge, while the two end segments each have a uniformly distributed
negative charge –Q. The rod is bent into a semicircle of radius R. Find the electric field at the
center of the semicircular arc.
57.(II) A particle of charge q and inertia m is placed above an infinite sheet with charge density σ.
The sheet carries the same sign charge as the particle. The particle is then released from rest. (a)
Describe the motion of the particle. (b) What is the acceleration of the particle? (c) What is the
change in kinetic energy of the particle after it has traveled a distance s?
58.(II) A uniformly charged thin rod lies along the x-axis from x=0 to x=+∞. Show that the electric
field along the positive y axis makes an angle of 45° with the rod and whose magnitude is
independent of the distance from the origin.
59.(III) A charged thin ring of radius R is prepared such that one half of the ring has a charge q1
uniformly distributed over it, and the other half has a charge q2 uniformly distributed over it. (a)
24
Find the component of the electric field parallel to the ring’s axis for any on it. (b) Find the
component of the electric field perpendicular to the axis for any point on it.
Section 27.8
60.(II) Microwave ovens, which permeate their chamber with oscillating electric fields, work well
at heating food with high water content. However, they do not heat thoroughly frozen food well, or
foods with a high oil content but low water content. Can you think of a reason for this? (Hint: Water
molecules, described in Problem 27.34, in ice are much more rigidly constrained than they are in
liquid, and oil molecules are nonpolar.)
61.(II) A small object with dipole moment p is released near its equilibrium orientation in a uniform
electric field of magnitude E. The rotational inertia of the dipole is I. a) Calculate the angular
frequency of small oscillations about the equilibrium orientation. b) What is the effective “spring
constant” of this simple harmonic system?
62.(II) A dipole that is free to move is placed along the perpendicular bisector of a fixed dipole, far
away compared to the size of the fixed dipole. Draw a diagram to illustrate the motion of the free
dipole if it is placed with its dipole moment in (a) the same direction as the fixed dipole, (b) the
opposite direction from the fixed dipole, (c) perpendicular to the orientation of the fixed dipole.
63.(I) A uniform external electric field exerts a torque of magnitude 10.0 × 10-9 Nm on an electric
dipole oriented at 30° from the field. The dipole moment of the dipole is 8.0 × 10-12 Cm. (a) What is
the magnitude of the external field? (b) If the two particles making up the dipole are 2.5 mm apart,
what is the magnitude of the charge on one of the particles?
64.(I) A dipole is to be released in a region with a uniform electric field where there are no damping
or frictional forces present. Describe the motion of the dipole if it released from rest in an
orientation (a) parallel to the electric field and (b) almost perpendicular to the electric field.
65.(II) In a particular region of space, the electric field is constant in direction but is uniformly
increasing in magnitude along that direction. A dipole is placed in this region and released from rest
perpendicular to the field. If no damping or frictional forces are present, give a description of its
subsequent motion, paying particular attention as to how it would differ from the case if the field
were uniform.
66.(II) A particle of charge 5.0 × 10-7 C is 30 cm from a small dipole along its perpendicular
bisector. The magnitude of the force on the charge is 10.0 × 10-6 N. Draw a diagram showing the
dipole with a direction for the dipole moment clearly indicated. On this diagram, show (a) the
direction of force on the particle, and (b) the direction of force on the dipole. (c) Determine the
magnitude of the force on the dipole.
67.(II) There are two possible alignments of a dipole in an external field where the dipole is in
equilibrium. (a) Draw a uniform field and show the two possibilities. (b) Are both of these
alignments stable (consider what would happen in each case if you gave the dipole a slight twist). (c)
Based on your answer to (b) and your experience in mechanics, in what orientation does the dipole
have the least potential energy?
25
68.(II) In some very rare cases, the polarizability of molecule may be negative. (a) Draw a picture
showing the resultant charge distribution in such a molecule, placed in the field of a positive point
charge. (b) Describe the motion that would result.
69.(III) Two identical particles, each with charge q, are a distance d apart. A dipole with moment p
is placed at the halfway between them and aligned with the axis joining the particles. Determine the
total force acting on the dipole. (Hint: you may wish to use a Taylor’s series expansion to
determine the small difference in the forces acting on the two separated charges that comprise the
dipole.)
26
Answers to review questions
1. The disturbance in the slinky (corresponding to a change in its springy “field") takes a nonzero
amount of time to propagate to the other end. Similarly, the electric field (and the associated force
on any charged objects in that field) takes time to propagate through space from its source.
2. A is the source of the electric field that B feels. B does not feel (that is, experience a force due to)
its own electric field.
3. The primary problem is that her map is a representation of a scalar field, but an electric field is a
vector field. There should be arrows that indicate direction as well as shaded bands that indicate
magnitude.
4. No, the units of the electrical field are [acceleration][mass]/[charge]. The peculiar accident that
happens with gravity is that gravitational “charge” m also has the unit [mass], leaving units
dimensionally equivalent to acceleration. Why inertial mass and gravitational “charge” mass are
identical is by no means obvious, and in fact is connected with the fundamental nature of that
interaction.
5. Nothing. The pellet contributes nothing to the electric field at the location of the pellet; that is, it
doesn’t feel its own field. Since nothing has changed with the source of the field, the field that the
pellet experiences doesn’t change. Note, however, the force exerted on the pellet would reverse
direction.
6. The behavior of the wheat visually mimics the field around a positively charged object, with the
electric field aimed radially outward around the charge with declining strength as one moves away
from the center. To mimic the field around a negatively charged object, you could imagine a big
vacuum hose poised over the ground.
7. No, the electric and gravitational fields have different units and can’t be added directly.
However, the forces exerted by both fields can be determined, and the total force acting on the
proton would be the sum of these two forces.
8. The answer has to do with whether the charge on the test object is big enough to disturb the
charge distribution on the other object, and hence the field. Generally, we mean by an insulating
object one whose net charge is not mobile except in a contact interaction. In this case, it doesn’t
matter (to a point) how much charge is on the test object. If, however, the other object is a
conductor, then the size of the charge on the test object might well affect the charge distribution on
the conductor, and so we generally assume that the charge on the test object is small enough that
doesn’t happen. Furthermore, we would have to worry about effects of induced polarization in the
test object if it were large and conducting.
9. The force is upward, and the electron is negatively charged, and so the electric field in that
vicinity must have a component that is vertically downward. (There might well be a component that
is parallel to the initial motion of the electron. This would change the speed of the proton, but not its
direction.) If it were deflected downward, we do not know if it were due to the electrostatic force
pointing down and adding to the force of gravity, or if the electric field is aimed upward but too
weak to produce an upward force strong enough to overcome the gravitational force downward.
10. No, because the two field contributions have to be added as vectors. In fact, the magnitude of
the total electric field at the pushpin could be anywhere from 100 N/C to 500 N/C.
11. The direction of all the vector arrows would reverse, but otherwise the figure would be identical.
12. Since 1 J = 1 N·m, 1 N = 1 J/m. The units of electric field (N/C) are then J/(C·m).
13. The gravitational field in the room is due to the earth, and that is not a uniform field because it is
radially directed from the center. However, the differences in the magnitude (from ceiling to floor)
and in the direction (from wall to opposite wall) of the field are generally so small that we treat it as
though it were uniform for all practical purposes.
27
14. No, because the force acting on the two charges of the dipole are equal in magnitude but
opposite in direction, making the net force zero. A net force non-zero on the dipole can only be
exerted by a non-uniform electric field since the forces on the two charges would not have the
exactly opposite one another.
15. The acceleration is not uniform (it is centripetal and always changing direction). Therefore
electrostatic force acting on the electron must be non-uniform, implying the electric field is also not
uniform.
16. Both magnitude and direction of the field at P would change. The contribution E2 would
increase, affecting both the x and y components of the field (and not in the same proportion).
17. The field decreases in proportion to the square of the distance. To reduce the field by a factor of
100, we’d have to increase the distance by a factor of 10, or twenty paces away from the sphere.
Moving the instrument another 18 paces out will do the trick.
18. The field from the small object Q would fall like 1/r2, while that from the dipole would fall like
1/r3. The ratio of the dipole and monopole fields would then be (Qd/r3)/( Q/r2) = (d/r) < 1. That is,
the dipole field is weaker at this distance.
19. No. The magnitude of the field varies as the angle ! varies. Compare the results of the
calculations at Equations 27.10 and 27.13. These two, which differ by a factor of 2, are the extreme
cases at a given distance r.
20. The radius of the annular ring is Rsin!, and its width is R d!. Thus the area of the ring is
2$(Rsin!)R d! = 2$R2sin! d!. Since the half-shell has an area of 2$R2, the ratio of the areas is
sin!d!. The ratio of the areas also represents the ratio of the infinitesimal charge to the total charge:
dQ/Q = sin! d!. Thus the infinitesimal charge is dQ = Qsin! d!.
21. Any time the field contribution from each infinitesimal piece is the same, then you can lump all
the charge together. However, this can only occur if the distance to the point is the small for all
infinitesimal pieces. Note this works for Figure 27.25, but it doesn’t work for Figure 27.26, where
the contribution from each infinitesimal ring is different. Thus for the latter, an integral is required.
22. Respectively, 1/r2, 1/r, and constant.
23. By definition, #E = p. The units of p are C·m, and those of E are N/C. Thus, the units of #= p/E
# are C2·m/N = C2s2/kg.
24. Because the two are an interaction pair, the magnitude of the force must be the same on both.
The field around Q falls like the distance squared, not the distance cubed, but the force on the dipole
falls off more steeply than the surrounding field does. This is because the force on the dipole is the
result of a near cancellation of the forces on the two ends of the dipole. This cancellation gets more
complete as the distance between the two ends of the dipole becomes small compared to the distance
between the dipole and the particle.
28
Chapter 28
Gauss’ law
Review questions
Answers to these questions can be found at the end of this chapter.
Section 28.1
1. Must all field lines in a region of space containing particles originate from a positively charged
object? Must all field lines terminate in a negatively charged object?
2. An isolated system contains two charged objects with charges that are equal in magnitude but of
opposite signs. Will all the field lines be contained inside some boundary for this system?
3. Would a positively charged particle released at rest follow the path of the field line passing
through its initial location? (Assume the electrostatic force is the only force acting.)
4. Can there be more than one field line passing through a given point?
Section 28.2
5. Which represents the intensity of the electric field at a given point: (a) the curvature of the field
lines, (b) the density of field lines, (c) the number of field lines passing through that point, (d) the
rate at which field lines are converging or diverging at that point?
6. A small charged ball is suspended at the center of a spherical balloon, which in turn is nestled
snugly inside a cubical cardboard box. On one side, the balloon touches the wall of the box. (a) At
this point of contact, is the electric field on the surface of the balloon equal to the electric field on
the surface of the box? (b) At this point, is the number of field lines per unit area on the surface of
the balloon equal to the number of field lines per unit area on the surface of the box?
Section 28.3
7. The net field-line flux crossing a certain closed surface is zero. Does that mean that there are no
charged objects inside the closed surface?
8. If the net charge inside a closed surface is zero, does that mean that there are no field lines
crossing the surface?
9. Eight electrons are the only charged particles inside of an isolated balloon, producing a flux
through its surface. If eight additional electrons are now set outside of the balloon to form the
corners of a cube, how will the field-line flux through the balloon change? How will the flux change
of the additional eight electrons were instead concentrated at some point outside the balloon?
Section 28.4
10. Recall that the density of electric field lines is proportional to the magnitude of the electric field
at any point. Look again at the field lines in Figure 28.14. The spreading of field lines with radial
distance from the axis tells you something about how the magnitude of the field falls with radius. (a)
Take two adjacent field lines penetrating the Gaussian surface and separated by a small distance
along the axis of the wire. Do these field lines separate with increasing radius? What is the
relationship between the separation and the radius? (b) Take two adjacent field lines penetrating the
Gaussian surface and separated by a small angular distance around a circle girding the cylinder. Do
1
these field lines separate with increasing distance? What is the relationship between the separation
and the radius? (c) What does this tell you about the density of field lines as a function of radius?
11. Consider the case of a charged dipole. Can you devise a Gaussian surface to compute the
electric field in a simple way for this case? If not, does this mean that Gauss’ law does not apply?
Section 28.5
12. Delicate electronics are protected from external electric fields by placing them inside a sealed
metal box. Your computer or calculator likely has some of its components shielded this way. Explain
what physical phenomena might cause this protection.
13. Two dabs of paint are placed on opposite ends of a metal soup can. One dab of paint is given a
positive charge, and the other is given a negative charge. What can you say about the path of the
electric field lines between the two dabs of paint?
Section 28.6
14. Why is it important for flux calculations for a closed surface that the area vector point outward,
away from the interior of the surface?
15. What is the relationship between electric field line flux, as defined in Section 28.3, and the
electric flux, as defined in Section 28.6?
Section 28.7
16. Would Gauss’ law hold if the electric field due to a charged particle had a 1/r dependence rather
than a 1/r2 dependence?
Section 28.8
17. You have a ball of radius R and charged uniformly throughout with a total charge Q. Suppose
you drew a concentric spherical Gaussian surface inside the ball with radius r. How would you write
an expression for the charged enclosed by the Gaussian surface?
18. A thin spherical metal shell of radius R has charge Q, and it is surrounded by a larger,
concentric, thin metal shell of radius 2R and charge –Q. For which of the following regions is the
electric field zero: r<R, R<r<2R, r>2R ?
Section 28.9
19. Would you apply Gauss’ law as it is diagrammed in Figure 28.23 if the charge on the long rod
were not uniform but slowly increasing from one end of the rod to the other?
20. In sketching a Gaussian surface for the case of the charged rod in Figure 28.23, we wanted to
know the field at a distance r from the axis of the rod. It made sense to make one of the dimensions
of the cylindrical Gaussian surface r. However, there was no guidance as to how long the cylinder
should be. How do you decide what that dimension should be?
Section 28.10
21. Is there reason why the Gaussian surface in Figure 28.24 should be a cylinder? Would the same
results apply if a thick jigsaw puzzle piece had been used instead, as shown here in Figure 28-0?
Figure 28-0
E
2
Developing a feel
Calculate or estimate the following quantities:
1. The electric flux through a pear-shaped closed surface placed entirely within a region of uniform
electric field. (Q, O)
2. The gravitational flux through the roof of a typical one-story house due to the Earth’s
gravitational field. (G, L, P)
3. The electric flux through a horizontal pie pan held 10 meters directly above a 100 µC charge. (C,
K, R)
4. The surface charge density that results when you place a 50 µC charge on a metal baking sheet.
(E, M)
5. The surface charge density required to “float” an electron 1 cm above a plastic cutting board. (D,
J, N, S, A).
6. The surface charge density on the ground required to create an upward electric field of 100 V/m a
few meters above the Earth’s surface. (T, H, B)
Hints:
A. What field strength is required?
B. What is the value of the constant εο ?
C. What is the area of a pie pan?
D. What forces must balance in order to “float” the electron?
E. What is the size of a baking sheet?
G. What is the floor area of a typical one-story house?
H. What should we use for the distance above the surface?
J. What is the size of a cutting board?
K. What is the magnitude of field at the location of the pie pan?
L. What do we do about the tilt of the roof surface?
M. How will the charge distribute itself?
3
N. What does the 1 cm distance tell us compared to the size of the baking sheet?
O. How much charge will be enclosed?
P. What are corresponding quantities between electric flux and gravitational flux?
Q. What do electric field lines look like near charged particles?
R. Is the field uniform over the pie pan?
S. What is the inertia of an electron?
T. Is spherical or planar geometry appropriate?
Key: A. ~6x10-11 V/m. B. ~10-11 in SI units. C. ~0.03 m2. D. The electric force due to the charged
board and the gravitational force due to the Earth. E. ~40 cm x 50 cm, ~0.2 m2. G. 150 m2. H. It is
not needed, as long as the Earth’s surface is approximately an infinite sheet. J. ~30 cm x 40 cm. K.
~9000 V/m. L. Nothing; the effective horizontal area, Acosθ, includes the slope of the roof. M.
~Uniformly on both sides of the conducting sheet. N. The cutting board might as well be infinite at
this close distance. O. None because
!" !" charged particles would create regions of radial lines coming
from a point. P. !E" !g and E " g . R. Because the pie pan is about 0.1 m across, the variation in
!"
angular direction of E over the surface is only diameter/distance = 0.1/10 = 0.01 radian # 1º, plenty
uniform. S. ~9x10-31 kg. T. Planar; the curvature of the earth is negligible for such distances. Q.
They radiate out from the charged particle.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any particular
section. Typically, an example that is worked out in detail is followed immediately by an example
whose solution you should work out by following the guidelines provided.
WP 28.1: Field lines of repulsing charges. What is electric field line pattern arising from two
particles with equal positive charges? What is the electric flux across the plane midway between the
two particles? How would the pattern change if they were, instead, negatively charged.
!Focus problem
We start with a simple representational sketch to start our thinking of the problem and its solution.
+
+
Field lines help us to visualize what the electric field looks like in the vicinity of charged objects.
The text has looked at the electric field lines arising from unlike charged particles in a dipole. We
need to figure out what the electric field vectors do at several points in space and connect these
points to draw the field lines.
!Plan approach
The electric field can be found by finding the force on a test charge located near the system for
which we want to trace the field lines. We could calculate this force at a large number of points and
from these vectors draw the field lines. For complicated distributions of charges this is the only way.
However, we can use the simple symmetry of the system in this case to figure out approximately
what the field lines should look like based on some simple inferences of the electric force acting on a
positively charged test particle. Once we have the field line diagram, the electric flux can be found
by
# #
ΦE = ! E • dA
!Execute plan
Near each of the charged particles, the field will be directed radially outward, perpendicular to the
“surface” of our charged particle.
We know that that the test particle will be repelled by both particles no matter where we put it. We
also know that it will be repelled more strongly by the closer of our charged particles. In the plane
midway the charges, the electric field cannot favor one charge over the other because they are
equally distant. Thus there can be no component of force parallel to the line joining the particles,
otherwise the test particle would move into the domain of one of the particles rather than keeping on
the boundary. Drawing in a few electric force (in blue) gives:
5
Using force vectors as a guide and the fact that field lines cannot cross each other, we can trace out
several field lines:
Notice that the electric field across the plane between the two particles is zero, meaning the electric
flux is zero. If the particles were negatively charged, the pattern would look the same except the
directions of the lines would be reversed.
!Evaluate result
We know that field lines between two particles that are oppositely charged look line bonds and
bridges. We would expect that the fields lines between two objects with same-sign charges should
look like they are pushing or keeping the particles apart. With a good field line diagram you can
really visualize the interaction of the charged particles in the system. Here, it’s very clear the
charges are repelling each other. physically plausible way if you vary the radius of the semicircle or
the magnitude of the charge?
WP 28.2: Uniformly charged square sheet. A square sheet carries a charge Q uniformly
distributed over its surface. Make a sketch of the sheet, and draw the electric field lines outside the
square in the plane containing the square. Based on this sketch, what is the electric flux across a
horizontal plane perpendicular to the sheet through its center?
!Questions and suggestions
1. After drawing your sketch of the sheet, describe the problem in your own words. What is the
problem asking you to find? Are you asked for a qualitative or a quantitative answer?
2. Are there symmetries you can make use of to simplify the problem?
3. What do you know about the direction of the electric field near the square's edges?
4. When you get very far away, what should the electric field approximate?
5. What equation will help you get the electric flux? Can you already guess what the flux will be?
6
6. Draw several field lines connecting the two limiting case close to the square and far away from it.
7. What can you say about the electric flux through any plan perpendicular to the square through its
center? physically plausible way if you vary the radius of the semicircle or the magnitude of the
charge?
WP 28.3: Flux of a proton through a cube. As you ponder your drink, you notice an ice cube
floating in your glass. Because you are studying electric flux, you ask yourself: what flux is
produced though each of the faces of the cube (of side length a) by a proton in a hydrogen atom at
one corner of the cube?
!Focus problem
We are asked to find the electric flux through different planes of a cube with a proton at one corner,
so we draw a simple sketch.
a
a
proton
Although there are many charged particles (protons and electrons) in the cube, we only concentrate
on the contribution of a single proton at a corner. Because of the spherical nature of the electric
field, the field across the face of any cube will be non-uniform. We suspect we will have to employ
brute-force integration over each face to get the flux.
!Plan approach
We know the flux is given by
# #
ΦE = ! E • dA
And, we know that the electric field due to a single proton is
#
E=
1
q
rˆ
4πε 0 r 2
where q = +e is its elemental charge. Although the electric field varies in strength over the three
faces with the proton at their common corner, we know that the direction of the electric field vector
on each face is parallel to the face with
no component perpendicular to the face. Since the electric
# #
field is parallel to these faces, E • dA = E cos 90°dA = 0 . Moreover, the fluxes through each of the
other three faces are equal because each vector area element on a face has a corresponding one on
the other two faces making the same angle with the electric field. Thus, we only need calculate the
flux through one face. The figure below should help us with setting up the integral.
7
!"
E
y
!"
d
A
a
x
0
r
z
R
θ
a
a
The area dA equal dxdy, with x and y in the directions indicated by the axes. r2 = a2 + R2 and R2 = x2
+ y2 with cos! = a/r. The flux through the small area dA is
# #
1 q
ΦE = ! E • dA = ! E cosθ dA = !
cosθ dA
4πε 0 r 2
a
=
a
! ! 4πε
0
1
0
qa
( a + x 2 + y 2 )3
2
0
dxdy
!Execute plan
That’s the end of the physics. Now for the math. Referring to standard integral tables, we find that
the integration over x yields
ΦE =
qa
4πε 0
qa 2
=
4πε 0
!
a"
a
%
$
' dy
2
2
2
2
2'
0$
+
y
)
a
+
x
+
y
(a
#
&0
x
a
! (a
0
dy
2
2
+ y ) 2a2 + y 2
If you cannot find the last integral in a reference book, the integration over y is not so easy. But, we
can use substitution of variables to get it in a form that most integral tables recognize. Use the
substitutions
u2 = a2 + y 2
u
dy = du
y
to transform the integral into
ΦE =
qa 2
4πε 0
qa 2
=
4πε 0
=
qa 2
4πε 0
!
!
!
2a
u / y du
u2 u2 + a2
a
2a
du
2
u u + a2 u 2 − a2
a
2a
a
du
u u4 − a4
8
You can find this integral in most tables
2a
qa 2 " −2
a2 %
arcsin
ΦE =
4πε 0 $# 4a 2
u 2 '& a
q
q "π π %
arcsin( 12 ) − arcsin(1) ] = −
=−
−
[
8πε 0
8πε 0 $# 6 2 '&
Using qproton = +e, the flux through each face of the cube not containing the proton in its corner is:
ΦE =
e
24ε 0
!Evaluate result
You might say (or scream), "There must be an easier way!" Sometimes there is, sometimes not.
Nonetheless, you should not be afraid to tackle integral problems like this. The important thing is to
set up the physics properly as we did in the Plan Approach step. Then, use the math tools you know
or can find to get to a final answer. We can check the units of the answer, charge/$0: C/[C2/(N·m2)]
= (N/C) ·m2, which are the proper units of flux. Although it might seem surprising that the flux
through the non-adjacent surfaces is independent of the size of the cube a, it is to be expected if one
thinks of the radial "rays" of electric field that emanates from the proton. Doubling the size of the
cube would still catch the same number of field lines on the surfaces and thus not affect the flux
through them.
WP 28.4: Flux of a charged rod through a cube. Consider an infinite line of charge with linear
charge density of " running through the center of your ice cube as depicted in Figure 28-0. Use
integration to determine the electric flux through each face.
Figure 28-0
!Questions and suggestions
1. In what sense in the problem similar to the previous problem? In what sense different?
2. What physical symmetry can you use to simplify the problem? What does the electric field look
like for an infinite line of charge?
3. Is the flux through any face identically zero? Why?
9
4. Which unknown quantities do you need to determine? Which equations allow you to express the
unknown quantities in terms of know ones? What information can help you find these?
5. The important thing is to set up the problem step by step. Begin with the definition of electric
flux. Determine the direction of the electric field at any point on the faces so that integration over
the surface can be done. Rewrite dA in terms of the coordinates of the point (probably x, y, and z).
The dot product will give you cos!. Can you relate this to the coordinate variables? What about the
distance from the wire to the area element dA?
6. Can you use symmetry or analogous arguments to infer the flux through the other faces once we
have it for one face?
7. Are the units and sign correct? Even though there are no numbers, check that the units of your
answer give units of flux (Nm2/C).
WP 28.5: Revisiting flux through a cube. We revisit the ice cube of WP 28.3. This time, we will
use Gauss’ law more cleverly to answer the question pounding in your head: what flux is produced
though each of the faces of the cube (of length a) by a proton in a hydrogen atom at one corner of
the cube?
!Focus problem
We begin with the same sketch we used for WP 28.3.
a
proton
a
This problem is identical to WP 28.3, except that we are now use Gauss’ law which says that the
flux through a closed surface is proportional to the charge enclosed
# #
ΦE = $
E
! • dA = qenc / ε0
!Plan approach
Because the proton is at the corner of the ice cube, the cube does not enclose the charge. However,
we can enclose the proton by surrounding it with eight ice cubes with the proton at the central
corners. It is then at the center of a single mega-cube with sides of length 2a. We can then apply
Gauss' law for the total outer surface of the giant cube and use symmetry to get the flux through any
face.
!Execute plan
If we surround the proton with eight ice cubes the flux through each of the six large faces must be
equal. But the each large face is composed of four of the smaller faces the size of our original cube.
10
a
a
a
a
Thus, but Gauss' law the flux through each of these smaller faces must be
ΦE =
1 qenc
e
=
6 ⋅ 4 ε0
24ε 0
This is the same answer we obtained in WP 28.3 by tedious integration!
!Evaluate result
This is much easier than the method we employed in WP 28.3 to do the same task. Gauss' law is
extremely powerful in cases where simple symmetry can be exploited. It takes practice and
experience to see how to construct the surfaces, but it is sweet when you can use it.
WP 28.6: Revisiting flux by a charged rod through a cube. Use Gauss' law to determine the
through each face of a cube by an infinite line of charge with linear charge density of " running
through the cube's center.
!Questions and suggestions
1. Because this problem asks the same question as WP 28.4, should you draw the same sketch?
1. In what sense in the problem similar to WP 28.5? In what sense different?
2. What physical symmetry can you use to simplify the problem? What does the electric field look
like for an infinite line of charge?
3. Is the flux through any face identically zero? Why?
4. Which unknown quantities do you need to determine? Which equations allow you to express the
unknown quantities in terms of know ones?
5. What is the amount of charge “enclosed” in this case?
6. Apply Gauss' law, using the symmetry of the system.
7. Should your answer agree with your answer from WP 28.4? Does it?
11
WP 28.7: Field of a uniformly charged cylinder. A very long solid Teflon rod has a volume
charge density (charge/volume) given by #(r)=#or; i.e., it increases linearly with radial distance r
from the symmetry axis. Calculate the electric field both inside and outside the rod near the
midpoint of the rod. Then, make a sketch of the field as a function of distance from the rod’s axis of
symmetry.
!Focus problem
We had better draw a simple sketch of a cylinder to start and keep track of our thinking.
a
The simple symmetry of the situation makes this a prime candidate for application of Gauss' law.
Since the rod is a cylinder, we’re after the electric field as the radial distance from the rod increases.
We’re told the rod is “very long”, so in that limit we won’t work about strange effects near the end
of the rod, and concentrate on calculating the electric field strength at distance R from the symmetry
axis in the region where the electric field is directed radially outward (i.e., not near the rod's ends).
!Plan approach
We’ll use Gauss' law to calculate the electric field, making use of the radial symmetry.
!Execute plan
Because of the symmetry we will use a cylinder as our Gaussian surface to surround a portion of the
charge in the rod. The surface depicted is inside the rod of radius R, extending some distance along
it’s length h. Because of the radial nature of the electric field of the rod, the #electric field is
#
everywhere perpendicular to the cylindrical surface. On the end caps, E • dA = E cos 90°dA = 0 .
a
r
R
h
By symmetry, the strength (i.e., magnitude) of the electric field is the same everywhere on the
Gaussian surface's wall. Thus, we need only integrate over the area of the cylindrical surface.
12
# #
ΦE = $! E • dA = $! E cos 0°dA
= E $! dA
= E (2π Rh)
Note that the math naturally leads us to an integral over the infinitesimal area scalar dA, rather than
!"
!"
d
the infinitesimal area vectors d A – which would have yielded $
! A = 0 ! (Can you explain why?)
Applying Gauss' law,
ΦE = E (2π Rh) = qenc / ε 0
E=
qenc
2π Rhε 0
To determine the total charge enclosed by Gaussian surface, we must take note that it varies with
radial distance r from the axis. We divide up the interior within the surface into a series of thinwalled cans of radius r, wall thickness dr and wall length h. The infinitesimal amount of charge
contained within the side walls of the "can" is dq = #(r)dV where dV = h(2π rdr ) is the volume of
the walls themselves. The total charge contained within the Gaussian surface of radius R is thus:
R
qenc = ! dq = ! ρ (r)dV =! ρo r[h(2π rdr)] = 2 3 ρoπ hR3
0
The electric field strength inside the rod (R<a) is thus:
Einside
ρoπ hR 3 ρo 2
R
=
=
2π Rhε 0
3ε 0
2
3
Outside the rod the total charge enclosed by the Gaussian surface is constant ending at the outer wall
at distance R=a; i.e., qenc = 2 3 π h ρo a 3 . This gives an electric-field strength outside the rod:
Eoutside =
2
ρoπ ha 3 ρo a 3 1
=
2π Rhε 0
3ε 0 R
3
!Evaluate result
A sketch of the electric field near the rod looks like the graph below. The field increases
quadratically as R2 up to the rod's surface (R=a), then falls off as 1/R.
E
R
a
13
The dependence of the electric field on radius makes sense. As you go radially outward more and
more charge is enclosed, the electric field gets stronger and stronger. Once outside the rod, though,
the field decreases as you get farther away just like in the example of the charged wire of Chapter
27. We also check to make sure our expressions for the electric field inside the rode and outside the
rod agree at the rod's boundary at R=a, and gratifyingly find that they both give the same result for
the limiting case of R%a from either side, i.e., E(a) = &#oa2/$0.
WP 28.8: Field of a non-uniformly charged sphere. A spherical distribution of charge has a total
charge Q within a distance a from the center. The charge density decreases with radius from the
center as 1/r, implying the charge concentrates toward the center. Find the electric field for points
inside and outside the sphere. Sketch a graph of the field.
!Questions and suggestions
1. Does the sketch below provide all of the necessary information and relevant quantities for us to
tackle this problem?
a
r
2. In what sense in the problem similar to the previous problem? In what sense different?
3. What physical symmetry can you use to simplify the problem? What does the electric field look
like for sphere of charge?
4. What would be an appropriate Gaussian surface?
5. Which unknown quantities do you need to determine? Which equations allow you to express the
unknown quantities in terms of know ones?
6. What is the amount of charge “enclosed” in this case?
7. Apply Gauss' law, using the symmetry of the system. The charge density is not constant in this
case, so finding the charge enclosed is not as simple as in the last example. Be attentive!
8. Even though there are no numbers, check that the units of your answer are correct.
9. Does the dependence of the electric field inside the sphere make sense to you? Explain in words
why this is reasonable.
14
Questions & problems
Section 28.1
1.(I) Can an electric field line ever have a kink in it? Explain your answer.
2.(I) Free electrons in the atmosphere near the surface of the earth generally experience an
electrostatic force upward when there isn’t a fluctuation due to a thunderstorm. Sketch the electric
field lines near the surface of the earth.
3.(III) Sketch a flow chart for a computer program that calculates field lines from a charged object.
4.(I) Explain whether you can, or cannot, draw a field line through a point in space where the
electric field is zero.
5.(I) Suppose a positively charged test particle were released from rest in an uniform electric field,
and that nothing else acts on it. (a) Describe the trajectory it follow. (b) How would the trajectory
change, if at all, if the particle were given a initial velocity in the direction of the electrical field, (c)
at an angle with respect to the electric field? (d) How would your answers to the questions above
change if the test particle were negatively charged?
6.(II) Figure 28-0 shows the electric field produced by a dipole. (a) How would a positively charged
test particle begin to move if it were released from rest at point A? Assume nothing else acts on it.
Repeat the above question points (b) B, (c) C, and (d) D. (e) How would your answer (a) change if
the particle were given an initial velocity at some angle with respect to the electric field at A?
D
Figure 28-0
7.(II) Two particles of charge +Q are situated at opposite ends of one diagonal of a square, and two
particles of charge –Q are situated at opposite ends of the other. Sketch the pattern of field lines
15
created, showing eight lines per charged particle in the plane containing the square.
8.(II) Does a line of force represent the actual trajectory, over time, of a test charge, released from
rest and only acted on by an electric field? Explain why or why not.
9.(II) There are many parallels between gravity and electrostatics. Think of a test object with mass
in a gravitational field as being analogous to a test object with positive charge in an electrostatic
field. (a) draw the gravitational field line pattern for the Earth-Moon system. Make sure you
include arrows on the lines to indicate direction. (b) What would be the signs of the charges in
electrostatic phenomena that would correspond to the Earth and moon? Do you see any differences
between electrostatics and gravitational phenomena?
10.(I) You and your friend are asked to draw the field line pattern for two charged objects located
near one another. The charge on the first object is +2q (with q positive) and the charge on the second
object is –q. Assume the situation is two-dimensional. Your diagram shows 32 field lines emanating
from the first charge and 16 field lines terminating on the second charge. Your friend’s shows 24
field lines emanating from the first charge and 12 terminating on the second charge. Which diagram
is correct?
Section 28.2
11.(II) Make a geometric argument, based on how field lines spread, that the magnitude of the
electric field surrounding a long, straight, charged wire falls like 1/r, where r is the distance from the
axis of the wire.
12.(II) Make a geometric argument, based on how field lines spread, that the magnitude of the
electric field surrounding a large, flat, charged metal plate is uniform and independent of the
distance from the plate as long as the distance is small compared to the size of the plate.
13.(II) Identically charged pellets are released from rest from starting lines a, b, c, shown in Figure
28-0, in a region of changing electric field. (a) Rank, from smallest to largest, the time it takes for
the pellet on line a to cross line b, the pellet on b to cross line c, the pellet on c to cross line d. (b)
How does your answer to (a) change if each of the pellets is given an identical initial velocity along
the starting lines, i.e., toward the top of the page?
a
b
c
d
Figure 28-0
14.(I) In the regions where field lines are shown in Figure 28.11, where does the electric field have
16
the largest magnitude? Explain.
15.(I) You decide to make a field line diagram in two dimensions, with 16 lines per coulomb of
charge. (a) How many lines should you draw from a point where a particle of charge 1.5 C is
situated? Should they enter or leave that point? (b) Answer the same questions for a particle of
charge –0.375 C. (c) Could you represent a particle of charge +0.8 C in this diagram? Explain.
16.(II) Consider the pattern of five field lines shown in Figure 28-0. Three points (A, B, C) are
marked on the diagram. Rank the points according to the magnitude of their electric fields, from
largest to smallest.
Figure 28-0
Section 28.3
17.(I) Imagine a single positively charged object in a system. Is it possible to have a closed surface
in this system with a negative field line flux through it? [Note: the charged object need not lie inside
the surface.]
18.(II) Three charged pellets are arranged along a line, and three different closed surfaces are
constructed, each with identical cylindrical walls but with different top caps, as shown in Figure 280. Rank each of the surfaces, from least to greatest, according to the total electric field line flux
through them.
–2Q
+Q
+Q
(a)
(b)
(c)
Figure 28-0
19.(II) A small grouping of charged objects with charges +Q, +2Q, +3Q, –4Q is shown in Figure 280. Assuming that a Gaussian surface surrounding only the object with the +Q charge has a field line
flux of +4, sketch Gaussian surfaces that contain at least the object with the +Q charge and have a
field line flux of (a) +24, (b) –4, (c) +8.
17
+Q
+2Q
–4Q
+3Q
Figure 28-0
20.(III) Make an argument based on Gauss’ law that a charged object cannot be held in static
equilibrium by electric fields alone. For a start, assume that there is no other charged object in the
region of interest, though of course the external electric field can be due to charged objects outside
this region.
21.(II) Two more small charged objects are added to the two-dimensional region of Figure S28.10,
creating the situation shown in Figure 28-0. The field pattern is now different, but twelve field lines
still emanate from the object with the +1 C. What is the field line flux through each of the closed
surfaces indicated?
Figure 28-0
22.(II) Consider the two-dimensional representation of the field-line pattern for a dipole in Figure
28.2. Could you draw a closed two-dimensional surface so that the field line flux through it is (a)
zero, (b) +16, (c) –16, or (d) +3? Explain.
23.(I) Consider three different closed surfaces. Surface A is a sphere of radius R, surface B is a
sphere of radius 2R; and surface C is a cube with side length R. At the geometric center of each
surface is a small ball carrying a positive charge +Q; each ball is completely enclosed by the surface.
Assuming there are no other charged objects inside any of the surfaces, rank the surfaces based on
the field line flux through them, from largest to smallest.
Section 28.4
18
24.(I) If the total field line flux through a spherical surface surrounding a charged ball is +$ball, then
what is the field line flux through a pair of square surfaces symmetrically surrounding the ball? The
square surfaces each have side b, and are separated by distance b from each other, as shown in
Figure 28-0.
b
b
Figure 28-0
25.(II) Let Eo be the magnitude of the Coulomb field at a distance R from a particle of charge Q.
Now consider a non-conducting sphere of radius R throughout whose volume a charge Q is
uniformly distributed. What is the magnitude of the electric field, in terms of Eo, at: (a) a distance R
outside the surface of the sphere; (b) at the surface of the sphere; (c) at a distance of ¼R from the
center of the sphere? [Hint: think of the sphere as a collection of charges.]
26.(III) You are given a uniformly charged spherical shell, of positive charge Q and radius R. In
order to measure the electric field inside, you drill a small hole, which removes 0.01% of the shell’s
material, and insert a probe into the center of the shell. What is the direction and approximate
magnitude of the measured field?
27.(I) Suppose Figure 28.12c represents the cross-section of an infinitely long, uniformly charged
cylindrical shell. How does the electric field strength vary with distance from the axis of the cylinder
(a) inside the shell; (b) outside the shell?
28.(II) Figure 28.15 shows a cylindrical Gaussian surface straddling a charged sheet. Are there any
other shapes that would be appropriate for the Gaussian surface in this situation? Explain your
reasoning.
Section 28.5
29.(II) A pellet with charge +Q is centered inside a spherical cavity in a conducting ball. The
conducting ball has a charge –2Q on it. The cavity in the ball is off-center, as shown in Figure 28-0.
(a) How much charge resides on the inner surface of the cavity? (b) Is the charge on the cavity
surface distributed uniformly or nonuniformly? Why? (c) How much charge resides on the outer
surface of the conducting ball? (d) Is the charge on the ball’s outer surface distributed uniformly or
nonuniformly?
19
-2Q
+Q
Figure 28-0
30.(I) Explain why the electric field at the surface of a conductor in electrostatic equilibrium must
point perpendicular to the surface at that point.
31.(II) A coronal discharge occurs when the electric field just above the surface of a conductor
exceeds approximately 3×106 N/C, strong enough to ionize the air by ripping electrons from atoms.
Coronal discharge is most likely to occur near a sharp tip of a conductor. This is why solder joints in
electronics are characterized as being of good workmanship if they don’t have sharp projections.
This is also why it’s easier to draw a spark from your fingertip than from your knee after having
scuffed your feet on the carpet. (a) Make an argument, based on the geometry of field lines emerging
from the surface of a conductor, that charge is more likely to concentrate on a convex region of the
conductor surface than a concave region. In fact, the more convex (sharper) the surface, the higher
the concentration of charge is likely to be. (b) Sketch a paring knife that has a net charge on it, and
shade the regions where charge is most likely to be found to concentrate on the surface.
32.(II) Two concentric, spherical, metal shells are insulated from each other and the surroundings.
The inner shell carries a total charge of +2Q, and the outer shell a total charge of –Q. In electrostatic
equilibrium, what is the charge on (a) the outer surface of the inner shell; (b) the inner surface of the
outer shell; (c) the outer surface of the outer shell? (d) The metal shells are now connected by a
metal wire. Answer questions (a)-(c) for the situation obtained after electrostatic equilibrium is
reached.
33.(I) The neutral metal object in Figure 28-0 contains particles, with charges +Q and –2Q, in
separate cavities (Q>0). What are the charges on the surfaces of the cavities and on the outer surface
of the object?
Figure 28-0
20
34.(II) A metal object has a single hollow cavity in which some charged particles are suspended by
insulated supports. It is known that the metal object has positive charge +Q distributed over its outer
surface and charge –2Q distributed over the surface of its cavity. (a) How much charge is suspended
within the cavity?
35.(II) A small ball carrying a negative charge –2q is placed at the exact center of a conducting
spherical metal shell that carries a charge of +q. (a) What is the sign and magnitude of the charge on
the inner surface of the shell? (b) What is the sign and magnitude of the charge on the outer surface
of the shell? (c) Do your answers change if the ball is not at the exact center of the shell? If so, how?
Section 28.6
36.(I) A cardboard box snugly contains a volleyball, which has become charged uniformly over its
surface in shipment. The electric flux through one side of the box is 5.2×102 Nm2/C. What is the
charge on the volleyball?
37.(II) A butterfly net hangs from a circular loop of diameter 40 cm. You hold the loop in the
horizontal plane in a region where the electric field is 150 N/C downward as shown in Figure 28-0.
(a) What is the electric flux through the netting? (b) Does the flux through the netting increase or
decrease if you turn the loop so that it lies in a vertical plane?
E
Figure 28-0
38.(I) A flat surface with an area of 3.0 m2 is placed in a uniform electric field with a magnitude of
10 N/C. While keeping the surface completely flat, can you orient the sheet so that the electric flux
through the surface equals (a) 6 Nm2/C , (b) 60 Nm2/C? Explain your answer.
39.(I) Figure 28.14 shows a cylindrical Gaussian surface enclosing part of a charged wire. Consider
using a spherical Gaussian surface enclosing the same segment of the wire. (a) Which surface has
the largest electric flux through it? (b) Which surface is more appropriate for calculating the electric
flux? Justify your answer.
40.(I) A small particle carries a 6.0 µC charge. Calculate the electric flux through a Gaussian surface
that is centered on the particle with a radius of (a) 0.04 m (b) 0.08 m.
41.(II) Using the dipole field line pattern shown in figure 28.2, use symmetry arguments to find the
electric flux through a Gaussian surface that encloses both charges. Does your value depend upon
the shape of the surface?
21
Section 28.7
42.(III) Take another look at the dipole shown in Figure 27.22. (a) Using Gauss’ law, calculate the
total electric flux through the x-z plane, the plane bisecting the dipole and perpendicular to the dipole
moment vector. (Hint: calculate the electric flux for each charge separately and apply the principle
of superposition.) (b) Calculate the flux explicitly by integrating the electric field in Equation 27.8
over the plane.
43.(II) Gauss’ law was originally developed for the case of gravity. (a) By analogy with Gauss' law
for electrostatics, argue that the equivalent expression of Gauss’ law for gravity is
" "
Φ grav = $! g • dA = −4π Gmenc ,
"
where g is the gravitational field as defined in Section 27.1, and menc is the mass enclosed by the
Gaussian surface over which the integral is taken. (b) Is menc analogous to positive charge or to
negative charge?
44.(II) A system of charges can be divided up into two or more groups. For example six positive
charges and two negative charges can be grouped that way, or as 4 positive charges and (two
positive and two negative charges). Or a region of empty space can be thought of as composed of a
uniform positive charge distribution superimposed on a uniform negative charge distribution of the
same magnitude of charge density. Use the superposition properties of electric field to argue that the
electric flux through any Gaussian surface is the sum of the fluxes through the same surface
produced by individual groupings of the enclosed charge.
45.(III) There is a Gauss’ Law for gravity analogous to Gauss’ Law for electric charges. (a) If the
electric flux through a closed surface is proportional to the enclosed charge, what would the
gravitational flux be proportional to? (b) Write an appropriate equation that states Gauss’ Law for
gravity.
46.(I) A particular closed surface has four sides. The electric flux through side 1 is +5.0 Nm2/C,
through side 2 is +8.0 Nm2/C, and through side 3 is -9.0 Nm2/C. (a) If there is no net charge
enclosed by the surface, what is the flux through the fourth side? (b) If the flux through the fourth
side were -8.0 Nm2/C, how much charge would be enclosed by the surface?
47.(I) A particular Gaussian cylindrical surface encloses no charge. 20 field lines pass into the
cylinder through one end cap and 16 emerge from the other end cap. What can you conclude about
the other field lines?
48.(I) Consider a line of charge with a finite length L and a uniform charge distribution. Comment
on whether Gauss’ Law is appropriate (and if so state the appropriate Gaussian surface) for
calculating the electric field at the following points: (a) a point r<<L from the midpoint of the line,
(b) a point r = L from the midpoint of the line, (c) a point r >> L from the line.
49.(II) The rectangular sheet of length L and width W shown in Figure 28-0 is positioned with its
center a distance ½W below an infinite line of charge that has a uniform linear positive charge
density '. (a) If the long sides of the sheet are parallel to the line of charge, what is the magnitude of
22
the electric flux passing through the sheet? (b) If the sheet is moved to a distance of ¼W from the
line does the magnitude of the flux increase, decrease, or stay the same?
Figure 28-0
50.(I) Three particles of charge +q are located at the corners of an equilateral triangle of side a. The
triangle is centered on the origin. (a) What is the electric flux through a spherical Gaussian surface
of radius a/4, also centered at the origin? (b) What is the electric flux through a spherical Gaussian
surface of radius 2a, also centered at the origin?
51.(I) Three particles of charge +q are located at the corners of an equilateral triangle of side a. The
triangle is centered on the origin. (a) Draw the electric field line diagram for this configuration of
charge. (b) What is the electric flux through a Gaussian surface containing one of the charges? (c)
What is the electric flux through a Gaussian surface containing two of the charges?
52.(I) The net charge on a conductor is zero. (a) Must the charge density be zero at every point on
the surface of the conductor? Why or why not? (b) Draw a figure illustrating your answer to (b).
53.(II) A small particle carries a -3.0 µC charge. A cubical Gaussian surface, 3 cm on each side, is
centered on the particle. (a) If you were to draw a field line map with 4 lines per micro-Coulomb,
how many field lines would pass through the surface? (b) Neglecting corners and edges, how many
field lines would pass through each face of the cube? (c) What is the electric flux through the cube?
(d) What is the electric flux through a single surface of the cube? (e) Which of these answers would
change if the particle were moved 1 cm in the positive x direction?
54.(II) In a cubical region of space located with one corner at the origin, and one edge along each of
!"
the positive axes, an electric field of the form E = bx 2 x% exists. (a) Draw a sketch of this region. (b)
How much charge is contained within this region?
55.(II) A particle carrying a positive charge +2Q is at the origin of a standard coordinate system. A
spherical surface of radius 5.0 cm has its center on the x-axis at 7 cm. (a) Sketch electric-field lines
for the particle. (b) Do any lines enter the spherical surface? (c) What is the net electric flux due to
the point charge through the spherical surface?
Section 28.8
56.(III) You are given the job of creating a uniform electric field for an experiment. You are
somewhat mystified because all have to work with is styrofoam ball that has a uniform charge
distribution throughout. However, inspiration strikes. You take the foam ball, cut it in half, and you
scoop out a hemispherical hole out of each half so that putting the two ball halves together results in
a spherical cavity off center inside the spherical foam ball. You know that the cavity can be
23
described as a region composed of equal amounts of uniformly distributed positive and negative
charges, giving it a uniform charge of zero. By using the principle of superposition, you are able to
calculate the electric field inside the cavity using vector addition of the superimposed fields.
c
a
b
Figure 28-0
57.(II) A solid nonconducting sphere of radius R, centered at the origin, has a total positive charge
+Q distributed uniformly throughout its volume. A particle with an unknown charge is located on
the x-axis at x = +2R. The net electric field from the sphere and particle is zero for the point x = +¼R
on the x-axis. (a) What is the charge of the particle? (b) At what other locations on the x-axis is the
net electric field equal to zero?
58.(II) A particle with an unknown charge is placed at the center of a conducting spherical shell. The
electric field outside of the shell is 3k|Q|/r2, directed toward the center. The total charge on the inner
surface of the shell is –2|Q|. (a) What is the net charge on the shell? (b) What is the charge on the
particle at the center of the shell?
59.(II) A point object with a positive charge of +Q is located at the center of one or more concentric
spherical shells. The shells are either conducting or nonconducting with uniform charge density.
Figure 28-0 is the graph of electric flux passing through Gaussian spheres, centered on the object, as
a function of the distance r from it. (a) What is the minimum number of spherical shells required to
produce the graph? (b) Describe each shell completely, specifying its net charge, its inner and outer
radius, and whether it is conducting or nonconducting.
Figure 28-0
60.(III) Can the graph in the previous question be produced by a system consisting only of charged
24
point objects? If not explain why not. If so give a description of the system containing the minimum
number of charged point objects.
61.(III) A point object with a charge of +4Q is located at the origin, and a uniformly charged
nonconducting solid sphere of radius R and total charge +Q is located at x = +6R on the x-axis. (a)
At what locations on the x-axis is the net electric field equal to zero? (b) If the nonconducting sphere
was replaced by a conducting sphere of the same radius and total charge, would the answers to part
(a) change? Explain.
62.(II) A charge Q is uniformly distributed throughout a non-conducting solid sphere of radius R.
(a) What is the electric field strength as a function of the distance r from the sphere's center for both
inside and outside the sphere. Do your expressions agree at r=R? (b) Graph the strength of the
electric field as a function or r.
63.(II) The volume charge density inside a solid non-conducting sphere of radius R is #(r)= #1r. (a)
Calculate the electric field as a function of the distance r from the sphere's center for both inside and
outside the sphere. (b) Repeat for a volume charge density of #o(1 –½R/r). (c) Repeat for a volume
charge density of #o(1–R/r).
64.(III) You are a crewmember of the space station orbiting Earth. To investigate some physics you
vaguely remember from your college days, you drill a small hole all the way through a large solid
non-conducting sphere of radius R that has a negative charge –Q evenly distributed throughout its
volume. You carefully make sure the hole goes right through the sphere's center. You go outside on
a spacewalk with it and release a small positively charged pellet at one entrance to the drilled hole.
You are gratified to find that the pellet oscillates with the frequency you calculated beforehand
based on Gauss' law and your knowledge of simple harmonic motion.
65.(I) A hollow sphere of radius 10 cm has a uniform surface charge density of 10 nC/m2. Find the
electric field (a) 2 cm, (b) 9 cm, and (c) 11 cm from the center of the sphere.
66.(I) A solid sphere of radius 10 cm is prepared such that it has a uniform volume charge density of
250 nC/m2. Find the electric field (a) 2 cm, (b) 9 cm, and (c) 11 cm from the center of the sphere.
67.(II) A solid sphere of radius R, centered on the origin of a coordinate system, is prepared so that it
has a uniform volume charge density of ρ. A spherical cavity of radius½R containing no charge is
located inside the sphere, with its center at x=¼R. Show the electric field in the cavity
is Ex = 112 ρ R / ε o . (Hint: use two spheres in superposition, one with +ρ and one with -ρ.)
68.(II) A small sphere carrying a charge of 30.0 nC is located at the center of a neutral conducting
spherical shell of inner radius 10 cm and outer radius 15 cm. (a) What is the surface charge density
on the inner surface of the conducting shell? (b) What is the surface charge density on the outer
surface of the conducting shell? (c) Find the electric field as a function of r, the distance from the
center of the shell.
69.(II) A solid conducting sphere 6 cm in radius carries a charge of 5.0 nC. A conducting spherical
shell of inner radius 10 cm and outer radius 12 cm carries a total charge of -4 nC, and is concentric
with the sphere. (a) Draw the electric field lines for the situation, clearly marking the location of all
charge on your diagram. (b) Calculate all charge densities. (c) Find the electric field as a function of
25
r, the distance from the center of the sphere.
70.(II) A nonconducting spherical shell of inner radius ½R and outer radius R, is prepared such that
it has a uniform volume charge density of ρ. Find the electric field as a function of r, the distance
from the center of the sphere.
71.(II) A conducting spherical shell of inner radius 10 cm and outer radius 12 cm has a particle
carrying a 3.0 nC charge placed at its center. (a) Draw the electric field line map for the situation,
clearly marking the location of all charge on your diagram. (b) Calculate all charge densities. (c)
Find the electric field as a function of r, the distance from the center of the sphere. (d) Draw an
electric field line map for the situation, if the particle is displaced by 3 cm from the center.
72.(II) A nonconducting sphere of radius b is prepared such that it has a charge density ρ0 for r<a,
and ρ0r, for a<r≤b. Find the electric field as a function of r, the distance from the center of the
sphere.
73.(III) A nonconducting solid sphere of radius R is prepared such that is contains a volume charge
density ρ given by ρ0 for r≤ ½R and 2ρ0(1-r/R) for ½R ≤ r ≤ R. (a) Determine the total charge Q on
the sphere in terms of ρ0 and R. (b) Find the electric field as a function of r, the distance from the
center of the sphere in terms of R and Q. (c) Show that the electric field is continuous at the
boundaries.
Section 28.9
74.(III) Electrical power lines are often not insulated, and the electrical field strength of the
surrounding air is about 3×106 N/C. Above that field magnitude, the air dissociates into charged
particles, causing a breakdown and a big spark. You’ve been given a sample of wire, and you
measure its diameter to be 1.66 cm. You figure you’d better build in a safety factor of 3 to account
for humid days and so forth, and you’re trying to figure out what the maximum linear charge density
the wire can safely support.
75.(II) An infinitely long nonconducting solid cylinder of radius R has a nonuniform but radiallysymmetric charge distribution. The volume charge density is given by ρ(r) = c/r, where c is a
positive constant and r is the distance from the axis of the cylinder. (a) What is the total charge in a
section of the cylinder of length L? Write an expression for the electric field for (b) r <R; (c) r >R.
76.(II) An infinitely long line of charge with a uniform linear charge density + is parallel to the yaxis and passes through the x-axis at x = –d as shown in Figure 28-0. The electric field at the origin
from this line of charge has a magnitude Eo directed to the right. A second infinite line of charge,
parallel to the first and with a uniform charge density, passes through the x-axis at x = +3d. The net
electric field at the origin from the two lines of charge has a magnitude of 2Eo. What is the charge
density of the second line of charge? Find all the possible answers.
26
Figure 28-0
77.(III) As drawn in Figure 28-0, an infinitely long line of charge with a uniform charge density is
placed along the y-axis, creating an electric field of magnitude E directed to the right at the point x =
d on the x-axis. A charged point object is then placed on the x-axis at x = 2d. Its contribution to the
electric field at x = d is a field of magnitude 2E directed to the left. (a) At what locations on the xaxis is the net electric field equal to zero? (b) What is the magnitude of the field produced by the line
of charge at these locations?
Figure 28-0
78.(II) Figure 28-0 shows a cross-sectional view through a system consisting of an infinite line of
charge in a non-conducting material with a uniform charge per unit length, surrounded by a
conducting cylindrical shell of inner radius R and outer radius 2R. The electric field associated with
the system is indicated by the field lines in the diagram. For a finite length L of the system there is a
total negative charge of –Q on the inner surface of the conducting shell. For that same finite length
determine the charge on (a) the line of charge, and (b) the outer surface of the shell. (c) In the same
finite length what is the net charge on the shell? (d) What is the ratio of the surface charge density on
the inner surface of the shell to the surface charge density on the outer surface?
27
Figure 28-0
79.(II) A system is composed of two infinite concentric cylinders. Each cylinder is either conducting
or nonconducting with uniform charge density. The outer cylinder is a cylindrical shell, and the
inner cylinder is either solid or a shell. The graph in Figure 28-0 shows the magnitude of the electric
field as a function of the distance r from the axis of a system with cylindrical symmetry. At all
values of r the electric field is either zero or points away from the axis. (a) Deduce whether the
cylinders are conducting or nonconducting and give the relevant radii. (b) What is the ratio of the
charge on the inner cylinder to the charge on the outer cylinder?
Figure 28-0
80.(II) An amount of charge is distributed throughout the volume of a long non-conducting solid
rod. The rod's radius R is very small compared to its length L. Calculate the electric field inside and
outside the rod as a function of the distance r from the central axis if the volume charge density is (a)
#o; (b) #o(1–2r/R) [hint: superposition].
81.(I) A cylindrical shell of length 10 m and radius 5 cm carries a uniform surface charge density
σ=+9×10-9 C/m2. (a) What is the total charge on the shell? (b) Find the electric field far from the
ends at r = 4.9 cm, and (c) r = 5.1 cm, where r is the distance measured from the axis of the shell.
28
82.(I) A cylindrical shell of radius a carries a total positive charge qa uniformly distributed on its
surface. A larger cylindrical shell of radius b is
!" concentric with the first and carries a charge qb
uniformly distributed on its surface. (a) Find E in all regions. (b) If qa = 5 nC, what should qb be for
the electric field to be zero for r > b? (c) Sketch the electric-field lines for this situation.
83.(II) A long thin wire runs down the axis of a cylindrical conducting shell of length 10 m, inner
radius 5 cm, and outer radius 7 cm. The wire carries a uniform linear charge density of 1.5 µC/m. (a)
Neglecting edge effects, what is the surface charge density on each surface of the shell? (b) Find the
electric field far from the ends at r = 4.9 cm, (c) r = 5.1 cm, and (d) r = 10 cm, where r is the
distance measured from the wire.
84.(III) How far must you be along the perpendicular bisector of a 25-cm line of charge with a total
charge of 30 nC before the infinite wire approximation gives you an answer with an error of more
than 5%?
85.(II) A solid nonconducting cylinder of length 10 m and radius 5 cm carries a uniform volume
charge density +9×10-9 C/m3. (a) What is the total charge within the cylinder? (b) Find the electric
field far from the ends at r = 4.9 cm, and (c) r = 5.1 cm, where r is the distance measured from the
axis of the shell.
86.(II) A think cylindrical shell of radius a carries a total negative charge qa uniformly distributed
on its surface. A larger cylindrical shell of radius b is concentric with the first and carries a charge qb
!"
uniformly distributed on its surface. (a) Find E in all regions. (b) If qa = -5 nC, what should qb be for
the electric field to be zero for r > b? (c) Sketch the electric-field lines for this situation.
87.(II) A power line can be approximated as an infinite line charge of uniform linear charge density.
Assume λ= - 1.5 µC/m and that the line is parallel to the earth's surface at 9 m. A point charge of 2.0
µC is located directly 3 m below the line. Find the electric field 1m below (-y) and 0.5 m to the right
(+x) of the charge. Ignore the electric field produced by the Earth itself.
Section 28.10
88.(II) Near the surface of the Earth, the electric field has a magnitude of 150 N/C and is pointed
downward. (a) Treating the surface of the Earth as a conducting sphere, calculate the surface charge
density necessary to produce this field. (b) Ignoring any other possible charge sources, how far
above the surface of the Earth must you go before the electric field would differ from the
approximation in (a) by 2%?
89.(III) Three nonconducting infinite planar sheets are parallel to the y-z plane. Each sheet has a
uniform surface charge density. The first sheet, with a negative surface charge density –σ, passes
through the x-axis at x=1 m. The second sheet has an unknown surface charge density and passes
through the x-axis at x=2m. The third sheet has a negative surface charge density -3σ and passes
through the x-axis at x=4 m. The net electric field due to the sheets is zero at x=1.5 m. (a) What is
"
the surface charge density on the second sheet? If the electric field at x=0 is Eo , what is the electric
field at (b) x=–2 m (c) x=3 m, and (d) x=6m?
29
90.(II) Consider the following four cases in an x-y-z coordinate system. Case A: a charged point
object at the origin; Case B: a charged conducting solid sphere of radius R, centered at the origin;
Case C: a uniformly charged nonconducting solid sphere of radius R, centered at the origin; Case D:
a nonconducting infinite sheet of uniform charge density in the y-z plane, shown in cross-sectional
view. In each case the electric field at x=R is the same. Rank the magnitudes of the electric fields for
the four cases from largest to smallest at (a) x=2R (b) x=½R.
Figure 28-0
91.(II) A nonconducting infinite sheet of uniform positive charge density σ lies in the x-y plane. As
shown in Figure 28-0, a circular piece of the sheet with radius R has been removed. (a) What is the
electric field at the center of the circle? (b) Use superposition to estimate the electric field at a
distance z = 4R above the center of the circle.
Figure 28-0
92.(II) An electron is fired toward a nonconducting infinite sheet of uniform charge density from a
distance of 3.00 m. The electron has an initial speed of 400 m/s, travels along a line perpendicular to
the sheet, and comes to rest momentarily at a distance of 1.00 m from the sheet before reversing
direction. (a) What is the charge density on the sheet? (b) Given the same initial velocity, from what
distance should the electron be fired if it is to just reach the sheet?
93.(I) Two infinite vertical planes of charge are parallel to each other and 5cm apart. (a) Find the
electric field in all regions of space when each plane has a uniform surface charge density σ = +3
nC/m2. (b) Find the electric field in all regions of space when the left plane has a uniform surface
charge density σ= +3 nC/m2 and that of the right plane is σ = -3 nC/m2. (c) Draw a diagram showing
the location of the planes, labeling the regions and drawing the field lines for each of the situations
above.
94.(I) An infinite planar charged sheet has a uniform surface charge density, σ = +4.0 nC/m2 and is
located in the y-z plane. Another infinite planar charged sheet is parallel to the first and carries a
uniform surface charge density, σ = -8.0 nC/m2. It is 4m to the right (+x) of the first sheet. (a) Find
the electric field in all regions of space. (b) Draw a diagram showing the location of the planes,
30
labeling the regions and drawing the field lines for each region.
95.(II) A charge of 6 nC is placed uniformly on one side of a square conducting slab of side 20 cm
and thickness 0.5 mm located in the y-z plane. (a) Draw a diagram showing the location of the
charge and the electric field close to and inside the sheet when it!" is in equilibrium. (b) What is the
surface charge density σ, neglecting edge
!" effects? (c) What is E just to the right and just to the left
of the conducting sheet? (d) What is E inside the conducting sheet?
96.(II) A planar nonconducting charged slab of thickness 2a has a uniform volume charge density
ρ0x/a. It is located centered on the y-z plane and parallel to it. (a) Find the electric field in all
regions of space as a function of x, the distance from the y-z plane. (b) A second nonconducting
charged slab of thickness 2t is now placed parallel to the first sheet such that it cuts the x-axis at x=d.
This second sheet has a uniform volume charge density ρ0. Find the electric field in all regions of
space as a function of the distance from the y-z plane.
97.(II) A particle of charge q and inertia m is placed above an infinite sheet with surface charge
density σ. The sheet carries the same sign charge as the particle. What is the acceleration of the
particle as a function of its distance d above the sheet.
98.(III) An infinite planar nonconducting sheet has a uniform surface charge density, σ1 = +130.0
nC/m2 and is located in the x-z plane. Another infinite planar nonconducting sheet carries a uniform
surface charge density, σ2 = 90.0 nC/m2, and intersects the x-z plane at the z-axis making an angle of
30° with the first sheet. (a) Draw a diagram showing the end view of the planes in the x-y plane. (b)
Find the expression for the electric field in the region between the planes. (c) Find the value of the
electric field at (3 m, 1 m, 0).
31
Answers to review questions
1. No, in both cases. Consider an isolated system with a positively charged particle in it. The field
lines will emanate from the positively charged particle and terminate nowhere inside the system.
Similarly, an isolated system with a negatively charged particle will have field lines that terminate at
the particle, but don’t necessarily originate inside the system.
2. Some lines can be totally inside a boundary, but there will be some field lines that loop to
infinity. Consider the dipole arrangement shown in Figure 28.2. If you make any Gaussian surface
enclosing the entire dipole, there are still field lines outside the boundary. But do notice that an
equal number of lines that penetrate outward equals the number that penetrate inward.
3. Only initially in most cases. Because it starts at rest, its initial motion will be in the direction of
its acceleration vector, which is tangent the field line. Once it is moving, the direction of the field
(and hence the force) only tells you the direction of the acceleration, that is, the direction of the
change in the velocity, not the velocity itself.
4. No. This is almost the same question addressed in Checkpoint 28.2. Certainly, field lines can’t
cross at a non-zero angle; see the answer to that Checkpoint. The remaining case is one where two
field lines are tangent to each other at a point. It’s possible by construction to have two field lines
apparently converge to a common point. However, it makes no sense to have the field lines diverge
from that common point again, even infinitesimally. The electric field vector at that place points in
one direction or the other, not both. Moreover, the convergence of two field lines at a point would
indicate an infinite field strength there, as is noted in Section 28.2.
5. (b).
6. (a) The electric field is identical at the point of contact. It must be so, because the electric field
can have only one (vector) value at a given point in space. (b) The electric field is equal to the
number of field lines per unit area on the surface of the balloon (or we can choose it to be so by an
appropriate scale of drawing lines), because the electric field lines are always perpendicular to this
surface. This, however, is not true for the box, because the surface of the box is not perpendicular to
the field lines.
7. No. The number of field lines penetrating the closed surface only tells you there is no net charge
inside the surface. There may well be several charged objects inside, but it’s entirely possible that
the charges have opposite sign and that the net charge is zero. A closed surface surrounding a
hydrogen atom, containing one proton and one electron, is a good example.
8. No. What it means is that every outward going field line corresponds to an inward going field
line, so that the total flux through the surface is zero. A good example you can sketch of this is any
finite closed surface surrounding a charge dipole.
9. Since the amount of charge enclosed by the balloon does not change, the total flux through its
surface does not change no matter what goes on outside of the balloon. The pattern of field lines
will change by putting electrons into position outside the balloon, but the flux will not.
10. (a) The field lines separated along the axis of the wire are parallel and don’t spread with radius.
That is, their separation is independent of radius. (b) The field lines separated around the circle
girding the cylinder spread with radius, and their separation is linear with the radius. (c) Combining
the answers from (a) and (b), we conclude that the density of field lines falls inversely proportional
to the first power of the radius, consistent with the Gaussian surface argument posed in the chapter.
Compare this with the spreading of field lines in a spherically symmetric case.
11. While it is in principle possible to find a Gaussian surface over which the electric field from a
dipole is constant in magnitude and perpendicular to its surface, this is harder in practice than just
solving for the field using Coulomb's law as we did in the previous chapter. However Gauss’ law is
still valid, even though not particularly useful in this case.
32
12. Within the metal of the box, the electric field must be zero. In the presence of an external
electric field, charges in the metal redistribute themselves on the outer surface of the metal to cancel
the electric field inside the metal. If you imagine a Gaussian surface within the metal of the box
itself surrounding the interior "cavity", there is no field anywhere on this surface giving a flux of
zero. This means there is no net charge inside this Gaussian surface. Because the external field
cannot "get through" the metal enclosure and because no net charge resides on the surface, it is
plausible that the field inside the cavity is zero. Note though: just because no net charge can reside
on the inner surface of enclosure does not mean that regions of opposing charge could not form in
some places on the inner surface, producing local electric fields. Why this does not happen requires
more physics than covered in this chapter.
13. Because the electric field inside of a metal must be zero (otherwise charges would freely move),
the electric field lines produced by the charged dabs of paint must lie completely exterior to the can,
traveling from one end of#&the
#& can to the other in the space around the can.
14. The scalar product, E ⋅ A , between the electric field and area vectors keeps track of the direction
of the electric field. We’ve already defined flux to be positive when field lines point outward from
the surface, and this scalar product would be positive in this case only if we defined the area vector
to point outward. For cases where the electric field points inward on the surface, cosθ is negative,
and the flux contribution is negative.
15. The two are proportional to each other. The electric field line flux has an arbitrary scale,
depending on how many lines we want to attribute to a certain amount of charge. The electric flux
has that scale fixed by physical definition according to Equation 28.2 (or more generally, by
Equation 28.4).
16. No. Checkpoint 28.20 makes a similar point. Gauss’ law is a direct consequence of the 1/r2
behavior of the electric field from a charged particle. Any observed deviation from this behavior
would render Gauss’ law invalid.
17. The ratio of the charges should be the same as the ratio of the volumes of the two spheres, since
the charge distribution is uniform. Therefore we can write qenc/Q = Venc/V = (4πr3/3)/(4πR3/3), or
qenc = Q (r/R)3.
18. The field is zero for the cases where a Gaussian sphere with radius r encloses zero net charge.
This happens for r<R and for r>2R.
19. The problem with using Gauss’ law here would be the error in assuming that a cylindrical
Gaussian surface reflects the symmetry of the electric field. Indeed, the topology of the field is not at
all obvious, although we can guess that it has a component along the axis of the rod, from the more
highly charged end towards the more weakly charged end. If the field doesn’t have the same
symmetry as the surface, then any of the usual assumptions we make in using Gauss’ law (for
example, that the field is constant and perpendicular to the surface over a portion of the surface, for
example) are invalid. Note Gauss’ law is still perfectly valid in this case, but its usefulness as a
calculation tool is lost here.
20. It doesn’t matter how long (h) the cylindrical surface is. This is apparent in the later stages of
the calculation, where the variable h cancels. This is a general feature. You have to define the
variable to completely specify the calculation, but if it is not important for the physics of the
problem (and details of an imaginary surface certainly should not be) then that variable should
cancel out in the end. If it does not, it’s likely you’ve made a mistake somewhere.
21. No, the puzzle piece works just as well. First of all, the sides of the puzzle piece (corresponding
to the walls of the cylinder in Figure 28.24) are still perpendicular to the charged sheet. The electric
field vectors on the walls of the!" puzzle piece lie in the surface, and so the electric flux on this portion
!"
of the surface is zero because E and d A are perpendicular. On the flat ends, the shape is irrelevant
and the area A of the ends eventually cancels from the calculation, as you can see in Equation 28.27.
33
In this case, the puzzle piece has just enough symmetry to reflect the symmetry of the field. The
cylinder used in Section 28.10 has more symmetry than needed, but more simple than the puzzle
piece to specify. Notice, too, that the thickness of the piece never enters into the calculation because
an infinite sheet looks the same from any height.
34
Chapter 29
Electrostatic work and energy
Review questions
Answers to these questions can be found at the end of this chapter.
Section 29.1
1. What is electrical potential energy?
2. In what way is electrical potential energy more complicated than gravitational potential energy?
3. How does the electrical potential energy of a dipole, near a stationary charged object, depend on
its orientation?
Section 29.2
4. What is electrostatic work?
5. When a charged particle moves along a path in an electrostatic field, on aspect(s) of its path does
the electrostatic work depend?
6. What is electric potential difference?
Section 29.3
7. What are equipotentials?
8. How are equipotentials spatially related to electric field lines?
9. How does the electric potential vary along electric field lines?
10. What characteristic of a charged particle determines whether it tends to move from a region of
higher potential towards a region of lower potential, or vice versa?
Section 29.4
11. What is the general mathematical expression for the electrostatic work done on a charged
particle, moved between arbitrary initial and final points, in the electrostatic field of a second
charged particle?
12. What is the simple mathematical relationship between electrostatic work and change in
electrostatic potential energy?
13. How can the potential energy of a system of two charged particles be defined, when only
differences in potential energy are physically meaningful?
14. What is the generalization of the electrical potential energy for any system of charged particles?
Section 29.5
15. What is the unit of potential difference?
16. What are two common ways of choosing a zero reference point so that electrostatic potential can
be assigned to all points?
17. What is the definition, in words, of the potential difference in an arbitrary electrostatic field?
18. What is a simple mathematical expression for the potential from a system of charged particles?
19. Is it possible to extract energy from an electrostatic field by going around a loop? Why?
Section 29.6
1
20. How can the expression for the potential from an object with a continuous distribution of charge
be obtained from the expression for the potential from a system of charged particles?
21. Is there any advantage to finding the potential from a distribution of charge, rather than the
electric field? Why?
22. Can the electrostatic field line pattern from charged objects be determined from a map of the
equipotentials? Why?
Section 29.7
23. What is the mathematical relation between the electrostatic field and the potential?
2
Developing a feel
Calculate or estimate the following quantities:
1. The electrostatic work done while assembling a hydrogen atom. (F, X, A)
2. The electrostatic work done by an external agent to bring two protons together to form a helium
nucleus. (F, M)
3. The potential change across the terminals of a car battery. (V)
4. The potential change needed to give a resting electron 1 Joule of kinetic energy. (Q, A)
5. The electric potential at a point located 3 cm perpendicular distance from the end of a glass rod
that has been charged by rubbing. (C, R, W)
6. The surface charge density on a manhole cover charged with a car battery. (D, I, N, S)
7. The potential change between floor and ceiling needed to “float” a proton in your room. (Z, U, P,
K)
8. The potential change required to stop an electron whose initial speed is one million m/s. (H, Q, A)
9. The initial speed of an α particle that can approach to within 7 x10-15 m of the center of a gold
nucleus. (B, G, L, Q)
10. The minimum work needed to assemble eight protons, one at each corner of a sugar cube. (E, J,
O, T, Y)
Hints:
A. What is the charge?
B. What is the potential change from the initial to final location?
C. What is the effective charged length of a glass rod in the laboratory?
D. How can this be done physically and what model charged object is it like?
E. How much work is required for the first proton, placed at the lower left front corner?
F. What is the initial separation?
G. What is the charge of a gold nucleus?
3
H. What is Kinitial ?
I. What equipotential is at 12 volts?
J. How much work is required for the second proton, if placed at the lower left rear corner?
K. Which surface must have positive charge?
L. What is the “final” speed?
M. What is the final separation?
N. What is the radius of a manhole cover?
O. How much work for the third proton, placed at the lower right rear corner?
P. What is the floor-ceiling distance?
Q. How are ∆V and ∆K related?
R. What value of charge will the rod have?
S. How does this relate to the charged disk example in the text?
T. Do you see a trend in the number of terms?
U. What value of E is required?
V. What value is the voltage of a car battery?
W. Where is the potential equal to zero?
X. What is the final separation?
Y. How many terms will be added together to get the total work done?
Z. What forces must balance to “float” the proton?
Key: A. Each particle has charge of magnitude 1.6x10-19 C. B. Zero, far from gold nucleus, to V =
kq/r at closest approach. C. ~0.2 m. D. Place cover on insulating stand, attach positive battery
terminal to cover, negative battery terminal to ground, producing a charged disk like Figure 29.20.
E. None, because there is no potential change to move through. F. Essentially infinite. G. 79e
~1.3x10-17 C. H. ~5x10-19 J. I. On the surface of the cover, in particular where z→0 in Figure 29.20.
J. q∆V = kq2/r, where r is cube edge and q is proton charge. K. The floor. L. Zero. M. nuclear
4
separation ~2x10-15 m. N. ~0.3 m. O. q∆V = kq2/r + kq2/√2r. P. 8 feet or ~2.5 m. Q. Assuming only
electrostatic work, q∆V = ∆K for this particle. R. ~10 µC. S. Both upper and lower surfaces will be
charged, so there are two disks to superpose. T. Each new charge added requires one term for each
charge already there. U. E = mg/q ~1x10-7 V/m. V. 12 volts. W. At a large (infinite) distance from
the rod. X. ~5x10-11 m. Y. Twenty-eight terms, counting each pair of charges once: twelve edge
terms, twelve face diagonals, and four body diagonals. Z. Gravitational and electrical interactions.
5
Worked and guided problems
These examples involve material from this chapter, but are not associated with any particular
section. Typically, an example that is worked out in detail is followed immediately by an example
whose solution you should work out by following the guidelines provided.
WP 29.1: Relocating a charge. Laser printers use static electric charges to adhere small particles
of toner to paper. The toner is subsequently melted to the paper, making a permanent image.
Consider a simple situation of moving a charge amongst other ones. Four charged particles form a
square with side length a=6.9 !m. Three of the particles have charge Q and the other has charge 2Q, where Q = 3.9x10-5 C. How much work must be done by an external agent on the particle
opposite the particle with charge -2Q to move it around the outside of all the other charges and
place it at the center of the original square?
!Focus problem
We will begin with a physical sketch of the charge arrangement and what we want to do
with the designated charge, which we will label as qm.
Q
-2Q
qm=Q
a
Q
We are not given much information about the path of the particle and the external force that causes it
to move. But we do know that the total change the particle’s total change in kinetic energy is zero
because it starts out with zero kinetic energy and will have zero again when it is place at the center.
Thus the desired work done by the external agent is the negative of the work done by the other
charges during the move, which is what has to be calculated.
!Plan approach
Because the change in kinetic energy is zero in this process, we can use
Wnet = Wexternal agent + Wother charges = ∆K = 0
Because the electrostatic forces are conservative, we know that the work they do on the moved
particle equals the negative of the change in electrostatic potential energy. Thus we can use:
6
Wagent = −Wcharges = −(− qmoved ∆Vdue to charges ) = qmoved ∆Vdue to charges
The electrostatic potential due to the three other charges is just given by
1
V=
4πε 0
3
qj
!r
j =1
j
We need determine the change in this potential between the start and end points of the particle’s
motion. We can use this to calculate the change in potential in two different ways: subtract the sum
of contributions of all the charges to the initial potential from the sum of contributions to the final
potential, or find the potential difference for each of the other charges and then sum these
differences. We will do the later because it will allow us to factor out common factors easily.
!Execute plan
The total work done on the particle q is then
qm
1 $ 3 qj 3 qj %
Wagent = qm $+V f − Vi %, = qm
'! − ! ( =
4πε0 +' j =1 rfj j =1 rij ,( 4πε0
=
Q
4πε 0
"& 3
$ 1 1 % #&
)! q j ' − ( *
&- j =1 +' rfj rij (, .&
"& $ 1
$ 1
$ 1
1%
1%
1 % #&
− ( + (−2Q) '
− (+Q'
− (*
)Q ' 1
1
1
+ 2 a/ 2 a,
+ 2 a / 2 a , &.
-& + 2 a / 2 a ,
Wagent =
2Q 2 $ 1
%
− 1(
'
4πε0 a + 2 ,
Putting in numbers
2(3.9 ×10-15C) 2
$ 1
%
− 1(
2
-6
−12 2
'
4π(8.85 ×10 C / nm )(6.9 ×10 m) + 2 ,
= −1.16 ×10−14 J
Wagent =
!Evaluate result
That the magnitude of the work is small is not unreasonable because the charges are small and the
particle was only moved a small distance. The decrease in electrostatic potential is expected because
the distance between the unlike charges decreases the more than between the positive charges in the
move. Consequently, the electrostatic charges do positive work on the particle as it journeys to the
center (i.e., a decrease in potential energy), which would it a net increase in kinetic energy if nothing
prevented it. The work done by the external agent must be negative particle from gaining this
kinetic energy.
7
WP 29.2: Forming and straightening a charged triangle. How much work must you do to bring
two particles of negative charge –Q and one of charge 4Q from large initial separation together to
form an equilateral triangle with side length L? How much work must you then do to straighten
these into a line with the 4Q-particle in the center with distance L between adjacent particles?
!Questions and suggestions
1. After drawing your sketch of the triangle forming, describe the problem in your own words. What
is the problem asking you to find? Are you asked for a qualitative or a quantitative answer?
2. How can you relate the work you need to do in forming the triangle to the potential energy of the
configuration?
3. What kind of drawing do you make to show the straightening process?
4. How much total work on the moved particle? How can you find the work you do compared to
that of the two other charges?
5. What formula gives you this work?
WP 29.3: Potential change in a non-uniformly charged rubber ball. A rubber ball of radius R
contains a total charge Q. The charge density increases linearly from the center. What is the
change in potential as one goes from the ball’s surface to a point a distance d from the center?
"Focus problem
As usual we begin with a sketch of the situation, drawing in the given elements. (This textbook has
the advantage in that the charge density can be represented with a color gradient that you should
probably not attempt yourself.)
Q
R
d
The change in potential is a line integral of the electrical field from the starting to the ending point.
Because of the symmetry of the situation, the potential will be the same for all pointe at any given
distance from the center, so we just need to do the line integral directly inward from R to d. Also
because of symmetry, the field will be directed radially. We know the strength of the electric field
will be affected by the charge distribution and must find a way to calculate it.
!Plan approach
The first thing to do is figure out what we need. The change in potential energy is given by:
8
f !
!
Vif = − / E (r )"dr
i
This tells us we need the electric field. Based on the work of the chapter, we know we can get the
electric field for at any distance r from the center by using the symmetry of the problem and Gauss’
law:
Q
R
r
!
!
#/ E (r )"dA =
!
E(r)
qenc (r )
εo
All we then need is the charged enclosed qenc(r) within any radius r. We can get this by noting that
the charge density increases linearly with distance x from the center, i.e., #(x)= #ox, and using
qenc (r ) = / ρdV
where dV is a thin spherical shell. The value of #o can be determined by noting that the total charge
Q must be within the ball’s radius R, i.e., Q= qenc(R). We just do these steps in reverse to get the
desired expression.
!Execute plan
First comes the charge enclosed within arbitrary radius r by adding up contributions from thin shells
of radius x and thickness dx.
r
qenc (r ) = / ρdV = / (ρo x)(4πx 2 dx) = πρo r 4
o
From this, we can express #o in terms of the given quantity Q:
qenc ( R) = πρo R 4 = Q
ρo =
Q
πR 4
Because the electric field must be radially directed by symmetry,
9
!
!
q (r )
"
E
(
r
)
dA
= E (r ) A(r ) = enc
#/
εo
E (r )(4πr 2 ) =
E (r ) =
πρo r 4 Qr 4
=
εo
εo R 4
Q
r2
4
4πεo R
The change in potential is then:
d !
d
!
V (d ) − V ( R) = − / E (r )"dr = − / E (r )dr
R
= −/
d
R
R
d
Q
Q
2
1 3
$
%
r
dr
=
−
r
3
4πε o R 4
4πε o R 4 + , R
V (d ) − V ( R) =
Q
$ R 3 − d 3 %,
12πεo R 4 +
!Evaluate result
If Q is positive, then the electric field will be directed radially outward. As we go into the sphere,
we would be going against the electric field, thereby increasing the electrostatic potential.
Consequently, the change in potential should be positive, which is what our answer says. The
opposite should be true if Q is negative, which is also consistent with our equation. Our equation
says the potential difference is linear in charge, and we would expect the effect to be larger if more
charge is involved. We note that if we make d equal to R, the potential change is zero as would have
to be the case.
WP 29.4: Charged plastic sphere. Negative charge is uniformly distributed through a plastic
sphere. What is the potential difference between the center and a distance r away from the center
inside the sphere? Outside the sphere?
!Questions and suggestions
1. After drawing a sphere, what quantities do you have to define yourself to characterize the physical
situation which were not explicitly given (e.g., radius R of the sphere).
2. How do you determine the potential differences inside the sphere? What quantities do you need
to calculate?
3. How do you determine the charge enclosed within any radius?
4. How do you connect the potential outside the sphere to that within it by matching expressions at a
boundary common to both regions?
5. Does the expression agree with that for a point particle at large distances from the center?
10
WP 29.5: Line ‘o Charge. A Lucite rod of length 2D is rubbed with fur to give it an electric charge
Q. It is then placed on an insulating stand centered on the x axis. Find the electric potential at the
point P located at an arbitrary position (x0, y0) from the rod’s center.
x0
P
y0
Q
2D
"Focus problem
This problem is somewhat similar to the example in this chapter of the Principles Volume for the
electrostatic potential for points along a line perpendicular to a rod’s end. However, we’re now
asked to find the potential at any point in space rod. W will need to determine the contribution to
the potential energy from each little charge segment of the rod because the distance to the designated
point is different for each segment. Then we just integrate the contributions to get the total potential.
!Plan approach
We will need to calculate the electrostatic potential for a charge distribution
V =/
1 dq
4πεo r
where r is the distance from infinitesimal charge element dq. We draw a diagram that shows the
necessary information.
x0
P
x
r
y0
Q
dq
2D
We need to find appropriate values for r and dq for each position x along the rod. The amount of
charge in a little segment of the rod of infinitesimal length dx will be dq = $dx, where $=Q/(2D) is
the linear charge density. We just then integrate from one end of the rod to the other to add up the
11
contribution of each dq to the total potential.
!Execute plan
The charge on the element dq is
dq =
and the distance from the element to P is
Q
dx
2L
r 2 = (x0 − x)2 + y02
Substituting gives the following expression for the potential at P
D
1 Q
dx
V=
/
4πε o 2 D − D ( x0 − x) 2 + y02
The integral can be put into a form that can be looked up with a simple substitution:
u = x0 − x
du = −dx
x −D
− du
1 Q 0
V=
/
4πεo 2 D x0 + D u 2 + y02
Note how the limits of integration have changed to match the change of variable to u. The integral is
identical to the integral in the related example in the Practice Volume, i.e.
V=
=
(
)
x0 − D
1 Q $
− ln u + u 2 + y02 %
,( x0 + D
4πε o 2 D +'
(
) (
)
1 Q $
ln x0 + D + ( x0 + D) 2 + y02 − ln x0 − D + ( x0 − D) 2 + y02 %
'
+
,(
4πε o 2 D
V=
2
2
1 Q 0 x0 + D + ( x0 + D) + y0
ln 2
4πεo 2D 2 x0 − D + ( x0 − D)2 + y02
4
1
3
3
5
Using this result, we can now plot the potential for the rod everywhere in the x, y plane. A contour
plot of the potential looks something like the graph below for a rod of length 1m.
12
!Evaluate result
If P is located at the end of the rod, as in the example, then x0 = D. You should check the above
expression to confirm that one does, indeed, get the same expression as Eq. (29.37) in the Principles
Volume, with the adjustment that the rod is now length 2D. A rather complicated expression – but,
you can now calculate the potential anywhere in the x,y plane. If we are very, very far from the rod,
the rod should look like a point particle and the potential should be that from a point particle of
charge Q: the potential should be zero at infinity. If we let either x0 or y0 or both become infinite in
our expression, the argument of the natural logarithm is unity, and ln(1)=0, yielding the expected
potential of zero.
WP 29.6: Charged ring. A non-conducting flat fiber washer-ring has inner radius RI and outer
radius RO. Its surface charge density %(a) decreases inversely with distance a from the center as
%(a)=%o/a. Determine the electrostatic potential at any point along the axis through its center.
!Questions and suggestions
1. First draw a sketch of the system.
!P
RI
RO
2. What expression can you use to determine the contributions from all the charge elements dq on
the ring?
3. What symmetry of the ring can you exploit to establish a group of charges that are all at the same
distances from the desired point? (See similar case in the Principles Volume.)
13
4. What mathematical quantities can you use to specify this infinitesimal charge group?
5. What distance is this charge group from the point P?
6. What variable due you integrate to account for all charges on the ring and what are the appropriate
limits of integration?
7. Where can you find an analytic expression for the value of the resulting integral?
8. Does the expression of electric field behave properly at large distances from the ring?
WP 29.7: Electric field around a charged rod. Determine the component of electric field that is
parallel to the rod in WP 29.5 for any point in space.
"Focus problem
The potential for any point in space for the charged rod was derived in WP 29.5. Rather than
calculating the electric field from Coulomb's law (as done for various cases in Chapter 27), we can
use the simple relationship between electric field and electrostatic potential:
!
∂V
∂V
E=−
xˆ −
yˆ
∂x
∂y
Using the diagram supplied with WP 29.5, we see that the y-component of the electric field is the
one parallel to the rod.
!Plan approach
We need to calculate Ey = -&V/&y. We have to take the partial derivative of the potential with respect
to y, while holding all other variables constant, i.e., we treat the position x as a constant. The
potential for arbitrary position (x,y) the potential is given by:
V ( x, y) =
1 Q 0 x + D + ( x + D)2 + y2
ln 2
4πεo 2D 24 x − D + ( x − D)2 + y2
1
3
3
5
Taking the derivative of the logarithm’s argument appears daunting, but we can simply things by
using ln(a/b) = ln(a)-ln(b).
!Execute plan
We just plow ahead with the partial derivative:
Ey = −
{
}
1
1
∂V
1 Q ∂
2
2
ln $ x − D + ( ( x − D)2 + y 2 ) % − ln $ x + D + ( ( x + D)2 + y 2 ) %
=
'
(
'
(,
,
+
∂y 4πεo 2D ∂y +
"$ 1 ( x − D)2 + y 2 − 12 2 y % $ 1 ( x + D)2 + y 2 − 12 2 y % #
)
)
1 Q &' 2 (
(−' 2(
( &*
=
)
1
1
4πεo 2D &' x − D + ( ( x − D)2 + y2 ) 2 ( ' x + D + ( ( x + D)2 + y 2 ) 2 ( &
, +
,.
-+
14
1
− 12
"$
2
2 −2
% $
%#
( x + D)2 + y 2 )
(
1 Q & ' ( ( x − D) + y )
(
'
( &*
Ey =
y)
−
1
1
4πεo 2 D & ' x − D + ( ( x − D) 2 + y 2 ) 2 ( ' x + D + ( ( x + D) 2 + y 2 ) 2 ( &
, +
,.
-+
This is rather complicated, but it was much easier to get the electric field component via the
potential than from superposition of electric as was done in Chapter 27. The x-component of
electric is even more complicated, but the procedure for obtaining it is the same. You may wish to
try it.
!Evaluate result
We can check if this result is not unreasonable by looking at some special situations. If the point is
along the y-axis, the electric field in the y-direction should be zero by symmetry. We see that setting
y=0, Ey(x,0)=0 as expected. If the point is along the perpendicular bisector of the rod (x=0), we
should recover Equation 27.23, but noting that the x and y axes are switched and L in that equation is
2D in ours. If we
1
1
1
"$
2
2 −2
% $ D2 + y 2 − 2 % #
" 2D D2 + y 2 − 2 #
(
)
(
)
1 Q &' ( D + y )
1
Q
&
&
&
(−'
(* =
E y (0, y ) =
y)
y)
*
1
1
2
4πε o 2 D & ' ( D 2 + y 2 ) 2 − D ( ' ( D 2 + y 2 ) 2 + D ( & 4πε o 2 D &
y
, +
,.
.&
-+
Q
Q
1
=
1 = k
1
4πε o y ( D 2 + y 2 ) 2
y ( 1 L2 + y 2 ) 2
4
This is, indeed, Equation 27.23. Whew! Success.
WP 29.8: Electric field of an atomic dipole. A thin disk of radius R carries a uniform surface
charge density %. Determine the electric field on the symmetry axis through its center, using
expression for its potential along that axis.
!Questions and suggestions
1. Find the expression for the potential for this system in the Principles Volume.
2. What is the relationship between this potential and electric field.
3. What component(s) of the electric field is zero by symmetry?
4. What quantities are considered constants when taking the appropriate partial derivative?
5. Is the result obtained the same as that derived in Chapter 27?
15
Questions & problems
Section 29.1
1.(II) A proton, a deuteron (a heavy hydrogen nucleus, consisting of a proton and a neutron), and an
alpha particle (a helium nucleus, consisting of two protons and two neutrons) are all accelerated,
from rest, in the same uniform electric field, through the same distance. Compare their final (a)
kinetic energies, (b) momenta, (c) speeds, and (d) times to cover the distance. (e) How do the
changes in electrical potential energy compare?
2.(II) A proton, a deuteron (a heavy hydrogen nucleus, consisting of a proton and a neutron), and an
alpha particle (a helium nucleus, consisting of two protons and two neutrons) are all accelerated,
from rest, in the same uniform electric field, for the same time intervals. Compare their final (a)
kinetic energies, (b) momenta, (c) speeds, and (d) distance traveled. (e) How do the changes in
electrostatic potential energy compare?
3.(II) You release three charged balls simultaneously from the same height above the floor. In
addition to the usual gravitational field from the Earth, there is also a uniform electric field directed
down. The balls have equal positive charges, but different masses. Assume that the balls are far
enough apart that they have negligible influence on one another, and that air resistance can be
ignored. (a) Which ball has the highest speed when it reaches the floor? (b) If all three were
negatively charged, which ball would reach the ground first?
4.(I) What orientation of an electric dipole in a uniform electric field has the greatest electrostatic
potential energy? What orientation has the least? Explain.
5.(I) Three objects are moved, one at a time, from point A to point B in a uniform electric field
(Figure 29-0). Object 1 has a charge of +Q (with Q a magnitude of charge) and a mass of m; object 2
has a charge of +Q and a mass of 2m; object 3 has a charge of –Q and a mass of m. Rank the objects
based on their change in electrostatic potential energy, from largest to smallest.
Figure 29.0
6.(II) Consider an isolated system of two identical electric dipoles as in Figure 29-0. For which
orientation is the electrostatic potential energy the least? Explain.
16
Figure 29.0
7.(II) A dipole, consisting of two objects with charges of +q and –q separated by a distance d, is
rotated by 180° in a uniform electric field as shown in Figure 29-0. What is the change in
electrostatic potential energy associated with this rotation?
Figure 29.0
8.(I) As sketched in Figure 29-0 three identical positively charged objects (A,B,C) are launched
with the same speed from the same point above a sheet of a negative charge that is producing a
uniform electric field. The directions of their initial velocities are shown in the diagram. Assuming
the objects do not interact with one another, rank the objects based on the speed they have when they
reach the sheet of charge.
Figure 29.0
Section 29.02
9.(I) In a region of electrostatic field, Path 1 between points A and B is twice as long as Path 2. If the
electrostatic work done on a negatively charged particle going along Path 1 is W1, how much work
17
is done going along Path 2?
10.(I) The electrostatic work done on a particle of charge Q, going from points A to B, is w. How
much electrostatic work is done on a particle of charge –2Q, going from B to A?
11.(I) An amount of electrostatic work Ws is done on a charged particle going on a straight path, one
centimeter in length, from points A to B. You next apply a force to move the particle from A to B
along a 2 cm-long, curvy path. How much electrostatic work is done compared to the straight path?
12.(II). Electrostatic work w is done on a charged particle, going from point A to point B, in a region
where no other fields are significant. You next apply a force to move the particle back to point A,
increasing its kinetic energy by an amount equal to 2w. How much work did you do?
13.(II) Three points (A, B, and C) form the vertices of a triangle, in a region of non-uniform
electrostatic field. The electrostatic work done on a particle of charge Q, going from A to B, is WAB,
and that done going from A to C is –WAB/2. How much electrostatic work is done on the particle
going from B to C?
14.(I) An electron travels from point A to point B in an electrostatic field, and gains kinetic energy.
(a) Is the electrostatic work done positive, negative, or zero? (b) Is the difference of potential
between points A and B positive, negative, or zero?
15.(I) A proton travels from point A to point B in an electrostatic field, and gains kinetic energy. (a)
Is the electrostatic work done positive, negative, or zero? (b) Is the difference of potential between
points A and B positive, negative, or zero?
16.(II) In the presence of an electrostatic field, you find out you must do positive work on an
electron to move it from point A to point B without changing its kinetic energy. (a) Has the electron
increased, decreased or left unchanged its electrostatic potential energy? (b) Is the difference of
potential between points A and B positive, negative, or zero?
17.(II) In the presence of an electrostatic field, you find out you must do positive work on an proton
to move it from point A to point B without changing its kinetic energy. (a) Have you increased,
decreased, or left unchanged, its electrostatic potential energy? (b) Is the difference of potential
between points A and B positive, negative, or zero?
18.(I) Points A and B are on the same electric field line. If the potential difference between points A
and B is positive is the field directed from A to B or from B to A?
19.(II) An electron moves from point A to point B, only under the influence of an electrostatic field.
If the potential difference between A and B is negative, (a) does the potential energy increase,
decrease, or remain the same? (b) Does the particle’s kinetic energy increase, decrease, or remain the
same?
20.(II). A proton moves from point A to point B, only under the influence of an electrostatic field. If
the potential difference between A and B is negative, (a) does the potential energy increase,
decrease, or remain the same? (b) Does the particle’s kinetic energy increase, decrease, or remain the
same?
18
21.(III) Points A, B, C, and D, sequentially define the corners of a square in an electric field (B is
adjacent to A, C is diagonal from A). The potential difference between A and C is the negative of
that between A and B, and twice that between B and D. In terms of that between A and B, what are
the potential differences between (a) B and C, (b) C and D, and (c) A and D?
22.(II). Consider an isolated, uniformly charged spherical shell, with charge positive charge Q and
radius R. Point A is on the shell, point B is a distance 2R from the center, point C is a distance R/2
from the center, and point D is at the center. (a) Is the potential difference between A and B positive,
negative, or zero? (b) What about the potential differences between A and C? (c) A and D?
23.(II) Rank the following situations based on their electrostatic work based on the positions A, B, C
shown in Figure 29-0. (a) an object with a charge +Q is moved from A to B; (b) an object with a
charge +Q is moved from A to C; (c) an object with a charge +Q is moved from B to C; (d) an object
with a charge –Q is moved from A to B.
Figure 29.0
Section 29.03
24.(I) Can an equipotential line (or contour line) ever cross itself? Explain.
25.(I) Can you draw an equipotential surface through a point where the electric field vanishes?
Explain.
26.(II) Sketch some equipotential lines for the field line pattern of a electric dipole.
27.(II) Sketch some electric field lines and equipotential lines associated with two identical charged
particles.
28.(I) Some equipotentials surrounding a negatively charged object are shown in Figure 29-0, with
equal potential differences for adjacent lines. (a) In what region is the electric field strongest, and (b)
What is its direction? (c) Which equipotential has the highest potential?
19
Figure 29.0
29.(I). Describe the equipotential surfaces associated with an infinite line of charge having a uniform
charge per unit length.
30.(II) You determine that it takes zero electrostatic work to move a charged object from point A to
point B along a particular path. (a) Can you conclude that points A and B are on the same
equipotential surface? (b) Can you say that the particular path is part of an equipotential surface that
includes A and B? Justify your answers.
Section 29.4
31.(I) Show that the units of electric field, N/C are the same as the units V/m.
32.(II) Two particles carrying a charge of 2 nC are separated by a distance of 2 m. If we imagine
these points make up the base corners of an equilateral triangle, (a) What is the potential relative to
zero at infinity at the third corner of the triangle, point P? (b) How much work is required to bring a
positively charged particle with a charge of 5 nC from infinity to point P if the other particles are
held fixed? (c) Answer parts (a) and (b) if one of the original two particles is replaced by a particle
with a charge of -3 nC. (d) Find the electrostatic potential energy for each of these charge
distributions.
33.(II) Two charged objects are held a distance r apart. The first object has a mass m and a charge
+2q, while the second object has a mass 2m and a charge +q. The objects are released from rest.
Assume that the only force acting on either charge is the electrostatic force from the other charge. (a)
When the objects are in motion what is the ratio of the first object’s kinetic energy to that of the
second object? When the charges are a distance of 4r apart what is the speed of (b) the first object?
(c) The second object? When the charges are very far apart, what is the speed of (d) the first object?
(e) The second object?
20
34.(II) Six particles each carrying a charge of 3 nC are equally spaced along the equator of a sphere
with radius 60 cm that has its center at the origin. (a) What is the electrostatic potential relative to
zero at infinity at the center of the sphere? (b) What is the electric potential relative to zero at
infinity at a pole of the sphere?
35.(I) A particle carrying a charge of 3 nC is at the origin. (a) What is the electric potential relative
to zero at infinity at a point 4 m from the origin? (b) How much work must be done by an outside
agent to bring another particle carrying a 3 nC charge from infinity to r = 4 m, assuming that the first
particle is held fixed? (e) How much work must be done by an outside agent to bring the first
particle from infinity to the origin if the second particle had been placed at r = 4 m first, then held
fixed?
36.(I) Four particles carrying a charge of 3.0 nC are at the corners of a square with 3-m sides. Find
the potential at the center of the square relative to zero at infinity if (a) all the charges are positive,
(b) three of the charges are positive and one is negative, and (c) two are positive and two are
negative.
37.(I) Four particles, each carrying a charge of magnitude 3.0 nC, are at the corners of a square with
3-m sides. Find the electrostatic potential energy for the charge distribution if (a) all the charges are
positive, (b) three of the charges are positive and one is negative, (c) two are positive and two are
negative with like charges at adjoining comers, and (d) two are positive and two are negative with
like charges at opposite corners.
38.(II) An electron and a proton are held in fixed positions on the x-axis, the electron at x = +1 m
and the proton at x = -1 m. (a) How much work is required to bring a second electron from infinity to
the origin? (b) If instead the second electron was given an initial velocity of 500 m/s toward the
origin from the point x = +20 m on the x-axis, would it reach the origin? If so, what would its speed
be at the origin? If not, how close would it come to the origin?
39.(II) A particle carrying a charge of 6 nC is released from rest at in a uniform electric field
!
E = 2 ×103 N/C xˆ (a) What is the change in the potential energy of the particle after it moves 4 m?
(b) What is the kinetic energy at that time?
40.(II) Four objects with charges of magnitude Q, two positive and two negative, are placed at the
corners of a square measuring d on each side. (a) Explain whether the electrostatic potential energy
associated with this set of charges positive, negative, or zero? (b) Demonstrate that your answer to
part (a) is correct by calculating the electrical potential energy.
Section 29.5
41.(I) Four possible paths for a positively charged object are shown below. Rank the paths based on
the electrostatic work associated with the motion of the object to reach their respective end points on
the 3-V equipotential.
21
Figure 29.0
42.(I) An infinite plane of charge has surface charge density 3.5 nC/m2. How far apart are the
equipotential surfaces whose potentials differ by 100 V?
43.(I) A particle carrying a charge +9 nC is at the origin. Taking the potential to be zero at infinity,
locate the equipotential surfaces at 20-V intervals from 20 to 100 V, and sketch them to scale. Are
these surfaces equally spaced?
44.(II) In the example on atomic potential change, we found that the potential energy of the
hydrogen atom was -4.3× 10-18 J. Let us compare this to the electrostatic energy within a uranium
nucleus in which protons are separated by about 2 × 10-15 m. (a) Find the electrostatic potential
energy of a pair of these protons. (b) Compare this energy to that found for the hydrogen atom.
45.(II) A point object with a charge of +Q is located on the x-axis at x = +d. A second point object
with a charge of -3Q is located on the x-axis at x = -7d . (a) If one chooses the potential to be zero at
infinity, at what locations on the x-axis is the electric potential equal to zero? (b) At what locations
on the y-axis is the potential zero? (c) Repeat these questions, replacing the object with the –3Q
charge by an object with a charge of +3Q.
46.(II) An infinite plane of surface charge density σ= +2.5 µC/m2 is at x=10 cm. (a) For x>10 cm,
!
what is E ? (b) What is the change in the potential V(20 cm) -V(50 cm)? (c) How much work is
required by an outside agent to move a particle carrying a charge of +1.5 nC from a point at x=50 cm
to x=20 cm? (d) Taking the potential to be zero at the plane, draw three equipotential surfaces at 20V intervals, to each side of the plane, labeling each with the appropriate value of potential.
47.(II) An infinite plane of positive surface charge density σ is at x = 0.5 m. Points a and b are on
the x-axis at x = 2 m and x = 7 m, respectively. (a) Is the potential change Vb - Va positive or
!
negative? (b) If the magnitude of the potential change is 15.0 V, what is E in this region? (c) If the
magnitude of the potential change is 15.0 V, what is σ?
48.(II) Figure 29.0 shows three configurations of charged point objects. All the objects are the same
distance from the origin. (a) Rank the configurations based on the electric potential at the origin,
highest to lowest. (b) Rank the configurations based on their electric potential energy, highest to
lowest.
22
Figure 29.0
49.(II) You are sketching equipotentials for a positively charged point object, having defined V = 0
at infinity. If you draw the 5-V equipotential as a circle with a radius of 5 cm what is the radius of
(a) the 10 V equipotential? (b) The 2-V equipotential?
50.(II) Two parallel conducting plates are 0.10 m apart and carry equal and opposite charge. They
are large enough compared to 10 cm that we can assume the electric field between them is uniform.
The potential difference between them is 500 V. An electron is released from rest at the negative
plate. (a) Draw the electric field between the plates, indicating the sign of the charge on each plate
and which plate is at the higher potential. (b) What is the magnitude of the electric field between the
plates? (c) What is the change in potential energy of the electron when it moves from plate to plate?
(d) Find the work done by the electric field on the electron as the electron moves plate to plate. (e)
What is its kinetic energy when it reaches the positive plate?
51.(II) Two parallel conducting plates carry equal and opposite charge. They are large compared to
their separation, so we can assume the electric field between them is uniform. The potential change
between them is 0.25 V. The magnitude of the electric field between the plates is 50 V/m. (a) What
is the charge density on the plates? (b) What is the separation of the plates? (c) Find the work done
by the electric field on an electron as it moves from plate to plate.
52.(I) The combination of the diagram and graph in Figure 29.13 shows why the spacing between
neighboring equipotential surfaces from a charged point object increases with distance from a
charged point object. (a) What would the dependence of the potential on distance have to be if the
spacing between equipotential surfaces was constant? (b) What would be the dependence of the
electric field on distance in that case?
53.(III) Two charged point objects are placed near the origin. The first has a positive charge of +Q
and is located on the x-axis at x = +d. The second has an unknown charge and an unknown location.
The magnitude of the net electric field at the origin is Q/(2'(od2), and the electric potential at the
origin, relative to V = 0 at infinity, is +3Q/(4'(od). (a) If the second object is located on the x-axis,
what is its charge and location? (b) If the second object is located on the y-axis, what is its charge
and location? Find all possible solutions.
54.(II) A point object with a positive charge of +Q is placed on the x-axis at x = -d. A second point
object with an unknown charge is located at some unknown location on the x-axis. The potential
energy associated with the charges is +Q2/(2'(od), and the potential at the origin due to the charges
is +Q/('(od). (a) What is the sign of the unknown charge? Justify your answer. (b) What is the
magnitude of the unknown charge and where is it located? Find all possible answers.
Section 29.6
23
55.(III) An electron orbiting a helium nucleus emits energy in the form of radiation as it jumps from
an orbit of 0.3 nm to an orbit of 0.2 nm. (a) What is the change in the kinetic energy of the electron
when it makes this jump? (b) Because energy is conserved in this process, what must the energy of
the emitted radiation be?
56.(I) Figure 29.0 shows three configurations. In case A there is a point object with a positive charge
+Q located a distance R from the origin. In case B the +Q charge is spread uniformly over a semicircle of radius R centered at the origin. In case C the +Q charge is spread uniformly over a circle of
radius R centered at the origin. (a) Rank these cases based on the magnitude of the electric field at
the origin. (b) Rank these cases based on the electric potential at the origin. (c) Setting the potential
set to zero at infinity, write an expression for the electric potential at the origin in case B.
Figure 29.0
57.(I) A disk of radius 6.25 cm carries a uniform surface charge density σ = 7.5 nC/m2. Find the
potential on the axis of the disk at a distance (a) 0.5 cm, (b) 3.0 cm, and (c) 6.25 cm from the disk.
58.(II) A point object with a charge +Q is placed at the center of a conducting shell of inner radius
R, outer radius 2R, and a charge of –4Q. A thin-walled conducting shell of radius 3R and a charge of
+4Q is concentric with the point object and the first shell. Defining V = 0 at infinity, find all the
distances from the center at which the electric potential is zero.
59.(II) The disk of radius R sketched in Figure 29.0 has positive charge uniformly distributed over a
central circular region of radius a, and negative charge uniformly distributed over the remaining
annular portion of the disk. The surface charge density on the inner region is +% and that on the
annular region is -%. The electric potential at point P, on the disk’s axis a distance R away from the
center of the disk, is zero. What is a in terms of R?
Figure 29.0
60.(II) The surface charge density on an insulating disk of radius R is given by % (r) = cr, with c>0.
24
The axis of the disk is the x-axis and the disk is centered at the origin. Derive an expression for the
electrostatic potential as a function of position x along the x-axis.
61.(I) A very long wire of linear charge density λ= 150.0 nC/m lies on the z-axis. Find the potential
at distances of (a) 2.0 m, (b) 4.0 m, and (c) 12 m from the line assuming that V = 0 at 2.5 m.
62.(II) A uniformly charged rod of length L and total charge +Q lies on the x-axis, extending from
the origin to x = -L. What is the electric potential due to this line of charge at a point on the positive
portion of the x-axis at a distance d from the origin?
63.(II) A charge of q = +10 nC is uniformly distributed on a spherical shell of radius 12 cm. (a)
What is the magnitude of the electric field just outside and just inside the shell? (b) What is the
magnitude of the electric potential just outside and just inside the shell, relative to zero at infinity?
(c) What is the electric potential at the center of the shell? What is the electric field at that point?
64.(II) Four uniformly charged insulating rods, each with length L and charge +Q, are arranged in a
square. (a) What is the magnitude of the electric field at the center of the square? (b) What is the
electric potential at the center of the square?
65.(II) A particle carrying a charge +10 nC is located at y = 0.03 m. A finite uniformly charged
insulating rod carrying a charge of -10nC is located along the x-axis, from x= 0.10 m to x= 0. Write
an expression for the electric potential for any point along the y-axis.
66.(III) A very long, nonconducting cylinder of radius r0 carries a non-uniform charge density,
ρ=+ar, where r is the distance from the axis of the cylinder. Find the change in potential from the
axis to the outer edge of the cylinder.
67.(II) A solid nonconducting sphere of radius R has a uniform charge distribution and a total charge
+Q. Assuming the electric potential is zero at infinity, and that r represents distance measured from
the center of the sphere, derive an expression for the electric potential at (a) r > R (b) for r < R. (c)
What is the electric potential at the center of the sphere?
68.(II) A solid sphere with a radius R is concentric with a conducting spherical shell with inner
radius 2R and outer radius 3R. The net charge on the shell is +Q. If the electric potential at the center
of the solid sphere is the same as the potential at infinity, what is the charge on the solid sphere if (a)
the solid sphere is conducting (b) the solid sphere is nonconducting and has a uniform charge
distribution throughout it?
Section 29.7
69.(II) Two particles carrying a positive charge q0 are on the y-axis at y = +a and y = -a. (a) Find the
potential for any point on the x-axis. (b) Use your result in part (a) to find the electric field at any
point on the x-axis.
70.(I) A particle carrying a positive charge of +3.00 nC is at the origin. (a) Find the potential V on
the x-axis at x = 3.00 m and at x = 3.01 m. (b) Does the potential increase or decrease as x increases?
Compute ∆V/∆x, where ∆V is the change in potential from x = 3.00 m to x = 3.01 m and ∆x = 0.01
25
m. (c) Find the electric field at x = 3.00 m, and compare its magnitude with -∆V/∆x found in part
(b). (d) Find the potential (to three significant figures) at the point x = 3.00 m, y = 0.01 m, and
compare your result with the potential on the x-axis at x = 3.00 m. Discuss the significance of this
result.
71.(II) Consider Figure 29.0 showing a two-dimensional slice through a set of equipotential surfaces.
(a) Considering points A through D only, at which of these points is the magnitude of the electric
field largest? (b) Considering points B and F only, at which of these points is the magnitude of the
electric field largest? Justify your answers. (c) Sketch some electric field lines to support your
arguments, clearly indicating the direction of the field.
Figure 29.0
72.(II) (a) Sketch V(x) versus x for a uniformly charged ring in the y-z plane as implied by Figure 2920. (b) At what point is V(x) a maximum? (e) What is Ex at this point?
73.(I) A particle carrying a positive charge of 3.00 nC is at the origin, and a second particle carrying
a charge of -3.00 nC is on the x-axis at x = 6.00 m. (a) Find the potential on the x- axis at x = 3.00 m,
relative to zero at infinity. (b) Find the electric field on the x-axis at x = 3.00 m. (c) Find the
potential on the x-axis at x = 3.01 m, and compute ∆V/∆x, where ∆V is the change in potential from x
= 3. 00 m to x = 3. 01 m and ∆x = 0.01 m. Compare your result with your answer to part (b).
74.(I) Imagine that the electrostatic potential in a particular xy-coordinate system is given by V(x,y)
= 3xy – 5y2. Find the expression for the electric field.
75.(I) In the following, V(x) is in volts and x is in meters. Find Ex when (a) V(x) = 4000 + 6000x; (b)
V(x) = 2000/x + 1500x; (c) V(x) = 2000x - 3000x2, and (d) V(x) = -2000, independent of x.
76.(I) The electric potential in some region of space is given by V(x) = A + Bx , where V is in volts, x
is in meters, and A and B are positive constants. Find the electric field in this region. In what
direction is the field?
77.(III) Consider a spherically symmetric situation where the electrostatic potential is given by: for r
> 2R, V(r) = -Q/(4'(or); for R < r < 2R, V(r) = -Q/(8'(oR); and for r < R, V(r) = -3Q/(4'(or) +
26
5Q/(24'(oR), where r is the distance from the center of the charged point particle(s) and/or
concentric sphere(s) responsible for the potential. (a) What is the electric field in the three regions?
(b) Describe a set of objects that could produce this electric potential.
27
Answers to review questions
1. Electrical potential energy is the potential energy associated with the relative positions of objects
carrying a charge.
2. The electrostatic interaction can be either attractive or repulsive, while the gravitational
interaction is only attractive. Therefore, for oppositely charged objects, electrical potential energy
(like gravitational potential energy) decreases with decreasing separation, but it increases with
decreasing separation for charged objects of the same sign.
3. The electrical potential energy of the dipole is a minimum when it is oriented in the direction of
the electric field of the object, & increases as the angle increases.
4. Electrostatic work is the work done by an electrostatic field on a charged particle. It is (the line
integral of) the product of the electric force on the particle & its displacement.
5. It depends, of course, on the amount of charge & the field, but it only depends on the endpoints of
the path & not the shape of the path.
6. The electric potential difference between points A & B is the negative of the electrostatic work
per unit charge done on a charged particle along any path from A to B.
7. Equipotentials are surfaces, or lines, along which the electric potential is constant.
8. Equipotentials & field lines are everywhere mutually perpendicular.
9. The potential decreases in the direction of a field line.
10. No. The electric force on positively charged particles is in the direction of the electric field, from
higher to lower potential, but the force on negatively charged particles is in the opposite direction,
from lower to higher potential.
11. This is given in Equation 29.5,
12. The change in electrostatic potential energy equals the electrostatic work done.
13. For a system of charged particles, one can choose the zero of potential energy when all have
infinite separation (which corresponds to zero force between particles).
14. For a system of charged particles, the electrical potential energy is the sum of the potential
energies for each pair of particles, as in Equation 29.14.
15. The SI unit of potential difference is the volt, which equals a joule per coulomb, and is named in
honor of Alessandro Volta (see following Equation 29.15).
16. For a system of charged particles, the zero of potential can be chosen for infinite separation. In
dealing with circuits, the connection to ground (the earth) is chosen as the zero of potential.
17. The potential difference is the negative of the work per unit charge done by the electrostatic
field.
18. From this, Equation 23.32 is the simple result for the potential from a system of charged particles
(with zero at infinite separation).
19. The answer is no, because the work done by an electrostatic field is path independent. It is
therefore zero for a loop because to go from start to finish for a closed loop has the same start point
and end point as just sitting there.
20. Treat each particle as an infinitesimal element of charge & integrate over the entire object,
obtaining Equation 23.36.
21. Potential is a scalar, which is easier to calculate than an electric field, which is a vector.
22. Yes, because the field lines are perpendicular to the equipotentials in the direction of decreasing
potential.
23. The component of the electrostatic field in any direction is equal to the negative of the derivative
of the potential in that direction.
28
Chapter 30
Charge separation and storage
Review questions
Answers to these questions can be found at the end of this chapter.
Section 30.1
1. What is the difference between electric potential difference and electric potential energy?
2. When charge is separated between to objects, on what quantities does the amount of stored
electrical potential energy depend?
3. In a system of objects, some of whose charge has been separated, where in space can one view
that the system’s electric potential energy is stored?
4. What does a Van de Graaff generator produce?
Section 30.2
5. What is a capacitor?
6. Describe a parallel-plate capacitor. What geometrical approximation is usually made in treating
parallel-plate capacitors, and what simplification for its electrical properties results?
7. How is the amount of charge on each plate of a parallel-plate capacitor related to the potential
difference between the plates?
8. For a given potential difference between its plates, how does the amount of charge on each plate
of a parallel-plate capacitor vary with the plate area and the plate separation?
9. In practice, what limits the amount of charge that can be stored on the plates of a parallel-plate
capacitor?
Section 30.3
10. What is a dielectric? How many general types of dielectrics are there, and what distinguishes
them? How do the general types behave when exposed to an external electric field?
11. From a macroscopic point of view, what happens to an object of dielectric material when it is
placed in an external electric field? For simplicity, assume that the external field is uniform.
12. Why is the electric field strength reduced inside a piece of dielectric material, when it is placed
in a uniform external field?
13. Does an electric field in a dielectric store less, the same, or more energy than a field of equal
magnitude in a vacuum? Why?
Section 30.4
14. What is a battery and what does it do?
15. What is emf?
Section 30.5
16. How is the capacitance of a capacitor defined?
17. On what properties of a capacitor does its capacitance depend?
18. What is the unit of capacitance? Is its size appropriate to most practical applications in electronic
devices?
1
Section 30.6
19. How does the energy stored in a capacitor depend on the amount of charge separated onto each
of its plates?
20. How does the energy stored in a capacitor depend on the potential difference between the
conductors?
21. What is the general expression for the energy density stored in an electric field in empty space?
22. How is the potential difference between the terminals of a charge-separating device related to its
emf?
Section 30.7
23. In Chapter 30, how is the dielectric constant defined?
24. Why is the dielectric constant of liquid water larger than that of other common materials used in
capacitors, like paper or Mylar?
25. For a parallel-plate capacitor, with a dielectric material filling the space between its plates, how
is the induced charge on a surface of the dielectric related to the surface charge on the adjacent
conducting plate?
26. Why is the energy stored in an isolated charged capacitor smaller with a dielectric slab inserted
between its plates than without the dielectric?
Section 30.8
27. How can Gauss’s law be applied to electric fields, in the presence of dielectrics, without prior
knowledge of the induced polarization charges?
2
Developing a feel
Calculate or estimate the following quantities:
1. The maximum potential change between two baking sheets separated by 10 cm of dry air. (A, D)
2. The maximum charge that can be placed on a basketball in dry air. (A, L, U)
3. The potential difference that causes a lightning strike. (A, H, S)
4. The dimensions of an air-filled parallel plate capacitor of 1 F capacitance. (A, F, N, T)
5. The capacitance of a raindrop in dry air and in very humid air. (B, G, M)
6. The plate area of a 50 femtofarad capacitor on a computer memory chip. (C, R, W)
7. The maximum electrical energy density achievable in dry air. (A)
8. The maximum electrical energy that could be stored in an electric field that fills your lab room.
(A, J)
9. The capacitance of the coaxial cable that connects your TV to the cable outlet in the wall. (E, K,
P, V)
10. The capacitance of a softball. (G, O)
Hints:
A. What is the breakdown strength of dry air?
B. What is the radius of a raindrop?
C. What is between the “plates”?
D. How does E relate to ∆V?
E. What is the length of the cable?
F. Is a large or small gap between plates desirable?
G. What is the radius of the “other” sphere?
H. Does lightning occur most dramatically in humid or dry air?
J. What is the volume of the room?
3
K. What are the radii of the conductors?
L. Where is the magnitude of potential largest?
M. What is the effect of humidity?
N. How small can a uniform gap between two large metal plates be made?
O. What is the radius of the Earth?
P. What about bends and loops in the cable?
R. What is the gap width?
S. What is the length of a lightning bolt?
T. What area is needed for this gap width?
U. What is the radius of a basketball?
V. What should we use as a dielectric constant?
W. What is “femto”?
Key: A. E ≈ 3.6x106 V/m. B. ~ 0.004 m. C. Silicon dioxide (κ ~5) is likely. D. ∆V = Ed . E. ~3 m.
F. A small gap is needed to reduce plate area. G. ~Infinite. H. Dry; see Key M. J. ~400 m3 for a
large lab room. K. R1 ~2x10-4 m, R2 ~2x10-3 m. L. Near the surface of the ball. M. The breakdown
strength of humid air is very low, ~0, so that no potential change can build up. N. No less than ~5
mm without trouble and expense. O. ~5 cm. P. Ignore bends and loops whose radii of curvature >>
R2 , treating the cable as a cylinder. R. Gaps are very small, ~1x10-8 m. S. ~2000 m. T. ~5x107 m2 .
U. ~0.15 m. V. With a plastic insulator κ ~3 . W. Femto = 10-15.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any particular
section. Typically, an example that is worked out in detail is followed immediately by an example
whose solution you should work out by following the guidelines provided.
WP 30.1: Common capacitors. Capacitors used in electronic circuits are
generally not simple parallel plates, but are a clever extension of this. The
parallel plates consist of thin metal foil with a dielectric sandwiched in
between. The whole thing is then rolled up like a jellyroll. A typical capacitor
of this type is shown above. Suppose that both the foil and the mylar
dielectric film have a thickness of 0.05 mm, and the capacitor is 1 cm high
and 8 mm in radius. Estimate the charge stored in the capacitor if there is a 25
V potential difference across the capacitor leads.
!Focus problem
As usual, we will begin with a physical sketch, looking down at the endcap of the jellyroll capacitor
with it curved plates separated by mylar. We label the endcap’s radius as R and the thickness of
both the plates and the mylar as t.
The problem asks how much charge can be stored in this object. Since capacitance is how much
charge you can store with a given potential difference, finding the capacitance should allow us find
the stored charge.
!Plan approach
If we unroll the capacitor, we will have two parallel plates separated by a sheet of mylar. The
capacitance of parallel plates with a dielectric is
κε A
C= 0
d
We're given the plate separation d=1 cm, and we can look up the dielectric constant of mylar
(!=3.1). All we need to do is estimate the area A of the parallel plates, and we know that area is
length L multiplied by width w. Unrolled, the height of the capacitor is the width of the parallel
plates, i.e., w=8 mm. We know that the how long the sheets are and their thickness will determine
the radius R=4 mm of the capacitor when they are rolled up. So we have to carefully infer the length
of the sheets from their thickness and the capacitor’s radius.
5
!Execute plan
We can estimate the plate’s length L by assuming the foil-mylar sandwich is rolled as tightly as
possible. But, we need to be careful - there's a trick here. Suppose the sandwich consists of foilmylar-foil. Then, when you roll it up the bottom foil will touch the top foil, connecting the two
plates and effectively nullifying your capacitor. You need to make a sandwich of foil-mylar-foilmylar to avoid this. The thickness of this sandwich is t=4d, where d is the thickness of each layer.
There are several ways to estimate the length of the unrolled foil. One is to use the area of the
capacitor’s circular top, which must be the same as the area of the edge of the foil sandwich, i.e.
A = πR 2 = Lt
L = πrR 2 / t
Alternatively, R/t gives the number of sandwiches across the radius, assuming their circles of
thickness t and radius r. However, each sandwich layer has a different length given by 2"r. The
average length of each layer is then 2"R/2, so the total length is
L = (number of layers)!2π( average r ) = ( R / t )2π( R / 2) = πR 2 / t
Reassuringly, both methods give the same length estimate! Putting it all together, the charge stored
on the capacitor is
Q = CV
κε A
κε Lw
= 0 V= 0 V
d
d
2
κε (πR ) w /(4d )
= 0
V
d
Q=
κε 0 πR 2 wV
4t 2
Substituting the values, we arrive at the numerical value for the capacitance.
Q = 34.5 nF
!Evaluate result
Our algebraic result implies that a bigger capacitor (larger R or w) means you can store more charge,
which makes sense. It also implies that if the sheets are packed more closely (smaller t), the stored
charge will also go up because you have packed more foil in the same volume. (But then you need to
worry about electrical breakdown of the mylar film). Although you probably haven’t developed a
feel yet for capacitors, this value is pretty reasonable for a capacitor this size. Typical capacitors in
most electric circuits have capacitances of a few picofarad (often called “puffs”) for the small ones
to several hundred microfarad for the very large ones about the size of a soft drink can.
6
WP 30.2: Odd capacitor. Let us say you are enamored with rectangular shapes, and you wish to
make a capacitor in that shape (with a square top) rather than the standard cylindrical “jellyroll”
(with a circular top). What would be the side length of the top if your capacitor is to have the same
capacitance, height, and plate/mylar thickness as the one described in WP30.1?
!Questions and suggestions
1. First draw of sketch of the capacitor, looking down on the top.
2. Label the relevant dimensions and quantities.
2. What is the number of (approximate) square layers within the capacitor, based on it outside side
length (the length you are tasked to find)?
3. Given this many layers, what is the total surface area of one continuous plate if rolled out?
4. How do you combine the information about the surface area and mylar thickness to the desired
value of capacitance?
5. Backtrack from this to the unknown side length.
WP 30.3: Thunderstorm capacitance. During a lightning strike, a charge on the order
of 10 C is typically transferred to the ground over a potential difference of 3x109 V. What
is the capacitance of the cloud-ground system? How much energy is stored in the system
just before the strike?
#Focus problem
We are asked to find the energy stored in the cloud-ground system when there is a specified stored
charge and potential difference between them. Capacitance is the ability of a physical system (in
this case the cloud-ground system) to store (or separate) charge. That stored charge has energy
because work was done to separate it out. Consequently, the capacitance also gives stored energy.
!Plan approach
The stored charge is Q=10 C and the potential difference involved is Vcap=3x109 V. The expression
for capacitance is:
Q
C=
Vcap
7
After determining the value for capacitance, other standard formulas can give the stored energy.
!Execute approach
The value for capacitance is:
C=
Q
10 C
=
= 3.33 nF
Vcap 3 ×109 V
The stored energy of the charge is:
1
1
U E = CVcap 2 = (3.33 × 10−9 F)(3 × 109 V)
2
2
10
= 1.5 ×10 J
!Evaluate result
Because we know that lightning strikes are dangerous and could do a lot of damage, we expect that
the energy delivered in a strike should be high. 10 billion Joules well satisfies that expectation.
WP 30.4: Laptop Power. Laptop computers require heavy and expensive batteries to keep the
computer running. Moreover, the lifetime of many rechargeable batteries is relatively short –
perhaps at best a few thousand chargings. Because of this, capacitors have been looked at to store
the required energy. Capacitors are quickly charged, and because no chemical reaction is involved
they have a much longer lifetime than do batteries. Let us say your ultra-low power laptop requires a
voltage of 16 V to run, and you’d like it to run for a minimum of three hours using an average power
of 1 W. Estimate the capacitance of such a capacitor required for typical laptop operation. If this
were a standard parallel plate capacitor, how large would the area of the plates have to be, assuming
a plate separation of 0.05 mm by a mylar layer? Does this seem a feasible alternative?
!Questions and suggestions
1. What is the relationship between the energy that needs to be stored and the power that needs to be
delivered for the specified lifetime?
2. What is the formula relating capacitance and this energy at the operating voltage?
3. What is the relationship between capacitance and surface area in a parallel-plate capacitor?
WP 30.5: Earth-sized capacitor. What is the approximate capacitance of the earth,
treating it as an isolated sphere?
#Focus problem
It is hard to think of Earth as an electronic device, in this case a giant capacitor. However, we know
there is formal definition of capacitance: the amount of charge that can be stored per unit volt
8
difference. But where is the other part of this capacitor from which to get charges? Because the
potential of an object is measured relative to the zero of potential at infinity, this gives us the clue
that we should bring charges in from infinity.
!Plan approach
We know that capacitance is charge stored per volt “across” the capacitor
C=
Q
Vcap
where Q is the charge on Earth. What we need to know is the potential Vcap between the Earth’s
charged surface and infinity. We have already worked this out in the last chapter where we found
that the potential at some radial distance from a charged sphere is:
V (r ) =
q
4πε 0 r
We then combine these equations for the case of Earth’s surface.
!Execute approach
The potential at a sphere’s surface of radius R when it is carrying a charge Q is
V ( R) =
Q
4πε 0 R
Substituting this in as the voltage across the Earth-infinity capacitor (noting R=RE) yields Earth’s
capacitance as:
Q
Q
= 4πε 0 RE
!
"
#
$
% 4πε 0 RE &
= 4π (8.9 ×10−12 C2 /Nm 2 )(6.4 ×106 m) = 0.71 mF
CEarth =
!Evaluate result
We see from the algebraic expression that the capacitance of Earth would increase if it had a large
radius. This is expected because the charges would be further apart and thus less difficult (less
work) to add for charge storage. It is also gratify to note that this is exactly the result one obtains
with a standard spherical capacitor (Eq. 30.15) when we make the outer sphere’s radius infinite.
WP 30.6: Tin can capacitor. A more complicated capacitor consists of a solid cylinder surrounded
by a close-fitting metal can with top and bottom metal caps. The radius of the top is R, and the
height is H. What at is the capacitance of this configuration if the separation between core and shell
is d?
9
!Questions and suggestions
1. As usual, we start by drawing a sketch of this object.
2. How is this capacitor similar to a coaxial capacitor? How does it differ?
3. What distances correspond to the two radii of a coaxial capacitor?
4. How can you account for the effects of the end caps on total capacitance?
WP 30.7: Thunderstorm capacitance. A variable capacitor can be constructed by inserting a slab
of dielectric constant of ! part way between a pair of parallel plates of length L and area A,
separated by distance d. The plates have a fixed charge Q. What is this objects capacitance as a
function of the insertion depth x?
#Focus problem
Before trying to figure out how to proceed, we need a sketch with which to think.
This looks like a parallel plate capacitor with a dielectric in only a portion of it, sort of like two
parallel plate capacitors butted together. Because of the dielectric, we know the electric field will not
be the same in the two regions. We will probably need to use the general definition of capacitance
and figure out how the two regions contribute to the total charge and potential difference of the
plates.
!Plan approach
The general defining equation for the capacitance of an object is:
Q
C=
V
10
But because there are two distinct sections of the capacitor, we have to be careful. We want to know
how much charge (that is free charge) store for a given external potential difference imposed from
an outside battery. We know that the potential is the same everywhere on each plate so that the
potential difference is the same across the plates, which is related to the electric field in the two
sections by Vbat = Ed. Consequently, the electric field must be the same in both regions. Normally,
a dielectric will decrease the field for a fixed free charge – but now the free charge density must
increase in compensation over the dielectric region, which we can get from Gauss’ law. The total
free charge from each portion of the capacitor will be the free charge density there multiplied by the
corresponding surface area. The total charge qfree will be the sum of these. Taking the ratio of qfree
and Vbat will give the capacitance.
!Execute approach
Since the entire area is A, the width of the plates is w=A/L. To keep track of things, let the section of
the capacitor section without the dielectric be labeled “1” (with area A1 = (L–x)w), and the section
with the dielectric labeled as “2” (with area A2 = xw). From Eq. 30.27, we see that the electric field
and surface free charge density in the section without the dielectric is:
E1 =
q
Vbat σ free,1
=
== free,1
d
εo
ε o A1
In the section with the dielectric, the electric field must be (Eq. 29)
E2 =
q
+q
Vbat σ 2
q
=
= 2 = free,2 bound ,2
d
ε o ε o A2
ε o A2
But from Eq. 30.36 (q = qfree+qbound = !qfree), we see that this becomes
Vbat q free ,2
=
d
κε o A2
The total free charge is:
q free = q free,1 + q free,2 =
=
Vbat
V
ε o A1 + bat κε o A2
d
d
Vbat
ε o ( ( L − x) w + κ xw )
d
We finally get the capacitance by:
C ( x) =
=
q free
Vbat
εow
d
=
εow
d
[ ( L − x) + κ x ]
[ L + (κ − 1) x] =
εo A '
x(
1 + (κ − 1) *
)
d +
L,
!Evaluate result
We know that dielectrics are used in capacitors to increase their capacitance. As we insert more of
11
the dielectric between the plates (increase x), the capacitance increases because (!–1) > 0. We
should also check the limit of x=0 and x=L. In the first case (x=0, no dielectric), our equation does
indeed agree with Eq. 30.5 for a dielectric-free capacitor. Likewise, if we fill the capacitor with
dielectric (x=L), then our equation becomes
C ( L) =
εo A '
ε A
L( ε A
1 + (κ − 1) * = o [1 + κ − 1] = κ o = κ C (0)
)
d +
L,
d
d
in agreement with Equation 30.26 giving the relationship between a capacitor with and without
dielectric Cd = !C.
WP 30.8: Vacuum Caps. In high-powered radio frequency (RF) applications highcapacitance capacitors are required in the electrical circuits that can withstand large
charge storage without electrical breakdown. Often “vacuum caps” are used, which
consist of conducting plates with a fairly large separation in a vacuum sealed canister.
The vacuum is not perfect, however. In addition, a strong enough electric field can
ionize the metal plates, ripping off an electron leading to electrical breakdown.
Suppose the effective dielectric strength is 10 times better than in air. What is the
maximum charge that can be stored on a 1.0 nF capacitor with a plate separation of 1
mm? What potential difference is required to store this much charge?
!Questions and suggestions
1. What basic capacitor geometry does this mimic: circular parallel plate or coaxial cylinder?
2. What is the capacitance for such an object with a dielectric present?
3. Where can you find the value for the maximum electric field of air and then for this dielectric?
4. How can you use Gauss’ law to determine the electric field within the dielectric for a given charge
on the plates?
5. Knowing this relationship between charge on the plates and electric field, how do you get the
maximum charge that such a dielectric capacitor can sustain without discharging?
6. What is the voltage across the capacitor when this free charge is stored on it?
12
Questions and problems
Section 30.1
1.(I) When two objects are charged by the process of charge separation, how are their charges
related?
2.(I) Could charge separation be applied to a system of three or more objects? If so, how would their
charges be related?
3.(I) What is special about charge separation if the objects involved are conductors?
4.(I) Does charge separation always lead to an increase in electrical potential energy? Explain why
or why not.
5.(I) To help yourself understand a new physical system or process, it is often useful to create a
model that has some analogy to what happens in a more familiar system. For instance, let us create
an elastic-band model to understand the behavior of the electric field when the two charged objects
are pulled apart in Figure 30.3. (a) Explain how treating the field lines as elastic bands could help
you to understand what happens to the potential energy when the oppositely charged objects are
moved further apart. (b) Can you think of any limitations of the elastic-band model?
6.(II) Given that charge separation increases the electrical potential energy, what can you conclude
about the stability criteria for a system of positively and negatively charged particles, in the presence
of electrostatic interactions only?
Section 30.2
7.(I) Why would two insulators, such as the rod-fur system in Figure 30.1, make a poor capacitor
compared to two conducting objects of the same shape?
8.(II) Explain why there must be some non-zero electric field outside the edges of a parallel plate
capacitor (as in Figure 30.7) so that the idealization in Figure 30.8 can never be exact. (Hint:
consider the potential difference between the plates for paths outside the edges.)
9.(II). A battery is connected to a capacitor with extendable plates, resulting in a charge +Q on one
plate and –Q on the other. (a) With the battery still connected, the area of each plate is doubled by
sliding the extension open. How much charge will be on each plate now? (b) Treating the field lines
as elastic bands (as in the previous problem) in which case is more force required to keep the plates
apart? (c) In which case is more pressure required to keep the plates apart? (d) If you think of a
battery as being similar to a pump, should you think of it as applying a certain force or a certain
pressure?
10.(II) Two parallel plate capacitors have the same plate area. The first has a plate separation twice
that of the second, and you place twice the amount of charge on the first as you do on the second.
13
How do the potential differences for the two capacitors compare?
11.(II) Two parallel plate capacitors have the same plate area. The first has a plate separation twice
that of the second, and you put twice the potential difference across the first as you do across the
second. How does the amount of charge stored on the two capacitors compare?
12.(II) Two parallel plate capacitors have the same plate separation. The first has a plate area twice
that of the second, and you place twice the amount of charge on the first as you do on the second.
How do the potential differences for the two capacitors compare?
13.(II) Two parallel plate capacitors have the same plate separation. The first has a plate area twice
that of the second, and you put twice the potential difference across the first as you do across the
second. How does the amount of charge stored on the two capacitors compare?
Section 30.3
14.(I) What are the two general types of dielectric materials, and what distinguishes them?
15.(I) Two parallel-plate capacitors have the same dimensions, but the space between the plates is
filled with air in the first, and with plastic in the second. The magnitude of the charge on the plates
of both capacitors is the same. (a) How do the magnitudes of the electric fields between the plates of
the capacitors compare? (b) How do the potential differences between the plates compare? Explain.
16.(II) How do the energies stored in the capacitors of the previous problem compare? Explain.
17.(II) Modify figure 30.11 by adding a graph showing the potential as a function of position with a
dielectric slab filling half the space inside the parallel-plate capacitor of Figure 30.10b instead of a
metal slab. Comment on how the graph is affected by the magnitude of the induced surface charge
on the dielectric slab.
18.(I) A parallel-plate capacitor is charged by connecting it to a battery. What happens to the
capacitance, potential difference, and charge on each plate when the plates are moved closer together
(a) if the capacitor is still connected to the battery? (b) if the battery is disconnected?
19.(I) Two parallel-plate capacitors have the same dimensions, but the space between the plates is
filled with air in the first capacitor, and with plastic in the second. The potential difference between
the plates of both capacitors is the same. (a) How do the magnitudes of the electric fields between
the plates of the capacitors compare? (b) How do the magnitudes of the charges on the plates
compare? Explain.
20.(II) How do the energies stored in the capacitors of the previous problem compare? Explain.
Section 30.4
21.(I) The term emf was originally an acronym for electromotive force. Why was this a misnomer?
14
22.(I) One way to view a battery is as a complete manufacturing cycle. Electrons, the product, are
produced by the chemical reaction at one electrode. The electrons are then recycled at the other
electrode. Examine the chemical reactions of the lead-acid battery and identify (a) the raw materials
used in manufacturing the product, (b) the raw materials used in recycling the product, and (c) the
waste products produced in the process.
23.(II) One electrode in an alkaline battery is zinc while the other is manganese dioxide. (a) The
chemical reaction taking place at the zinc electrode is:
Zn + 2OH- $ ZnO + ____ + 2e-. Fill in the missing piece. (b) The chemical reaction taking place at
the other electrode is: 2MnO2 + H2O + ____ $ Mn2O3 + 2OH-. Fill in the missing piece. (c) Which
electrode is positive and which is negative?
Section 30.5
24.(II) Two conducting objects, one with a charge of +Q and one with a charge of –Q, are placed on
the x-axis at x = -4 m and x = +4 m, respectively. The electric field on the x-axis between the
charges is given by E(x) = aQ(x2 + b), where a and b are constants. (a) What are the units of the
constants a and b? (b) What is the capacitance of this configuration of two objects?
25.(II) In a parallel-plate capacitor with a charge +Q on one plate and –Q on the other, each plate
contributes equally to the electric field between the plates and the field outside the capacitor is
essentially zero if the plates are large compared to their separation. Consider a spherical capacitor
consisting of two concentric spherical shells, one smaller than the other, with a charge of +Q
uniformly distributed on one shell and a charge of –Q uniformly distributed on the other. (a) Do the
spheres contribute equally to the field in the region between them? If not, what fraction of the field
comes from the smaller sphere? (b) What is the field outside the capacitor?
26.(II) To double the capacitance of a parallel-plate capacitor the separation between the plates can
be reduced by a factor of two. (a) If you had a spherical capacitor consisting of two concentric
spherical shells of radius a and b (a < b) you could double the capacitance by changing the radius of
the outer shell to a value d. What is d in terms of a and b? (b) What would you change it to if you
doubled the capacitance by changing the radius of the inner shell instead? (c) Repeat the question
using a capacitor consisting of two long concentric spherical shells of radius a and b (a < b) instead.
27.(II) A particular spherical capacitor consists of two concentric conducting spherical shells of radii
R and 2R. (a) How long would a coaxial cylindrical capacitor, made of two concentric cylindrical
conductors of radii R and 2R, be to have the same capacitance as the spherical capacitor? (b) What
separation between the plates would be required for a parallel-plate capacitor, made of two circular
conductors of radius R, to have the same capacitance as the spherical capacitor?
28.(III) The expression for the capacitance of a parallel-plate capacitor having plates of area A
separated by a distance d is given in equation 30.5. (a) Show that equation 30.15, for a spherical
capacitor, reduces to equation 30.5 in the limit that R1 approaches R2, with d = R2 – R1. (b) Show that
equation 30.10, for a cylindrical capacitor, reduces to equation 30.5 in the limit that R1 approaches
R2, with d = R2 – R1. [Hint: you might find the Taylor series expansion of ln(1+x) useful.]
15
Section 30.6
29.(II) (a) A spherical capacitor consists of two concentric conducting spheres of radius R and 2R. If
the two spheres have charges of +Q and –Q, what is the average energy density inside the capacitor?
Express your answer in terms of R and Q. (b) A cylindrical capacitor of length L consists of two
concentric conducting cylinders of radius R and 2R. If the two cylinders have charges of +Q and –Q,
what is the average energy density inside the capacitor?
30.(II) Two parallel-plate capacitors, A and B, are identical except that capacitor A has a charge +Q
on one plate and –Q on the other, while B has a charge of +2Q on one plate and –2Q on the other.
Compare the two capacitors based on their (a) capacitance (b) potential difference (c) field between
the plates (d) energy stored.
31.(II) A parallel-plate capacitor has a charge of +Q on one plate and –Q on the other. The plates
have an area A. (a) How much force, in terms of Q and A, does one plate exert on the other? (b) In
terms of the electric field strength E inside the capacitor, what is the pressure on one of the plates?
32.(II) A parallel-plate capacitor has two identical plates of area A. The plates are initially separated
by a distance D but this can be varied. If the capacitor is connected to battery what should the plate
separation be if you want to quadruple (a) the capacitance (b) the charge stored (c) the energy stored
(d) the energy density?
33.(II) Repeat the previous question but for an isolated capacitor that had been charged by a battery
and then had the battery connections removed.
34.(I) Three parallel-plate capacitors are separately connected to identical batteries. Capacitor 1 has
a plate area A and a plate separation L. Capacitor 2 has a plate area 2A and a plate separation L.
Capacitor 3 has a plate area A and a plate separation 2L. Rank the three capacitors, from highest to
lowest, based on their (a) capacitance (b) charge stored (c) electric field between the plates (d)
energy stored (e) energy density.
35.(I) Three parallel-plate capacitors each store the same amount of charge. Capacitor 1 has a plate
area A and a plate separation D. Capacitor 2 has a plate area 2A and a plate separation D. Capacitor
3 has a plate area A and a plate separation 2D. Rank the three capacitors, from highest to lowest,
based on their (a) capacitance (b) potential difference (c) electric field between the plates (d) energy
stored (e) energy density.
36.(III) A parallel-plate capacitor has an initial charge Q and a plate separation d. How much work
do you have to do, in terms of Q and d, to increase the plate separation to 3d if (a) the capacitor is
isolated (b) the capacitor is connected to a battery. (c) The work you do is positive in both cases.
Explain why in one situation the potential energy increases and in the other it decreases.
Section 30.7
37.(I) (a) Does it make sense to talk about a dielectric constant for a conductor? If so, what value
does it have? (b) What is the breakdown field for a conductor?
16
38.(II) When you remove a dielectric slab from between the plates of a charged isolated capacitor
what happens to the energy stored by the capacitor? Explain why this happens.
39.(II) Two parallel-plate capacitors are identical except that the first has vacuum between the plates
and the second has a dielectric slab of dielectric constant ! filling the space between the plates. The
capacitors are connected to identical batteries. Compare the two based on their (a) capacitance (b)
charge stored (c) energy stored (d) electric field (e) energy density.
40.(II) Two parallel-plate capacitors are identical except that the first has vacuum between the plates
and the second has a dielectric slab of dielectric constant ! filling the space between the plates. Each
capacitor is isolated and the capacitors store equal charges. Compare the two based on their (a)
capacitance (b) potential difference (c) energy stored (d) electric field (e) energy density.
41.(I) When a dielectric slab of a particular material completely fills the space between the plates of
a parallel-plate capacitor the magnitude of the bound charge is one-fourth the magnitude of the free
charge. (a) What is the dielectric constant of the material? (b) If the width of the slab is instead equal
to half the distance between the plates, would the magnitude of the bound charge change? If so,
how?
42.(II) The potential difference across the plates of a parallel-plate capacitor is gradually increased.
(a) What happens when the breakdown field is exceeded? (b) If you wanted to increase the potential
difference before breakdown occurred, what could you change about the capacitor?
43.(II) If you want to maintain a potential difference of 6000 V between the plates of a parallel-plate
capacitor, what is the minimum value the plate separation can be if the space between the plates is
completely filled by (a) a slab of mica (b) a slab of barium titanate? (c) What is the ratio of the
maximum charge you can obtain with the barium titanate slab to the maximum charge possible with
the mica?
44.(I) (a) Write out a short list of ways to increase the capacitance of a parallel-plate capacitor. (b)
Do “increase the charge” or “decrease the potential difference” belong on the list? Explain why or
why not.
45.(II) A dielectric slab fills the space between the plates of a parallel-plate capacitor. The
magnitude of the bound charge on the slab is 75% of the magnitude of the free charge on the plates.
The capacitance is 480%oL and the maximum charge that can be stored on the capacitor is 240%oEmax,
where Emax is the breakdown field. (a) What is the dielectric constant for the dielectric slab? (b)
What is the plate separation, in terms of L? (c) What is the plate area, in terms of L?
Section 30.8
46.(II) A solid conducting sphere with a radius R and a charge +Q is embedded in a neutral
insulating spherical shell with an inner radius of R, an outer radius of 2R, and a dielectric constant of
!=3. (a) Relative to an electric potential of zero at infinity, what is the electric potential at the center
of the conducting sphere? (b) If the dielectric constant of the shell is increased what happens to the
potential at the center of the sphere?
17
47.(II). A solid conducting sphere with a radius R and an unknown charge is embedded in a neutral
insulating spherical shell with an inner radius of R, an outer radius of 2R, and an unknown dielectric
constant. Inside the shell the electric field has a magnitude E = 3Q/(4"%oR) and, relative to an
electric potential of zero at infinity, the electric potential at the surface of the conducting sphere is
+15Q/(16"%oR) . (a) What is the charge on the conducting sphere? (b) What is the dielectric constant
of the shell?
18
Answers to review questions
1. Potential difference is a measure of the work done by the electric field in moving a unit test
charge (not part of the system) between two parts of the system. The potential energy is determined
by the work done in establishing the configuration of separated charges that make up the system.
2. The potential energy depends on the amount of charge separated & the distance separating the
objects.
3. Electric energy is stored in the electric field, in any region of space where such a field exists.
4. A Van de Graaff generator produces a very large separation of charge, between an insulated metal
sphere & the ground, & hence a very large potential difference. (See box on p. ??? for details.)
5. A capacitor is a system of two objects (almost always conductors) which is capable of storing
electrical charge & potential energy.
6. A parallel-plate capacitor is an arrangement of two parallel conducting plates, of the same area, A,
separated by a distance, d. When d is small compared to the lateral dimensions of the plates, the
electric field is approximately localized & uniform between the charged plates, & is zero outside
them.
7. The magnitude of the charge on each plate is proportional to the potential difference between the
plates. For a parallel-plate capacitor, this is because the potential difference equals the product of the
field strength & the plate separation, & the field strength is proportional to the magnitude of charge
on a plate per unit area.
8. For the reasons stated in the previous solution, the amount of charge on a plate varies directly with
the area & inversely with the separation of the plates.
9. Dielectric breakdown occurs when the charge on the plates is great enough to produce an electric
field which ionizes the air (or other dielectric material) between the plates.
10. A dielectric is a nonconducting material, one in which charge carriers are not free to move
around. There are two types, polar & nonpolar, depending on whether or not the constituent atoms or
molecules possess a permanent electric dipole moment. In an external field, the centers of positive &
negative atomic or molecular charge are displaced slightly, producing an induced dipole moment. In
a polar dielectric, the randomly oriented molecular dipoles become aligned with the external field,
producing an additional polarization, which is generally much larger than the induced polarization in
nonpolar dielectrics.
11. In a uniform external field, there is no net charge induced on any volume that lies entirely inside
a piece of polarized dielectric, but uncompensated induced charge densities appear on the surfaces,
which account for the macroscopic differences between polarized & unpolarized dielectrics.
12. Inside a polarized dielectric, the induced surface charge densities give rise to an induced field
which is in the opposite direction to the external field, so the total field inside is reduced.
13. Work must be done by the external field to produce an induced polarization, so the potential
energy stored in the field inside a dielectric material is greater than that for an equal field in vacuum.
14. A battery is an assembly of one or more voltaic cells, which separate charge carriers by
converting potential energy released in chemical reactions into electrical potential energy.
15. Emf is the work done per unit charge, by nonelectrostatic interactions, inside a device which
maintains a separation of positive & negative charge carriers between its terminals.
16. The capacitance is the ratio of the magnitudes of the charge on one conductor & the potential
difference between the conductors.
17. Capacitance depends on the geometrical properties of a capacitor only, i.e. on the shape &
separation of the conductors.
18. The unit of capacitance is the farad, which is one coulomb per volt. Since one coulomb is an
enormous amount of charge, compared to the net charge on conductors in electronic devices,
19
capacitances in the range of microfarads to picofarads are more common practically.
19. It is proportional to the square of the charge (see Equation 30.17).
20. It is proportional to the square of the potential difference (see Equation 30.18).
21. It is given by Equation 30.20.
22. The potential difference is less than or equal to the emf, because some of the work done by
nonelectrostatic interactions may be dissipated inside the device. (See the discussion in the
paragraph containing Equation 30.22.)
23. The dielectric constant is the ratio of the potential difference across an isolated parallel-plate
capacitor, without & with the dielectric material completely filling the space between its plates. (See
Equation 30.24.)
24. The induced polarization in liquid water (which is composed of polar molecules that can align
themselves) is much stronger than in paper or Mylar (which are nonpolar dielectrics).
25. The bound surface charge is opposite in sign & smaller in magnitude than the adjacent free
surface charge that causes it, depending on the dielectric constant as in Equation 30.33.
26. The energy is smaller because the capacitor does positive work on the dielectric slab when it is
inserted. (See the Example: Capacitor with Dielectric.)
27. In the presence of dielectrics, Gauss’s law may be applied using only the free charge enclosed by
the Gaussian surface, provided, at points on the surface, one multiplies the electric field by the
dielectric constant.
20
Chapter 31 (version 03/12/06)
Magnetostatic Interactions
Review questions
Answers to these questions can be found at the end of this chapter.
Section 31.1
1. What is a magnet and what is a magnetic material?
2. What is a magnetic pole? How many different types are there & how do they interact
with one another?
3. What are elementary magnets & how does a model based on them explain
magnetization?
Section 31.2
4. How can the concept of a magnetic field be used to describe the action of a magnet?
5. What information about a magnetic field is conveyed by its magnetic field line
pattern?
6. How does the magnetic field line flux through a closed surface depend on the number
and strength of the magnetic poles inside?
Section 31.3
7. This section describes another source of magnetic fields besides a magnet. What is it
and how is it defined?
8. Describe the magnetic field produced by a straight current carrying wire.
9. Does a magnetic field exert a force on an electric current, & if so, how is the direction
of the force related to the directions of the field & the current?
Section 31.4
10. Why should one expect the magnetic interaction to depend on the frame of reference
of the observer?
11. If magnetic forces arise as relativistic corrections to electric forces, why are they so
readily observable under ordinary circumstances?
12. To an observer at rest with respect to a current carrying wire, how does the charge
density of the fixed ions in the wire compare to the charge density of the moving
charge carriers? How do these charge densities compare according to an observer
moving along with the charge carriers?
Section 31.5
13. When is the magnitude of the magnetic force, on a given length of current carrying
wire placed in a given external magnetic field, a maximum & when is it a minimum?
1
14. On what factors does the magnitude of the magnetic force on a current carrying wire
depend?
15. How can the magnetic force on a straight piece of current carrying wire be used, in
principle, to determine a magnetic field (assumed uniform in the region of space
containing the piece of wire)?
Section 31.6
16. How is the magnetic flux through a surface defined?
17. What is Gauss’s law for magnetism & what does it imply about magnetic sources?
18. How does the magnetic flux through one surface bounded by a loop compare with the
flux through a different surface bounded by the same loop?
Section 31.7
19. The net force on a current carrying wire is caused by what?
20. If a charged particle is moving in a magnetic field, what is the direction of the
magnetic force on it?
Section 31.8
21. Why is the density of ions greater than that for electrons to an observer moving along
with the electrons in a current-carrying wire?
22. What physical law, in conjunction with relativity theory, explains the magnetic
attraction of two current-carrying wires?
2
Developing a feel
Calculate or estimate the following quantities:
1. The current needed to produce a magnetic field equal in strength to that of the Earth
at a point 1 m to the side of a long straight wire. (C, Y)
2. The flux of the Earth’s magnetic field through the top of your lab table. (C, K, X)
3. The magnitude and direction of magnetic force on an electron moving west along the
Earth’s equator at 105 m/s. (C, L, G)
4. The maximum value of B needed to deflect electrons in your TV picture tube. (A, E,
I, M, Q, U)
5. The strength of uniform magnetic field required for a proton of speed 106 m/s to
move in a circular orbit the size of a basketball. (L, P, T)
6. The strength of uniform magnetic field needed to orbit an electron through one
revolution each microsecond. (D, H, L)
7. The minimum electric current needed to “float” a piece of 18 gauge copper wire in a
horizontal 0.5 T magnetic field. (V, R, N, J, B)
8. The maximum acceleration of a 10 cm length of copper grounding rod which carries a
current of 100 A in a uniform 1 T magnetic field. (F, O, J, S, W)
Hints:
A. What potential change value is used to accelerate the electrons?
B. What is the gravitational mass of a length L of this wire?
C. What is the strength of the magnetic field near the surface of the Earth?
D. What is the inertia of an electron?
E. What is the speed of the electron toward the screen?
F. What is the diameter of this rod?
G. In what direction is the magnetic field aimed for points along the equator?
H. How is the period related to the magnetic field strength B?
I. What is ratio of maximum sideways deflection distance to the distance from launch to
screen?
J. What is the density of copper?
K. What is the “dip angle”, the angle between the direction of magnetic field and the
horizontal?
L. What is the magnitude of electric charge carried by this particle?
M. Through what fraction of a complete revolution must the electrons be deflected?
N. What is the volume of a length L of this wire?
O. What is the volume of this rod?
P. What is the radius of a basketball?
Q. What is the time available for magnetic deflection?
R. What is the diameter of 18 gauge wire?
S. What is the inertia of the rod?
T. What is the inertia of a proton?
3
U. How is this deflection time related to a possible electron orbital period in a magnetic
field?
V. What forces must balance to “float” the wire?
W. What maximum magnetic force is available?
X. What is the area of your lab table top?
Y. How are magnetic field strength B and current I related for a long straight wire?
Key: A. ~10 kV. B. ~7×10-3 L kg/m. C. ~5×10-5 T. D. ~9×10-31 kg. E. Using q∆V = ∆K,
v ~6×107 m/s. F. Assume ~1 cm. G. ~Horizontal and North. H. T = 2πm/qB. I. ~20 cm /
25 cm ≈ 0.8. J. ~9000 kg/m3. K. This varies; in the continental US, from ~50°-75°, so
let’s say ~60°. L. e=1.6×10-19 C. M. ~tan-1(0.8) or ~0.1 rev. N. πr2L ~[8×10-7 m2]L. O.
~8×10-6 m3. P. ~0.1 m. Q. ~20% of the travel time, or 8×10-10 s. R. ~1×10-3 m. S. ~0.07
kg. T. ~1.7×10-27 kg. U. Time for 0.1 rev is T/10. V. Gravitational and magnetic. W. ILB
~10 N. X. ~2 m2. Y. B = 2kI/c2r.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any
particular section. Typically, an example that is worked out in detail is followed
immediately by an example whose solution you should work out by following the
guidelines provided.
WP31.1. Rail Gun
B
10 cm
i
.
A rail gun is a device used to accelerate projectiles without the use of explosives. The
sketch shows a view from above (down is into the page). A cross bar slides on two rails.
Current flows along one rail, into the cross bar, and back through the other rail. A
uniform magnetic field points up out of the page. (a) Which direction is the crossbar
accelerated? (b) A constant current of 10 A flows through the system as shown. What is
the coefficient of kinetic friction if the crossbar moves at constant speed? The mass of the
crossbar is 2 kg and the magnetic field strength is 0.1 T
!Focus problem
A wire with current flowing through it will feel a force from a magnetic field. This force
will act perpendicularly to the wire, accelerating it. In this case, however, a frictional
force balances the magnetic force. We’re asked to find the coefficient of friction.
!Plan approach
We learned in this chapter, a magnetic field exerts a force on a current-carrying wire
according to the relation
"! "! "!
F! = LI × B = LIBsin θ
Using the right-hand rule, with the current flowing up and the magnetic field out of the
page you should verify that the crossbar feels a force to the right. That answers part (a).
As with all problems involving forces, it’s a good idea to draw a free body diagram.
Taking the crossbar as our system I’ll draw the forces as we look at the cross bar from the
side, end on. We can see that the magnetic and frictional forces cancel, since the cross bar
moves at constant velocity. There is also a gravitational force pointing down and two
normal forces pointing up, one for each rail.
y 2F N
rc
Frcf
FBc B
x
0
FEc g
5
The frictional force is related to the normal force of the bar pressing down on the rails by
f
N
F = µkF
where !k is the coefficient of kinetic friction. These are all the pieces we need to find
what we’re looking for.
!Execute plan
Because force is a vector we can break the problem into x and y directions. In the x
direction Newton’s 2nd laws becomes "!
"!
! F = ma
B
f
FBc − Frc = 0
N
ILB
! − µ k Frc = 0
Since the magnetic field and the direction of the current are at right angles, sin 90°
becomes 1. In the y direction we get
g
2FrcN − FEc
=0
FrcN = mg / 2
Combining the final two equations and rearranging I get
µk =
2ILB
mg
Plugging in numbers
µk = 1.02x10 -2
!Evaluate answer
This seems like a very small coefficient of friction – but that’s what you want. The idea is
that at much larger currents (1000 A) you can accelerate the cross bar to very high
speeds. Notice how free body diagrams – skills learned way back in Chapter 9 – are still
very useful.
WP31.2 Magnetic Scale
I
B?
10 cm
m
.
A sensitive scale might be constructed as shown above. The object hangs from a wire
attached to a sliding rod, which is part of a simple electronic circuit. The current is
adjusted so the sliding rod and object are motionless. Which direction must the magnetic
6
field be for the magnetic force on the rod to balance the mass? What current should be
used for a 157 g object (and rod) with a magnetic field of 0.15 T? In what situations is
this scale practical?
!Questions and suggestions
1. Describe the problem in your own words. How is the problem similar to other
problems and examples in this chapter?
2. What assumption will you need to make about the sliding rod?
3. Are there forces involved? (a rhetorical question). Draw a free body diagram!
4. Which equations allow you to express the unknown quantities in terms of known ones?
5. Work through the algebra and solve for the desired unknown quantity. Substitute
values you know to get a numerical answer.
6. Has the question been answered?
7. Are the units and sign correct?
8. Is this a large or small value? To get an idea, you might note how big (i.e. the
maximum current) the circuit breakers or fuses in your home are.
WP31.3 Torque on a Wire
I
B
θ
A straight wire 79 cm long is oriented at an angle of 37° to a uniform magnetic field.
Which direction is the magnetic force? If the bottom end of the wire is fixed but the rest
of the wire is free to move, what is the torque about the end of the wire? Which direction
will it rotate? The field strength is 0.01T and the current in the wire is 10A.
!Focus problem
A magnetic field will exert a force on a current carrying wire. The force is perpendicular
to the wire, described by the right-hand rule. That force will tend to want to pivot the
wire about the end point. We want to calculate the magnitude and direction of this torque.
!Plan approach
The force on the wire due to the magnetic
"! "! field
"! is given by
F! = LI × B = LIBsin θ
This force acts along the entire length of the wire and is constant. Thus, for the purposes
of calculating the torque, we can treat the problem as if the force acted only at the wire’s
center – just as the force of gravity acts at an object’s center of mass. The torque is given
by
"! "! "!
τ! = r × F
7
!Execute plan
First we need to determine the direction of the force. From the right-hand rule, pointing
your fingers along I and curling them in the direction of B you should find that your
thumb points into the page, which is the direction of the force. Thus, the wire will tend to
rotate into the page. Curling you fingers in the direction of the rotation now, your thumb
will point in the direction of the"!torque, which is perpendicular to the wire. You should
"!
get the same answer using !!
r × F.
rotation
I
I
F
F
B
r=L/2
τ
Solving for the torque
τ = r sin φ ( LIB sinθ )
L
(sin 90°)LIB sin θ
2
= 12 L2 IB sin θ
=
Be careful with your angles. In the torque equation sin is the angle between the force
(into the page) and the vector r pointing from the axis to the point of application of the
force. This angle is 90°. It is not the same as the angle between the current and the
magnetic field.
Using the numbers given I get the magnitude of to be
τ = 1.88x10-2 Nm
!Evaluate answer
This isn’t a very large torque, though the current if fairly sizable. The magnetic field is a
reasonable value. As discussed in the solution to the previous problem, this field might be
the strength of a good horseshoe magnet. As you should be beginning to understand,
magnetic fields generally exert very weak forces.
WP31.4 Torque on a Current Loop
B
y
z
0
y
x
0
x
I
B
A square loop of wire with a current I = 1.7 A flowing in it is oriented parallel to a
uniform magnetic field as shown. Each leg of the loop has a length of 1 cm while the
8
strength of the magnetic field is 0.23 T. What forces are on the loop? Does this lead to a
net torque? If so, which direction is the rotation? Calculate the torque.
!Questions and suggestions
1. Put the problem in your own words. How does the magnetic field and current affect
the loop of wire?
2. What equations can help you determine the unknown quantities?
3. Apply the right-hand rule to find the directions of these quantities. Draw a free body
diagram indicating the forces and torques.
4. Obtain an algebraic expression for the forces and the torques. Then substitute
numbers to get a numerical answer.
5. Has the question been answered?
6. Are the units and sign correct?
7. Is the answer unreasonable? Do you expect a large or small force? Large or small
torque?
WP31.5. Bad Picture?
FB
B
electron gun
•
•
•θ •
y
0
•
•
•
•
screen
x
z
You television works by shooting electrons at a phosphor screen to create the image you
see. The electron gun aims the electrons at various parts of the screen using an electric
field to steer the charged particles. However, as we’ve seen, magnetic fields can deflect
charged particles, too. Can Earth’s magnetic field distort the image you see? Estimate the
maximum deflection of an electron aimed at the center of the screen 30 cm away from the
electron gun. The electrons have a speed of 3x107 m/s. Assume Earth’s magnetic field is
directed northward, oriented at an angle of 45° to the horizontal, and has a strength of
3.0x10-5 T. Does the orientation of the television matter?
!Focus problem
Magnetic fields can bend the path of moving charged particles. We’re asked to estimate
how much an electron is bent by Earth’s magnetic field.
!Plan approach
The force on a moving charged particle due
field is
"! to"!a magnetic
"!
F! = qv × B
9
This force is always perpendicular to the direction of the particle, so as the particle turns
the direction of the force also changes. To simplify the problem, we’ll assume the particle
isn’t deflected by very much. We can then make the approximation that the force is
constant and points perpendicular to the electrons initial direction. In this sense, the
magnetic field acts like gravity. Then we can use Newton’s second law and kinematics
with constant acceleration to find the particle’s deflection.
We’ll ignore the effects of gravity. First, however, we need to find the orientation of the
television to give maximum deflection.
!Execute plan
If the television is aligned so the electrons shoot northward, the magnitude of the
magnetic force is
F = qvB sin45°
while if the television points east or west
F = qvB
So the maximum deflection of the electrons will occur when the television points east or
west. The direction of the deflection can be found from the right-hand rule. If v points
west then B points up and to the left as you look at the screen. v × B then gives a force
down and to the left. However, since electrons are negatively charged, they will feel a
force up and to the right.
From Newton’s 2nd law we get
evB
a=
me
As in our Plan Approach, we’ll assume that the deflection is small and the force, and
consequently the acceleration, is constant. Thus, we can use the kinematic equations. I’ll
choose x along the direction of the electron path and y along the direction of the
deflection.
∆ x = v xt
∆ y = vy t + 12 at 2
The initial velocity in the y direction is zero. I’ll solve for t in the x equation and
substitute into the y equation. You should check me to see that you get an expression for
the deflection y of
evx B " ∆ x %
∆y =
$ '
me # v x &
Using the numbers provided, the deflection is
2
1
2
∆ y = 7.9 mm
!Evaluate answer
That seems like a pretty big deflection, one easily big enough to notice if you turn your
television. In fact, when I did this calculation I assumed I did something wrong – until I
tried it. I used my computer monitor. If you look closely, as you rotate the screen right or
left you can actually see the picture shift. The shift is not quite as large as we’ve
calculated. What approximation did we make that leads to this overestimate? You might
10
check the final velocity of the electron in the y direction to confirm that the
approximation is nonetheless pretty good. You should also calculate the effect of gravity
on the electron to check this assumption.
WP31.6 Fusion Energy
plasma inside Alcator C-mod
One topic of modern physics research is controlled fusion energy. Fusion, the process by
which the sun creates energy, involves superheating a gas of electrons and protons to
temperatures exceeding 106 degrees called a plasma. At high enough energy the protons
will combine to create helium and at the same time liberate energy. In order to trap these
hot charged particles strong magnetic fields are used. Alcator C-mod at the
Massachusetts Institute of Technology can achieved magnetic fields of over 8 T. Assume
that the containment vessel has a circular cross-section with a uniform magnetic field.
What is the minimum radius of the vessel so that both electrons and protons are trapped?
What is the period of the orbit for both protons and electrons? At 10 million degrees the
average velocity of the particles is about 2x107 m/s.
!Questions and suggestions
1. Describe the problem in your own words.
2. Which unknown quantities do you need to determine?
3. Which equations allow you to express the unknown quantities in terms of know ones?
4. Work through the algebra and solve for the desired unknown quantity. Substitute
values you know to get a numerical answer.
5. Has the question been answered?
6. Are the units and sign correct?
7. Is the answer unreasonable? If it is, you probably made an error somewhere. Go back
and check your work.
11
WP31.7. Going with the flow
Suppose in the situation described in Section 31.8 that the current in the wire is 5 A and
the charge density is 1022 electrons per meter. What is the velocity of the reference frame
such that only an electrostatic force is observed? What electrostatic force would an
observer in this reference frame measure on the proton q located 1 cm from the wire?
What acceleration will this produce on the proton?
!Focus problem
The situation describes how different observes will measure a different types of
interaction depending on their reference frame. An observer moving with the moving
electrons and single proton would observe a purely electrostatic interaction. We’re asked
to find the velocity of this reference frame, which is the same as the velocity of the
electrons flowing in the wire. We are also asked to calculate the magnitude of the force
and the acceleration the proton feels.
!Plan approach
From the text we know we can relate the current to the charge density and the velocity of
the charges through
I = λ0 v
We can then use the value of v we calculate to find the magnitude of the electrostatic
force. The electrostatic force is given by "!
"!
e
F! = qE
where the electric field for an observer moving with the electrons is given by
2kλ 0 2
E′ =
γβ
r
with "= v/c and γ = 1 / 1 − β 2
!Execute plan
We can find the velocity of the electrons with straightforward substitution
v = I / λ0 = I / (en) = 3.13 mm / s
Note that e is the charge on an electron and n is the number density of electrons, 1022/m.
The electrostatic force is then
2ke2 n (v / c)2
2ke 2 n v 2
e
F =
≈
r
r c2
1 − (v / c)2
12
F e = 5.02x10-26 N
This will give an acceleration of the proton of
a = F e / mp = 3.00 m / s
!Evaluate answer
Notice that the speed of the electrons is really pretty small. Did you expect this? Well
then, how is it that when you flip a light switch the light comes on almost instantly?
When we discuss current flow, it will become clear how this can be. The force we
calculated is extremely small, but the acceleration of the proton is not, due to its small
mass.
WP31.8. Half speed
Suppose in the situation described in the previous problem that the proton is moving at
only half the speed of the electrons. If you are an observer at rest with respect to the wire
what electrostatic force would you observe? Would you observe a magnetic force? If so,
of what magnitude?
!Questions and suggestions
1. Describe the problem in your own words.
2. Which equations can help you determine the two types of interaction?
3. In the relativistic situation described, what equations can help you calculate the
electric and magnetic fields?
4. For the situation described, what important simplification can you make?
5. Work out the needed quantities algebraically and then substitute numbers to get a
numerical answer.
6. Has the question been answered?
7. Does your answer make sense? Based on the previous problem, what do you expect
the value to be? Is your suspicion confirmed?
13
Questions and problems
Section 31.1
1.(I) Which way does the north pole of a compass needle point in the Southern
Hemisphere? Explain.
2.(I) What type of magnetic pole is located near the Earth’s south geographic pole?
3.(I) How are the elementary magnets aligned in the magnet illustrated in Figure 31.1(b)?
4.(II) Could a magnet have more than just two opposite magnetic poles? Explain.
5.(III) Figure 31.8a shows the polarity of a bar-shaped piece of uniformly magnetized
material. Describe the polarity of a spherically shaped piece of uniformly magnetized
material.
Section 31.2
6.(I) Is there a field line pattern that could everywhere represent either a magnetic field or
an electrostatic field? Explain.
7.(I) Which of the following field line patterns could represent a magnetic field? Explain.
8.(I) Are the field line patterns from a magnetic dipole and an electric dipole the same?
Explain why or why not.
9.(II) Suppose the non-uniform magnetic field in Figure 31.16c were weaker towards the
bottom left and stronger towards the top right. Describe what happens to a small bar
magnet, placed in this field, as shown below.
14
10.(III) Describe what happens to the small bar magnet in Figure 31.16c, if the nonuniform magnetic field lines go in the opposite direction (as redrawn below).
Section 31.3
11.(I) A long, straight, current-carrying wire lies in the plane of the page, as shown. The
magnetic field it produces at position P points out of the page. (a) In what direction does
the magnetic field point at position Q? (b) In what direction does the current flow in the
wire?
15
12.(I) A long, straight, current-carrying wire runs perpendicular to the plane of the page,
as shown. It produces a magnetic field which points to the right at position P. (a) In what
direction does the magnetic field point at position Q? (b) In what direction does the
current flow in the wire?
13.(II) A wire is coiled in the shape of a spiral spring, with closely-spaced turns. When
current is passed through it, does the coil tend to lengthen, shorten, or stay the same?
Does your answer depend on the direction of the current? Explain.
14.(II) A conducting rod is suspended from a spring in a region where a uniform
magnetic field points out of the page. The rod can be supplied with current by two thin
wires connecting its ends to the terminals of a battery (not shown). Does the tension in
the spring depend on how, or whether, the wires are connected? If so, which end should
be connected to the positive terminal of the battery in order to increase the tension in the
spring?
16
15.(III) Two rods, carrying currents, are perpendicular to one another, as shown. Describe
the net magnetic force and torque, if any, each rod exerts on the other.
Section 31.4
16.(I) A positive charged particle is at rest on the positive z-axis in system S. System S’
is moving along the positive x-axis, system S’’ is moving along the negative x-axis, and
system S’’’ is moving along the y-axis, all with speed v relative to S. The axes of all the
systems are oriented the same way and their origins coincide at time zero, when
observers in each system make measurements. In which systems do observers measure an
electric field?
17.(II) What can you say about the magnitudes and directions of the electric fields
measured in the previous problem?
17
18.(I) In which of the systems in Problem 31.04.01 do observers measure a magnetic
field?
19.(II) What can you say about the magnitudes and directions of the magnetic fields
measured in the previous problem?
20.(I) If two charged particles are at rest relative to one another, there is no magnetic
force between them. If one of the particles were to start moving, would a magnetic force
be exerted? Explain.
Section 31.5
21.(I) You measure a current of 2.5 A flowing through a wire to the right. (a) If the
charge carriers are only electrons, how many electrons pass through a cross sectional area
of the wire each second? (b) Which way are the electrons moving? (c) If this wire is in an
external magnetic field of 0.2 T directed upward, what is the force per unit length on the
wire? (d) Draw a diagram (remember the instructions in Fig. 31.15 for denoting into and
out of the page) showing the direction of the current, the field and the force on the wire
due to these.
22.(I) A 70-cm long wire carries a current of 1.4 A. The wire is at an angle of 53° to a
uniform external magnetic field. If the force on the wire is 0.2 N, what is the magnitude
of the magnetic field?
23.(II) An 8-cm long wire segment is free to slide along so that its ends just touch two
conducting rails (see figure). The rails allow a current to be maintained through the wire
segment as it slides along them. The rails are aligned vertically, with a magnetic field of
magnitude 0.25 T directed into the page. If the segment has an inertia of 0.004 kg, what is
the current in the wire if it doesn't fall under gravity?
Segment
B into page
24.(II) When 1 m long wire carries 20 A in the positive x-direction in a uniform magnetic
field, the force on the wire is measured as (3 ˆj + 2kˆ) N . For the same wire and magnetic
field, when the wire is rotated until the current flows in the positive y-direction, the force
on the wire is measured as (−3iˆ − 2kˆ) N . What is the magnitude and direction of the
magnetic field?
25.(II) A 7-cm long wire is bent in a right triangle such that the wire starts at the origin
and goes in a straight line to x= 3 cm, then in another straight line from x= 3 cm to x=3
#
cm, y=4 cm. The wire is in an external, uniform magnetic field B = 0.5 kˆ T . (a) Find the
force on the wire. (b) If the first wire is replaced by a straight wire 5 cm long carrying the
same current between the same endpoints, find the force on this new wire.
18
26.(II) Current I flows through a semicircular wire of radius r lying in a plane which is
#
perpendicular to a uniform magnetic field B . What is the total magnetic force on the
wire?
27.(II) A current carrying wire is bent into a circular loop of radius R, which lies in the x#
y plane. A uniform magnetic field B = Bkˆ exists over the plane of the loop. Find the
magnetic force on the loop.
28.(III) An arbitrary shaped tangle of wire is connected such that is carried a current Io
#
#
#
from position r1 to position r2 in a region occupied by a uniform magnetic field Bo .
#
# # #
Show that the force on the wire, F = I o (r2 − r1 ) × Bo .
29.(II) Assume you have the same set up as in Example 31.3. The position of the wire is
measured, and then a 5.0 mg piece of plastic is placed on the center of the rod. How
much current must be run through the rod to return it to its original position?
30.(III) Two long parallel conducting rods are connected such that a conducting crossbar
free to slide along them will have a constant current I0 running through it (see figure).
The parallel rods are separated by a distance $, and are set up in a region of uniform
magnetic field, B0 directed as shown. The crossbar has an inertia/length λ. (a) In which
direction will the crossbar move? (b) If there is a coefficient of static friction µ between
the rods and the crossbar, what is the minimum current I0 necessary for the wire to move?
Section 31.6
31.(I) At the Earth’s equator, B is horizontal and to the north, with a magnitude of
approximately 7×10-5 T. (a) What is the magnetic flux through a loop of radius 0.1 m,
lying flat on the ground? (b) If the same loop is balanced on its edge such that its axis
19
points northwest, what is the magnetic flux through the loop? (c) If the loop is now
rotated such that its axis points north, what is the magnetic flux through the loop? (d) If
the loop is now rotated such that its axis points west, what is the magnetic flux through
the loop?
32.(I) A square loop 10 cm to a side is placed on a wooden table in a uniform magnetic
field of magnitude 0.25 T. It is found that the biggest magnetic flux through the loop is
measured when the loop is flat on the table. What is the flux through the loop when it is
tilted such that the plane of the loop makes a 60° angle with the table? Draw a good
diagram indicating all necessary vectors.
33.(I) A circular loop of radius 10 cm is placed in a magnetic field of magnitude 0.03 T.
If the flux is measured to be 3.00×10-4 T⋅m2, what is the angle between the plane of the
loop and the field?
34.(II) A hemispherical bowl of radius r is placed in a uniform magnetic field. The open
#
top of the bowl is in the xy-plane. The magnetic field is given by B = Bo kˆ . Find the
magnetic flux through the surface of the bowl.
Section 31.7
35.(I) A proton moves in a circular orbit 15 cm in radius that is perpendicular to a
uniform magnetic field. The magnitude of the field is 0.25 T. (a) Find the angular
frequency and period of motion of the proton. (b) Find the speed of the proton. (c) Find
the kinetic energy of the proton.
36.(I) An electron with kinetic energy 7.5×10-15 J moves in a circular orbit perpendicular
to a uniform magnetic field of magnitude 0.35 T. (a) Find the radius of the electron's
orbit. (b) Find the angular frequency and period of motion of the electron. (c) Find the
speed of the electron.
37.(I) In Example 31.5, calculate the time required for the oxygen ion to complete the
semicircular path.
38.(I) An alpha particle has approximately four times the inertia (m= 6.65×10-27 kg) and
twice the charge of the proton. It is moving in a circle of radius 0.75 m perpendicular to
a uniform magnetic field of magnitude 0.75 T. (a) Find the angular frequency and period
of motion of the alpha particle. (b) Find the speed of the alpha particle. (c) Find the
kinetic energy of the alpha particle.
39.(II) A deuteron is a particle with approximately twice the inertia of a proton, but the
same charge. An alpha particle has approximately four times the inertia and twice the
charge of the proton. For this problem, assume the inertia ratios are exactly two and four.
If these three types of particles move in a uniform magnetic field at the same radii,
compare their (a) velocities, (b) kinetic energies and (c) angular momenta.
20
40.(II) A small particle has an inertia m and a charge q. The particle moves in a circle of
radius r when in is in a uniform magnetic field of magnitude B. Show that (a) its
momentum is given by Bqr and (b) its kinetic energy is given by B2q2r2/2m.
41.(II) Mg24 (inertia 3.983×10-26 kg) and Mg26 are to be separated using a mass
spectrometer. The magnetic field in the spectrometer is 0.577 T. What is the minimum
value of the potential difference through which these ions must be accelerated so that the
separation between them is 2.6 cm? (Assume the mass ratio is 26/24.)
42.(II) A small particle has an inertia m and a charge q. The particle enters a region of
uniform magnetic field of magnitude B, directed along the x axis. The initial velocity of
#
the particle is v = vix iˆ + viy ˆj . (a) Describe the motion of the particle. (b) Find how the
particle's position has changed from when it enters the field at t=0 to the time t=2πm/qB
after it enters the field.
43.(II) A proton moves in the +x direction through a velocity selector, undeflected, at
6.67×105 m/s. You measure the electric field and find it to be 2.0×105 N/C in the positive
#
z direction. (a) What is the magnetic field B ? (b) If the proton moves with twice this
speed, in which direction will it be deflected?
44.(II) The set up in Example 31.6 is modified. The region occupied by the fields is now
only 4 cm long in the x direction, where x is taken to be the original direction of motion.
The electrons then travel an additional 30 cm in the x-direction before hitting a
phosphorescent screen, which will "light up" where the electrons strike it. The electron
pass undeflected through the region of crossed fields when the electric field is tripled
(3.0×103 N/C) and the magnetic field is reduced by a factor of 10-3. (a) Find the speed of
the undeflected electrons. (b) If the magnetic field is turned off, so that the electrons
travel between the plates only acted on by the electric field, what is the total deflection of
the electrons when the reach the phosphorescent screen?
45.(II) Blood contains charged ions. For a large artery the blood flow speed is measured
at 0.6 m/s. The person in which this measurement is made is placed in a magnetic field
such that the artery is in a region of uniform magnetic field of magnitude 0.2 T. If a
potential difference of 1×10-3 V is measured across the diameter of the artery, what is the
diameter of the artery.
46.(II) A metal strip is 0.1 cm thick and 2.0 cm wide. It carries a 20-A current in a
uniform magnetic field of magnitude 2 T directed as in Fig. 31.38. The change in
potential across the strip is measured to be 4.27 µV. (a) If the left side of the wire, as
shown in the figure, is found to be at higher potential, what is the sign of the mobile
charge carriers? (b) What is the velocity of the charge carriers in the strip? (c) Find the
number density of charge carriers in the strip.
47.(II) The cross section of a copper strip is 0.1 cm thick and 2.0 cm wide. A 10 A
current is flowing through this cross section, down the length of the wire. The wire is
21
placed in a uniform magnetic field of magnitude 2 T directed as in Fig. 31.38. If the
number density of free electrons in copper is 8.47×1022 e/cm3, find (a) the velocity of the
electrons in the strip and (b) the change in potential across the width of the strip.
Section 31.8
48.(II) On average, the number density of free electrons in copper is 8.47×1022 e/cm3. (a)
Calculate what the charge density λ for a copper wire 1 mm in radius would be if this
quantity of electrons were missing. (b) Now, assume this copper wire is neutral in the lab
frame. Calculate the charge density λ′ in the frame moving with the electrons if the
electrons are moving at a drift velocity of 4.7×10-4 m/s (this corresponds to a current of
about 20 A for this size wire).
22
Answers to review questions
1. A magnet is an object that attracts iron, nickel, cobalt, & certain alloys, that are called
magnetic materials. (In particular, a magnet itself is composed of magnetic material.)
2. The places on the surface of a magnet where the attraction of iron is strongest are
called poles. There are two types, which are always paired, designated north or south
poles, depending on how they align relative to a compass needle (which is itself a small
bar magnet, with a pair of opposite poles at the ends). Opposite poles attract; like poles
repel.
3. Some atoms act like tiny bar magnets. These atoms (or groups of them) are called
elementary magnets, & magnetic materials are made up of them. In a magnet, all the
elementary magnets are aligned, & the places on the surface where predominantly one
kind of pole of the elementary magnets is exposed are the poles. In an unmagnetized
piece of magnetic material, the elementary magnets are randomly oriented, so their
magnetic effects cancel.
4. A magnet is surrounded by a magnetic field, which exerts a force on the poles of
another magnet.
5. Magnetic field lines point in the direction of the magnetic field, that is, in the direction
the north-pole of a freely suspended compass needle would point. The field line density is
proportional to the magnetic field strength.
6. It doesn’t; the magnetic field line flux through a closed surface is always zero, since
magnetic field lines always form closed loops.
7. An electric current in a conductor, defined as the rate at which charge flows past a
fixed point in the conductor, causes a magnetic field.
8. The magnetic field lines are circles, in planes perpendicular to the wire & centered on
it, with the field line density (proportional to the field strength) decreasing with distance
from the wire.
9. An electric current in a magnetic field experiences a force, unless the directions of the
current & the field are parallel or anti-parallel. The direction of the force is perpendicular
to both the current & the field, according to the right hand rule preceding Example 31.2.
10. One expects this because electric currents, which both cause & interact with magnetic
fields, depend on velocity, & velocity must be relative to an observer.
11. The huge electric forces between the charged constituents of different objects of
ordinary matter balance to zero. However, even a small relativistic correction to such a
huge force is of measurable size, so the magnetic force (if the objects are magnets or
carry currents) is observable.
23
12. The observer at rest with respect to the wire, Si in Figure 31.27(c), sees equal charge
densities for the ions & the charge carriers, because the wire appears neutral in this
system. An observer moving with the carriers, Se in Figure 31.27(b), sees a greater charge
density for the ions, & thus the wire does not appear neutral in the moving system.
13. The magnetic force on a current carrying wire in a magnetic field is a maximum when
the current is perpendicular to the field; it is zero when the two are parallel or antiparallel.
14. The magnitude of the magnetic force depends on the magnitude of the field, the
current, the length of wire in the field, & the sine of the angle between the directions of
the current & the field (see Equation 31.4).
15. Make measurements for conditions when the force is a maximum & zero. The
magnitude of the field is the maximum force per unit length, per unit current (see
Equation 31.3). The direction of the field is along the direction of the wire, when it is
oriented so that the force is zero, consistent with the right hand rule for the direction
when the force is a maximum.
16. The flux is the surface integral of the normal component of the field (see Equation
31.9).
17. Gauss’s law for magnetism states that the magnetic flux through any closed surface is
always zero. This implies that there is no equivalent of electric charge as a source for
magnetic fields (i.e. “magnetic charges”, “magnetic monopoles”, or isolated magnetic
poles do not exist).
18. The flux through any surface bounded by the loop is the same (see Equation 31.14),
provided the normal to surfaces are in the same direction.
19. The sum of the forces on each of moving charges that make up the current in the
segment of wire
.
20. The force is directed perpendicular to the magnetic field’s direction and the particle’s
velocity
21. The observer see the distance between ions decrease because length contraction as the
ions move relative to this observer.
22. Coulomb’s law.
24
Chapter 32 (version 03/15/06)
Magnetic fields of particles in motion
Review questions
Answers to these questions can be found at the end of this chapter.
Section 32.1
1. What are the sources of magnetic fields?
2. Are magnetic forces central?
3. Is a single magnetic pole one of the sources of a magnetic field?
Section 32.2
4. What general shapes can a magnetic field line have?
5. How are the patterns of magnetic field lines from a current loop and a bar magnet
related?
6. What does the magnetic field from a spinning charged particle look like?
Section 32.3
7. What is the direction of the magnetic dipole moment vector?
8. What are three different situations where a right hand rule for magnetism is used?
9. Describe the magnetic interaction experienced by a current loop placed in a uniform
external magnetic field.
10. What does the commutator do in an electric motor?
Section 32.4
11. What properties of electric and magnetic field lines make the electric and magnetic
fluxes through closed surfaces different?
12. How do the line integrals around closed paths differ, for electrostatic and
magnetostatic fields?
13. What is Ampere’s law?
14. How does one determine whether a current encircled by an Amperian loop makes a
positive or negative contribution to the line integral of the magnetic field around that
loop?
Section 32.5
15. What is the proportionality constant in Ampere’s law?
16. What kind of symmetry is displayed by the magnetic field of a long straight wire,
carrying a steady current? How does this symmetry affect the choice of Amperian loops
to determine the magnitude of this magnetic field using Ampere’s law?
17. How does the magnitude of the magnetic field from a long straight wire, carrying
steady current, depend on the distance from the wire? Suppose the wire had a radius R
1
and uniform current density. Is the spatial dependence of the magnetic field different in
different regions? From the point of view of Ampere’s law, why is this?
18. Describe the magnitude and direction of the magnetic field produced by a large flat
sheet of uniform current.
Section 32.6
19. What is a solenoid? Describe the magnetic field (magnitude and direction) produced
by a long solenoid carrying a steady current.
20. What is a toroid? Describe the magnetic field (magnitude and direction) produced by
a toroid carrying steady current.
21. Does the magnetic field inside a very long solenoid differ from that inside a tightly
wound toroid, carrying the same current in the same number of turns? If so, how?
Section 32.7
22. Compare and contrast the form of the infinitesimal magnetic field from a small
current-carrying segment, in the Biot-Savart law, with that of the electrostatic field from
a small element of charge, in Coulomb’s law.
23. How is the proportionality constant in the Biot-Savert law related to the definition of
the ampere?
24. Comment on the range of applicability of the Biot-Savart law, relative to that of
Ampere’s law, for finding magnetic fields.
Section 32.8
25. How could the magnetic field from a uniformly moving charged particle be expressed
in terms of the electric field of the of the particle?
26. What is the origin of the two cross products in the expression for the magnetic force
between two moving charged particles?
27. What is the value of the product of physical constants that occurs in the ratio of the
magnitudes of the magnetic and electric forces between two moving charged particles?
Does this value suggest anything?
2
Developing a feel
Calculate or estimate the following quantities:
1. The maximum magnetic field you would likely be exposed to due to current in the
electrical wiring in your house. (E, L, Q)
2. The straight-wire current needed to reverse the deflection of a compass needle on
your lab table. (H, A, P, X)
3. The maximum magnetic field strength 10 m from a typical lightning bolt. (G, S)
4. The maximum value of B you can produce in an open-air-core solenoid on your lab
table. (D, J, O, R, V)
5. The maximum force per meter between the anti-parallel (oppositely flowing) currents
in your household wiring. (Q, E)
6. The electric current around the equator that would be needed to produce the Earth’s
magnetic field at the North pole. (B, I, N, T)
7. The magnetic field strength at the nucleus of a hydrogen atom caused by the electron
in orbit, using the Bohr model. (C, K, U)
8. The magnetic field strength at the North pole that might be attributed to rotation of
Earth, assuming a uniformly charged surface. (F, M, I, W)
Hints:
A. What is the horizontal component of Earth’s magnetic field in your neighborhood?
B. What is the magnetic field strength at the North pole of the Earth?
C. What is the radius of the circular orbit?
D. What are the crucial variables to maximize?
E. What is the current configuration?
F. What is the typical value of E near the Earth’s surface?
G. How can we model the current?
H. How should the wire be oriented?
I. What is the radius of the Earth?
J. What is the limiting factor that you have least control over?
K. What is the speed of the electron?
L. How close do you come to the wiring?
M. What surface charge density could create the desired E field?
N. What is the straight line distance from a point on the equator to the North pole?
O. What maximum current is reasonable?
P. What magnetic field strength is needed?
Q. What is the maximum likely current in each conductor?
R. What minimum wire diameter can handle the current?
S. What is the peak current?
T. How can you compute the vector sum of B due to each tiny element of the current
loop?
U. What is the charge carried by the electron?
3
V. What is the maximum number of turns?
W. How can the current be modeled?
X. How close to the wire can you place the compass?
Key: A. ~2×10-5 T. B. ~7×10-5 T. C. ~5×10-11 m. D. B=µonI, so both I and n must be
maximized (note the conflict in desirable wire diameter). E. Paired conductors with
equal, opposite currents separated by ~10-2 m. F. E~100 V/m. G. ~ a vertical line current.
H. The compass will detect horizontal B field, so place the wire vertically. I. ~6000 km.
J. Lab tables are powered with 20 Amp, 120 V circuits, so the useful power limit is ~2
kW . K. Electrical attraction provides centripetal force, so v~2×106 m/s. L. ~0.1 m if you
stand near a wall. M. From Coulomb’s Law, σ ~10-9 C/m2. N. ~9000 km. O. Even with
thick wire, R~1Ω, so I~40 A will approach the power limit. You may be able to scrounge
a power supply that can deliver this. P. Enough to more than cancel Earth’s field, say
~4×10-5 T . Q. ~20 A for large appliances. R. Plan on #10 solid copper wire, diameter
~2.5 mm. S. ~105 A. T. Use the Biot-Savart rule and the argument in section 32.2. U. e =
1.6×10-19 C. V. In order to fit the resistance limit (and to fit on a lab table), better keep
n~500 turns. W. Consider a stack of horizontal current loops of varying radius (careful;
current depends on radius). X. ~0.02 m.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any
particular section. Typically, an example that is worked out in detail is followed
immediately by an example whose solution you should work out by following the
guidelines provided.
WP 32.1: Electrical wiring. An electrician while wiring a new house lays two wires in
the wall designed to carry the current indicated in the figure. She worries that the
magnetic field near the wire might be a problem. To assuage her, calculate the direction
and magnitude of the magnetic field at the point x=0, y =4 m. (Each tick mark is 1 m.)
B?
i =10 A
i =16 A
60°
!Focus problem
Currents can generate magnetic fields. Here two current-carrying wires will each create a
magnetic field at the point indicated. We are asked to calculate the value of the field.
!Plan approach
In this chapter we calculated the magnetic field generated by a long, straight wire to be
B=
µ0 I
2π r
Since magnetic field is a vector we will need to add the vector components of the field to
get the total field magnitude and direction. We also know that near a wire the magnetic
field lines form concentric circles. This should help determine the direction.
!Execute plan
Just by observation we can get an idea of the direction of the magnetic field. Since the
field lines form closed loops around the wire the magnetic field from the right wire (10
A) will point out of the page at B, based on the right-hand rule (Point your thumb in the
direction of the current. Your fingers will curl in the direction of the magnetic field).
Using the same logic, the field from the left wire will also point out of the page. Great –
5
we’ve found the field’s direction already. It’s out of the page. Now for the magnitude.
The total magnitude of field will be
B=
µ 0 ! Il Ir $
# + &
2π " rl rr %
r is the minimum perpendicular distance to the point of measurement. In this case rr = 4m
sin 30° =2 m, while rl = 4 m. Putting in the other numbers, the magnitude of the magnetic
field is
"!
-6
!!B = 1.30x10 T, out of the page
!Evaluate answer
This is not a very strong magnetic field, so I guess her worries were groundless. Very
large currents are required to create strong magnetic fields, usually of the order of several
thousand amps. Be sure when you calculate magnetic fields to include the direction; it is
a vector quantity.
WP 32.2: Speaker wire. A student runs a very, very long speaker wire around his room
to the speaker around the corner. The wire is bent at right angles with a circular arc of
radius 1 cm. Calculate the magnetic field at point P if the current in the wire is 540 mA.
/
r=1 cm
P•
!Questions and suggestions
1. Describe the problem in your own words.
2. What assumptions will you need to make?
3. Can you break the problem into parts?
4. Which equations will help you get the magnetic field for the various parts?
5. Can the equations specific to a slightly different situation be modified for the
situation at hand?
6. Magnetic field is a vector. How will you determine the vector direction?
6
7. Work through the algebra and solve for the desired unknown quantity. Substitute
values you know to get a numerical answer. Be sure and find the vector direction and
include it in your answer.
8. Has the question been answered?
9. Are the units and sign correct?
10. Does your answer behave as you would expect with curvature radius?
WP 32.3 Galvanometer. A galvanometer is an old fashioned device used to measure
current. In the simple model show in the figure it consists of a coil of wire in a magnetic
field with a spring attached to one end. With no current in the loop the spring is relaxed
and the needle points straight up. When current flows in the coil a torque is induced and
the needle swings right or left, indicating the magnitude of the current on a scale. As long
as the angle is not too large, the spring is stretched only vertically, as shown. Viewed
from above, which direction is the current flowing in the coil in the figure? Derive an
algebraic expression for the current in the coil as a function of the angle φ. The coil is
square with sides of length a. Calculate the current when the needle makes an angle of
5.7°. The coil has N=100 turns, the magnet has a uniform field of 0.01 T and the spring
constant is 2.0 N/m.
φ
N
S
∆x
!Focus problem
The poles of the magnet produce a magnetic field that is directed from the north to the
south poles, i.e., left to right in the diagram. We have discussed how a loop of current
will experience a torque when placed in a magnetic field. In the last chapter we
calculated, in Worked Problem WP 31.4, what the magnitude of such a force might be. A
galvanometer makes practical use of this property. We’re asked to find the angle the
needle makes with the vertical as a function of the current in the loop.
!Plan approach
Newton’s second law tells us that the angular acceleration of an object is proportional to
the sum of the torques.
"!
"!
'
! τ = Iα
7
We have two torques in this system: the torque exerted by the spring and the torque due
to the current in the loop. These two torque balance to give zero acceleration of the
needle.
The torque on an object can be found from the applied force using
"! "! "!
τ! = r × F
The spring’s force is exerted downward with magnitude F s = k | ∆x | . The force on the
left segment of the current-carrying wire that is perpendicular to the page is directed
upward, while that on the right segment is directed downward as shown in Figures 32.11
and 32.14. The magnitude of each force is
" "
F = | LI × B |= LIB
From WP 31.4 we already calculated for a single loop of wire in a horizontal orientation.
We can borrow from those results here. We just need to put these pieces together to get φ
as a function of I. The direction we can find from the right hand rule.
!Execute Plan
From the figure we need an upward force"!on the
"! left arm of the current loop and a
I
×
B
I find the current must flow counterdownward force on the right arm. Using !
clockwise when viewed from above. The other two side legs will have an outward
directed force on them, which does not produce a torque. This is the same result we found
in WP 31.4.
Since the needle does not accelerate once a steady current is applied, the sum of the
torques is zero
τ ϑB + τ ϑs = τ B + (−τ s ) = 0
τ B =τ s
"! "!
The torque due to the magnetic field on each segment is r! × F B , the cross product of the
position vector from the loop center to the wire segment ! and the force due to the
magnetic field. The magnitude of the cross product is rFBsin!, where !=½"-φ is the
angle between the position vector and the force. The total current through each segment
is NI, where N is the number of turns in the loop. Because there are two wire segments,
the torque on the loop because of the magnetic field is
"
"
τ B = 2|r ×FB |
= 2 [ 12 a(aNIB) sin( 12 π − φ ) ]
= a 2 NIB cos φ
The torque exerted by the spring can be approximated, for small angles, by
8
"
"
τ s =| r ×Fs |
= 12 a | k ∆x | sin( 12 π − φ )
= 12 a | k ( 12 a sin φ ) | cos φ
≈ 14 ka 2 sin φ cos φ
We have used the substitution ∆x = 12 a sin φ . This is an approximation, since we have
assumed that the spring stretches only vertically, whereas in reality it will also be pulled
slightly to the right.
Setting the two torques magnitudes equal to each other, I solve for the current to get
I=
k
sin φ
4NB
For small angles sin! # ! (in radians!) so the angle if deflection is linear with current.
I=
k
φ
4 NB
Putting in the numbers provided, we get
I = 49.7 mA
!Evaluate result
Looking at the equation we derived, notice how the angle does not depend on the size of
the loop we use; only the number of turns is important and the strength of the magnetic
field. Can you figure out why the loop size doesn’t play a roll? Hint: look at the
dependencies of the two opposing torques. Be careful with your angles. Go through the
problem and make sure you fully understand each cross product and where each angle
comes from. Torque can be a fairly complex subject because of all the angles involved.
The magnitude of the current is pretty reasonable. Galvanometers will often have a
switch to allow you to measure different ranges of current to insure that the angle indeed
stays small. Moreover, a real galvanometer will have a more sophisticated construction.
A spiral spring is used instead, and the magnets are designed so that the magnetic field
points radially in and out towards the center of the coil. This problem is a good example
of how to solve seemingly complex problem by putting several smaller pieces together.
You need to reach back to Chapter 9 to recall how springs work. Chapter 13 discussed
torque.
9
WP 32.4: Force on a rectangular current loop. An integrated circuit board has
rectangular loop of wire is placed next to a very long straight wire a distance x away.
Both have a current of 39 mA flowing in them as shown. What is the force exerted on the
loop? The rectangle is 5.7 mm long, 0.9 mm wide and is located 0.3 mm from the wire.
I
x
I
a
L
!Questions and suggestions
1. Put the problem in your own words.
2. Can you break the problem into parts?
3. What equations can help you determine the unknown quantities?
4. What simplification can you make immediately just by looking at the problem? What
assumptions do you need to make.
5. Apply the right-hand rule to find the directions of the magnetic field and force.
6. Get an algebraic expression for the total force. Then substitute numbers to get a
numerical answer.
7. Has the question been answered?
8. Are the units and sign correct?
9. Is the answer unreasonable? Do you expect a large or small force? What does this say
about your observations in the real world about current-carrying wires?
WP 32.5: Wire with uniform current density. A long straight wire with a circular
cross section has a current I flowing through it with a uniform distribution across the
surface. Derive and expression for the magnetic field as a function of radius r inside and
outside the wire. Plot the magnetic field as a function of radius.
a
!Focus problem
10
r
Currents create magnetic fields with field lines as closed loops around the wire. We are
asked to calculate the magnetic field for a current distributed across the area of the wire,
rather than just outside an infinitely thin current channel.
!Plan approach
Because of the circular symmetry, this problem makes for a good application of
Ampere’s law
"! "!
B
(! ⋅ ds = µ0Ienc
Ampere’s law is to currents what Gauss’s law is to charges. Enclosed charges create an
electric flux through a surface, enclosed currents create a magnetic field around a path.
We should have no trouble applying Ampere’s law in this case of a uniform current
distribution, just as we applied Gauss’s law for a uniform charge distribution in the
Worked and Guided Problems of Chapter 28.
!Execute Plan
We’ll draw an Amperian loop at a radius r inside the wire. Because of the symmetry, the
magnetic field along this circular path is constant, and we can bring it outside the
integral.
B( ds = µ0 Ienc
B(2π r) = µ0 Ienc
The uniform distribution of the current means, the current enclosed is
Ienc = I
πr 2
2
2 = I(r / a)
πa
Therefore, the magnetic field as a function of radius is given by
B(2π r) = µ0 I(r / a)2
B=
µ0 I r
2π a2
Outside the loop the enclosed current remains constant, so the magnetic field is the same
as that for a straight wire
B=
µ0 I
2π r
A sketch of the magnetic field looks like the following
11
B
r
a
!Evaluate result
Because of the uniform current distribution, as you go out in radius more and more
current is enclosed in your Amperian loop. Thus, the magnetic field increases linearly
until you reach the edge of the wire. Once outside the wire the current enclosed remains
constant so the field drops off as 1/r. Notice the similarity between this problem and
Worked and Guided Problem 28.7.
WP 32.6: Magnetic field of a coaxial cable. A coaxial cable consists of an inner
conductor and an outer stranded shell typically separated by some dielectric material.
One reason for using “coax” is that the magnetic field is “trapped” in the cable when
current flow through the wire. Show why this is so by calculating the magnetic field
outside the inner conductor both inside and outside the cable, assuming a current flow in
one direction in the center conductor and in the opposite direction in the outer shell.
!Questions and suggestions
1. Describe the problem in your own words.
2. How is this problem similar to the previous problem? What approach seems like the
simplest way to attack this problem?
3. Which equations allow you to express the unknown quantities in terms of know ones?
4. Work through the algebra as done in the previous problem and solve for the desired
unknown quantity. Sketch a graph of the magnetic field as a function of radius to
check the behavior inside and outside the cable.
5. Has the question been answered?
6 Are the units and sign correct?
7. Is the answer unreasonable? If it is, you probably made an error somewhere. Go back
and check your work.
12
WP 32.7: Equivalence. Using the equation derived in Section 32.8, show that the force
due to the magnetic field of a current-carrying wire on a moving charged particle is equal
to the force due to the magnetic field of the charged particle on the wire. Assume that the
charged particle moves parallel to the wire at a distance a away from it.
v
FwcB
FcwB
I
!Focus problem
Basically, we’re asked to prove the physics always gives the right answer no matter how
you do the problem. Here we want to show that a charged particle moving in the
magnetic field of a wire feels the force as the wire does due to the magnetic field of the
moving particle.
!Plan approach
We know from the previous chapter that a charged particle will feel a force due to a
magnetic field of
"! "! "!
F! = qv × B
Here, B is created by the wire. We also know that a current-carrying wire feels a force
due to an outside magnetic field given by
"! "! "!
F! = LI × B
In this case B is created by the moving charged particle. We’ll substitute for the
appropriate fields and see if we can get the two forces to be equal.
!Execute Plan
The first part we can simplify by just substituting the magnetic field of a long straight
wire
F = qv
13
µ0 I
2πa
The direction of the force is toward the wire, since the magnetic field due to the wire is
out of the page at the location of the charged particle (confirm this!).
The second step isn’t quite as straightforward. As the figure below shows, the moving
particle exerts a magnetic force on a segment of the wire dx proportional to the distance r
to the segment. The magnitude of this force is
dF = IBcdx
v
a
θ
r
x
I
dx
The magnetic field due to a moving charged particle is
"! µ0 qv"!× rˆ
B=
2
! 4π r
Substituting this into the force equation gives
dF = I
µ0 qv sin( 12 π − θ )
dx
4π
r2
Again, the cross product tells us that the wire is attracted towards the charged particle.
There are two methods of approaching this integral. In the first method you might note
that r 2 = a 2 + x 2 . Substituting for cos! = a/r yields the integral
∞
µ qva
dx
F= I 0
(
4π −∞ ( x 2 + a 2 )3
This integral is “look-upable”, and you can get and expression for the force.
Suppose, however, you’re on a desert island, or worse, trapped in an exam. There is an
approach that leads to a much simpler integral to solve: We’ll choose our variable of
integration to be !. We can change the differential dx by first noting that x = a tan!. Since
a is constant we can take the derivative of both sides
14
d
1
x=a
2
dθ
cos θ
dθ
dx = a
cos 2 θ
Again, substituting for cos! = a/r yields the much simpler integral
π/ 2
µ qv
F= I 0
cos θdθ
4π a −π(/2
Note how I’ve changed the limits of integration appropriate for the new variable !.
Doing the integral I get
µ0 qv
[sin θ ]π−π/ 2/2
4π a
µ0 qv
F= I
2π a
F= I
The force is identical to the force found in the first part.
!Evaluate result
Hurray! Physics is saved. Such checks are very invaluable for confirming the validity of
equations we derive. If the checks don’t pan out, it’s an indication that our derivation is
suspect.
WP 32.8: Magnetic field of an orbiting electron. In the classical model of the
hydrogen atom an electron orbits a single proton at the Bohr radius of r = 0.53 Å. What is
the magnetic field at the location of the proton produced by the orbiting electron? Which
direction must the electron be orbiting to produce the field depicted?
B
ve
!Questions and suggestions
15
1. Describe the problem in your own words.
2. What quantities do you need to find? Can you break the problem into parts to help
you find them? Are there previous problems (from other chapters) that might help
you?
3. Which equations can help you determine the quantities you need? What force keeps
the electron in orbit? Can you use this information to help you find the electron’s
orbital speed?
4. Work out the needed quantities algebraically and then substitute numbers to get a
numerical answer.
5. Has the question been answered?
6. Does your answer make sense? Do you expect a large or small value? Are you
surprised?
16
Questions and problems
Section 32.1
1.(I) Consider two charged particles, capable of spinning. In which of the following
circumstances is there a magnetic interaction between them?
(a) Both are at rest and not spinning.
(b) Both are at rest and only one is spinning.
(c) Both are at rest and spinning.
(d) One is at rest and not spinning; the other is moving and not spinning.
(e) One is at rest and not spinning; the other is moving and spinning.
(f) One is at rest and spinning; the other is moving and not spinning.
(g) One is at rest and spinning; the other is moving and spinning.
(h) Both are moving and not spinning.
(i) Both are moving and only one is spinning.
(j) Both are moving and spinning.
2.(II) At a certain instant, two positively charged particles are moving in perpendicular
directions in a plane, as shown, where the second particle is along the direction of motion
of the first. Describe and sketch the magnetic forces between the pair.
3.(I) A negatively charged particle is at rest at the midpoint between two bar magnets, as
shown. The magnet on the left is twice as strong as the magnet on the right. What is the
direction of the magnetic force on the particle?
17
4.(I) A negatively charged particle is a rest in a region where a uniform magnetic field is
pointing in the positive x-direction and a uniform electric field is pointing in the positive
y-direction. Describe the direction of the combined electromagnetic force on it.
5.(II) In the Bohr model of the hydrogen atom, an electron orbits a nucleus, consisting of
a proton. Given that the electron and proton are both elementary particles with spin, what
types of different magnetic fields and interactions would you expect might be associated
with the hydrogen atom? Explain.
Section 32.2
6.(I) Explain where you expect the magnitude of the magnetic field due to a circular
current loop to be the greatest.
7.(I) Sometimes, rectangular geometry is more convenient than circular. How would the
magnetic field from a small square current loop compare with that from a small circular
current loop?
8.(I) Many bar magnets have a rectangular cross-section, rather than circular. Could Fig.
32.6(b) represent their magnetic fields?
9.(II) A negatively charged particle, at rest at the origin, is spinning clockwise around an
axis in the x-direction (when looking from the positive x-axis). What is the direction of
its magnetic field at points on the positive or negative (a) x-axis, (b) y-axis, and (c) zaxis?
Section 32.3
10.(I) The main component of the Earth’s magnetic field (outside the interior) is a dipole
field, like that shown in Fig. 32.6(c). In what direction is the Earth’s magnetic dipole
moment?
11.(I) The Earth’s magnetic field is thought to be generated by currents in the liquid outer
core. In what direction (clockwise or counterclockwise, when viewed along the rotation
axis from north to south) does this net current circulate?
12.(I) A current loop lies in the xy-plane, with current circulating counterclockwise when
viewed from the positive z-axis. Is there a torque on the loop, and if so, in what direction,
if a uniform magnetic field is applied in the (a) x-direction, (b) y-direction, and (c) zdirection?
13.(II) A current loop, lying in the xy-plane, experiences a clockwise torque around the
y-axis (viewed from the positive y-axis) due to the presence of a uniform magnetic field
in the x-direction. How does the current circulate around the loop?
14.(III) A current loop lies in the xy-plane, with current circulating counterclockwise
when viewed from the positive z-axis, It experiences a counterclockwise torque around
18
the x-axis (viewed from the positive x-axis), due to a uniform magnetic field. Describe
the direction of this field.
15.(III) A negatively charged particle is a rest in a region where a uniform magnetic field
is pointing in the positive x-direction and a uniform electric field is pointing in the
positive y-direction. The particle is spinning clockwise around an axis in the z-direction
(viewed from the positive z-axis). (a) Would the combined electromagnetic force be
different than if it were not spinning? Explain. (b) Would there be any other changes, and
if so, explain?
Section 32.4
16.(I) A current loop, carrying current I as shown, lies in the xy-plane. For each of the
Amperian loops indicated, is the line integral of the magnetic field positive, negative, or
zero?
17.(I) Wires numbered 1 to 5 carry currents into or out of the page, as shown. What is the
magnitude of the net current encircled by the Amperian loop indicated? Is the line
integral of the magnetic field around this loop, in the direction indicated, positive,
negative, or zero?
19
18.(II) Wires numbered 1 to 3 carry steady currents perpendicular to the page. The line
integrals or the magnetic field around the three Amperian loops shown all have the same
positive value. How do the magnitudes and directions of the three currents compare?
19.(II) A positively charged particle, located at the origin, is spinning counterclockwise
around the z-axis (looking from the positive side). Is the line integral of the magnetic
field positive, negative, or zero for an Amperian loop going (a) clockwise around the
origin, in the xy-plane (looking from the positive z-axis), (b) clockwise around the origin,
in the yz-plane (looking from the positive x-axis), (c) clockwise around the origin, in the
xz-plane (looking from the positive y-axis)?
20.(III) The diagram shows magnetic field lines in a certain region. Are there any
locations where there must be currents flowing perpendicular to the diagram? (Consider
locations near the numbers on the axis shown.) What types of current are present and in
what directions? Explain.
20
Section 32.5
21.(I) Two current carrying wires separated by a very small distance run parallel to each
other, one with a current of 3.0 A to the right and the other with a current of 4.0 A to the
left. Give the approximate value for the magnitude of the magnetic field a large distance
R from the two wires.
22.(I) The magnitude of the magnetic field due to a current carrying wire 2.5 cm away is
2.0×10-5 T. Calculate the magnitude of the current in the wire.
23.(II) A moving particle of charge e traveling to the right with a velocity of 2.5×107 m/s
initially feels no magnetic force. When a long current-carrying wire is placed parallel to
the line of motion of the particle 3.0 µm away, the particle instantaneously feels a force
of 7.0×10-13 N. Give the magnitude of the current running through the wire.
24.(II) Current is uniformly distributed throughout a long, straight wire of diameter 5 cm.
If the total current through the wire is 6 A, calculate the magnitude of the magnetic field
at a point (a) 2 cm away from the center of the wire and (b) 5 cm away from the center of
the wire. (c) What would the current through the wire need to be to create a magnetic
field of 1 T at the point 5 cm away from the center of the wire?
25.(II) Two large, flat current-carrying sheets are placed parallel to each other so that one
sheet is above the other. The top sheet carries a current 2.0 A per unit length to the left
and the sheet underneath it carries a current 5.0 A per unit length to the right. Calculate
the magnitude of the magnetic field (a) between the two sheets, (b) above the top sheet,
and (c) below the bottom sheet. (d) Draw a diagram showing the direction of the
magnetic field in each of these regions.
26.(II) A proton travels to the right with velocity 3×106 m/s between two large, flat sheets
that are parallel to each other and the proton’s line of motion. If a current of 8 A to the
right through the top sheet and 8 A to the left through the bottom sheet is then generated,
21
what is the magnitude of the magnetic force exerted on the electron? Draw a diagram
showing the direction of the magnetic field in between the two sheets and the direction of
the magnetic force on the electron.
27.(III) A particle of mass 9.11×10-31 kg and unknown charge is shot at a velocity of
2.0×104 m/s to the right and enters the magnetic field generated by a large, flat currentcarrying sheet that is parallel to its own line of motion. The particle exits the magnetic
field 9 cm above the point where it enters. If the sheet carries a current of 3 A per unit
length, what is the magnitude of the charge of the particle?
28.(II) A wire carrying current of 1.5 A to the left is placed above a large, flat sheet
through which current is uniformly distributed at 3.0 A/m to the left. What is the
magnitude and direction of the magnetic force exerted on the wire?
Section 32.6
29.(I) A long solenoid with 3 windings per cm has a current of 1A running through it.
Calculate the magnetic field inside the solenoid.
30.(I) You need to produce a magnetic field of .07 T using a long solenoid. If the
maximum current you are able to run through the windings is 20 A, what is the minimum
number of windings per cm the solenoid must have?
31.(II) A long, straight wire carries a current of 2.5 A to the left. A solenoid with 10
windings per cm and a current of 4.5 A running through it is placed directly above and
parallel to the wire, such that the distance from the wire to the center of the coils of the
solenoid is 5 cm (See diagram). Calculate the magnitude and direction of the magnetic
field at the center of the solenoid.
32.(II) Calculate the magnitude of the magnetic force exerted on a 2.0 cm long wire
carrying a current of 4.0 A when it is placed inside a solenoid at an angle of 45° to the
magnetic field. The solenoid has 7 turns per cm and carries a current of 3.0 A.
33.(II) A toroid with square windings of 50mm on each side and inner radius 120 mm has
250 windings carrying a current of 3.0 mA. Solve for the magnitude of the magnetic
field (a) at the center of the square coils, (b) at the midway point between the center of
the toroid and the inner radius of the coils, and (c) 30 mm beyond the coils.
34.(III) For what distance from the center of a toroid of 200 windings does the magnetic
field equal the magnetic field inside a solenoid of 5 turns per cm if the current through
both the toroid and the solenoid is the same?
35.(II) A toroid with circular windings of radius r and inner radius R has n windings per
unit length (as measured from the inside edge of the windings) and carries a current I.
Derive an expression for the magnetic field of such a toroid at any point inside the
circular loops.
22
Section 32.7
36.(II) What is the magnitude of the magnetic field of a wire carrying current 3 A and
extending from x=0 to x=10 at a point P that is a perpendicular distance 2.0 m away from
the bottom of the wire at x=0?
37.(II) A 3.0-m wire of mass .010 kg/m is fixed in place with a current of 10 A running
through it to the right. Another wire is placed parallel to, and 5.0 mm directly above, the
first wire. When the second wire is released, it stays in place. What must the magnitude
and direction of the current through the second wire be?
38.(I) Calculate the magnitude of the magnetic field at a point at the center of a circular
arc of radius 2.5 cm spanning an angle of ½" and carrying a current 3.0 A.
39.(II) Use the Biot-Savart Law to derive an expression for the magnetic field at the
center of a circular current loop. Compare the magnitude of the magnetic field at the
center of a circular current loop of radius 1 cm and width 1 mm to the magnitude of the
magnetic field at the center of the loops of a solenoid of the same radius and with 1 turn
per mm. Assume the current is the same through the current loop and the solenoid.
40.(I) Two parallel wires of length 2m and separated by a distance of 25cm carry currents
of 2A to the right and 3.0 A to the left, respectively. Determine the direction and
magnitude of the magnetic force exerted by wire 2 on wire 1.
41.(I) A 5.0-m wire carrying a current of 3.0 A experiences magnetic force of 4.0×10-7 N
per when placed 9.0 cm away from a wire running parallel to it. What must the
magnitude of the current in the second wire be?
42.(II) A wire of mass 10 g and length 1.0 m with 1.5 A of current running through it is
initially at rest on a table. When another wire is placed parallel to the first wire a distance
2.0 mm away, the first wire begins to move across the table away from the second wire.
If the coefficient of static friction between the first wire and the table is .05, then what
must be the minimum current through the second wire?
43.(II) Find the magnitude and direction of the magnetic field for the point P as shown in
the diagram. The distance from P to the large semi-circle is 7.0 cm and the distance to
the small semi-circle is 2.0 cm. The current through the wire is 3.0 mA.
Section 32.8
44.(I) An electron moves in a straight line at a speed of 6×107 m/s. Calculate the
magnitude and direction of the magnetic field at some point 5mm behind the electron and
15mm below its line of motion.
44.(I) Two electrons separated by a distance of 10 nm are traveling parallel to each other
with velocities -4×107 m/s and 7×106 m/s, respectively. What is the magnitude of the
magnetic force that the second electron exerts on the first electron?
23
45.(II) A proton traveling in the -x direction with some velocity v1 is directly below
another proton traveling with velocity v2 so that its path makes an angle of 45º with the
positive x-axis. If the two protons are separated by some distance r, calculate the
magnitude of the magnetic force that each proton exerts on the other.
46.(II) An electron is shot out of an accelerator with a velocity of 2.0×103 m/s. 1 µs later,
a second electron is shot out of an accelerator on a path parallel to, and 10 mm below, the
path of the first electron. The velocity of the second electron is 5.0×103 m/s. (a) What is
the magnitude of the magnetic field at a point 3.0 mm ahead of the first electron and 5
mm below it 3 µs after the first electron was shot out? Assume that the magnetic forces
the electrons exert on each other for the 2.0 µs before you make your calculations does
not alter the electrons’ paths significantly. (b) Show whether or not it is valid to assume
that the magnetic forces the electrons exert on each other before the 2 µs do not alter the
electrons’ paths significantly.
47.(II) In Chapter 31, you learned that a charged particle traveling through a uniform
magnetic field perpendicular to its line of motion will feel a magnetic force that causes it
to take circular path. Give an expression for the magnitude of the magnetic field due tot
he charged particle at the center of the circular path in terms of the uniform magnetic
field, the charge of the particle, and the mass of the particle.
48.(II) An electron traveling to the right at 1.5×106 m/s is accelerated downwards at 9.0
m/s2 due to the magnetic force of another electron directly beneath it traveling with
velocity 4.0×106 m/s to the left. What must the distance between the electrons be?
24
Answers to review questions
1. The source of all magnetic fields is charged particles in motion, either moving or
spinning (including currents).
2. Magnetic forces are not central, i.e. they do not point along the line joining two
source elements.
3. No, because there are no single magnetic poles; magnetic poles always occur in pairs.
4. Magnetic field lines always form closed loops, possibly extending to infinity.
5. The magnetic fields from a current loop and a bar magnet both resemble a dipole
field. In fact, the field of a bar magnet is like the field from a coaxial configuration of
current loops.
6. The magnetic field from a spinning charged particle looks like the field from a small
(infinitesimal) dipole magnet.
7. The magnetic dipole moment vector is in the direction of the magnetic field through
the center of the dipole. This is from S to N in a small bar magnet, or according to the
right hand rule in Fig. 32.9(c) for a current loop.
8. Three applications of right hand rules in magnetism (to find the circulation of
magnetic field lines around a current element, or the force on a current element in an
external field, or the magnetic dipole moment of a current loop) are illustrated in Fig.
32.9.
9. The net force on a current loop in a uniform field is zero, but there is a net torque
tending to keep the loops magnetic moment aligned in the direction of the external field.
10. The commutator reverses the direction of the current in the moving coils of a dcmotor, so that the direction of the magnetic torque always causes rotation in the same
direction.
11. Electrostatic field lines originate or terminate on charged particles; magnet field lines
form closed loops. Therefore, the electric flux through a closed surface is proportional to
the amount of charge inside, while the corresponding magnetic flux is always zero.
12. The line integral around a closed path of an electrostatic fiend is always zero; that for
a magnetostatic field depends on the amount of current encircled.
13. Ampere’s law states that the line integral of the magnetic field around a closed path is
proportional to the net current encircled by the path.
25
14. If the Amperian loop encircles the current in the same direction as the magnetic field
from that current, the current makes a positive contribution to the line integral; otherwise,
it makes a negative contribution. (If you turn a right hand screw in the direction of the
Amperian loop, it would advance in the direction of a positive encircled current.)
15. The proportionality constant in Ampere’s law, between the line integral of the
magnetostatic field around a closed path and the amount of current encircled, is called the
permeability constant (of free space) $0, and has the value 4" × 10-7 T m/A.
16. The long straight wire has cylindrical symmetry, so the closed magnetic field lines are
concentric circles with constant magnitude at each radius. In using Ampere’s law to
determine this magnitude, one must choose an Amperian loop coinciding with a field
line.
17. The magnitude of the magnetic field at any point outside the wire is inversely
proportional to the perpendicular distance to the axis of the wire. This is because all the
current in the wire is encircled (see Equation 32.4). At points inside the wire, the
magnitude is proportional to the distance from the axis, because only a fraction of the
current is encircled, which for uniform current density, depends on the ratio of the area
encircled (see solution to Checkpoint 32.12).
18. The magnitude is uniform, on either side of the sheet, given by Equation 32.10. The
magnitude is the same on both sides of the sheet. The direction of the field is parallel to
the sheet and perpendicular to the current, as given by the right hand rule shown in Fig.
32.23(b). The direction is diametrically opposite the two sides.
19. A solenoid is a long coil of wire used to produce a magnetic field. The field pattern is
actually like that of a similarly shaped bar magnet, and much stronger inside the solenoid
than outside. In the limit of a thin tightly wound solenoid, the field is approximately
uniform inside, proportional to the current and the number of turns per unit length (see
Equation 32.15), and approximately zero outside.
20. A toroid is a doughnut-shaped coil of wire, as would result from bending a solenoid
into a circle so it ends meet. If the toroid is tightly wound, its magnetic field is
approximately confined to the interior of the coils, with magnitude depending inversely
with the distance to the center of the toroid, and direction in accordance with the right
hand rule for the direction of current circulating around the coils. See Equation 32.18 and
Fig. 32.28(b).
21. The field inside the solenoid is approximately uniform, but the field inside the toroid
is not – it depends inversely on the distance to the center of the toroid.
22. Both fields are directly proportional to the strength of the source and inversely
proportional to the square of the distance, but the electric field is along the unit vector to
the field point, while the magnetic field depends on the cross product of the current
segment with this unit vector. (See Equations 27.14 and 32.22.)
26
23. The magnetic field from a long straight current-carrying wire depends on the BiotSavart law constant, and so does the force per unit length exerted on a similar parallel
wire. This underlies the practical definition of the ampere.
24. The Biot-Savart law is a prescription for calculating the magnetic field produced by a
steady current, circulating in any closed circuit, provided, of course, that the integral is
amenable to analytic or numerical evaluation. Ampere’s law, on the other hand, is limited
to a few special situations, where the direction of the field pattern is known from
symmetry and only the magnitude must be determined.
25. Use of Equations 27.4 and 28.11 allow the magnetic field in Equation 32.42 to be
#
# #
expressed as B = ε o µo v × E .
26. One cross product comes from the magnetic field produced by particle 1, the other
from the interaction with it of particle 2, which “feels” this field.
27. The product of these constants is %0$0 = (8.85 ×10-12)(4" ×10-7)(m/s)-2 = 1.11 ×1017
(m/s)-2 = 1/(8.99×016 m/s) = 1/(3.00 ×108 m/s)2. It is no accident that this is the
reciprocal of the square of the speed of light.
27
Chapter 33 (version 04/16/06)
Changing magnetic fields
Review questions
!"#$%&#'()'(*%#%'+,%#(-)"#'./"'0%'1),"2'/('(*%'%"2')1'(*-#'.*/p(%&.
Section 33.1
1. What are two possible results that could occur due to the motion of a conducting
object relative to a magnetic field?
2. What is the difference between an induced current and a regular current?
3. Do the effects produced on moving conductors in a magnetic field depend on the
direction of the motion of the conductor relative to the field?
Section 33.2
4. How does changing the frame of reference affect your explanation of induced current
in a conductor moving relative to a uniform magnetic field?
5. What is Faraday’s law?
6. What does electromagnetic induction mean?
Section 33.3
7. Must there be a magnetic force on the charge carriers in a conducting loop in order
for an induced current to flow?
8. Does the nature of the force that causes electromagnetic induction depend on the
choice of reference frame, and if so, how, or if not, why?
9. What is the difference, if any, between induced electric field lines and electrostatic
field lines?
Section 33.4
10. What does Lenz’s law say happens in circuits?
11. What general physical principle underlies Lenz’s law?
12. What are eddy currents?
Section 33.5
13. What is the significance of the negative sign in Faraday’s law?
14. If the magnetic force on charge carriers never does work on them, where does the
energy for induced currents come from, for example, in a situation where part of a
circuit is moving in a magnetic field (as in Fig. 33.19)?
15. How does induced emf differ from potential difference?
16. What kind of emf is induced in a flat coil, rotated at a constant angular speed in a
uniform magnetic field?
Section 33.6
1
17. How is the induced electric field related to the induced emf?
18. How does the induced electric field vary, in magnitude and direction, in a circular
region of uniform magnetic field?
Section 33.7
19. Changing the current in one circuit will induce an emf in a second nearby circuit. Will
changing the current in a circuit induce an emf in the circuit itself, even when no
other circuits are nearby?
20. What characteristic of a circuit does its inductance describe. On what properties of
the circuit does it depend?
21. What is the SI unit of inductance?
Section 33.8
22. Work must be done against the induced emf in an inductor to change the current in it.
What happens to this energy?
23. How does the magnetic energy in an inductor depend on the current through it? Does
it depend on other properties of the inductor?
24. How does the magnetic energy in an inductor depend on the magnetic field strength
within it? Is this relationship applicable to magnetic fields other than those in
inductors?
2
Developing a feel
4/5.,5/(%')&'%#(-6/(%'(*%'1)55)$-"7'+,/"(-(-%#8'
1. The potential difference you can generate by rotating a metal clothes hanger? (P, Q,
O, B)
2. You separate the two wires leading to your reading lamp. How fast is the magnetic
field changing at a point a few cm from one of the wires? (H, T, F)
3. The potential difference generated around a metal clothes hanger held near one of the
wires in the previous problem (so that the wire is in the plane of the clothes hanger).
(P, H, T, F)
4. The magnetic energy stored in 1 m3 of air from the Earth's magnetic field? (O)
5. The potential difference generated around the periphery of your key ring when you
walk under a residential power line? (A, S, G, U, L, F)
6. The maximum magnetic energy stored in the volume of a doghouse placed under the
power lines leading to your house? (A, E, R, U, C, I)
7. The induced electric field around the outside of your car when you park it under the
high tension power lines leading from a nuclear powered electric generating plant. (A,
D, N, V)
8. The magnetic field needed to have the same energy density as gasoline (3*107 J/liter).
(I)
9. The amount of wire needed to make a 1 Henry inductor in the form of a toroidal coil
with a ring radius of 20 cm and a loop radius of 5 cm. (J, M)
10. The maximum potential difference between the two ends of a 2-m metal rod dropped
from the top of a 10 story building so that it falls horizontally. (Q, K, O)
Hints:
A. How much current does the power line carry?
B. How rapidly can you rotate the loop?
C. What is the volume of a dog house?
D. What potential difference do they carry?
E. What potential difference is the power coming in to a typical house?
F. How often does it change?
G. What potential difference is it at?
H. What is the current in the wire?
I. How much energy is stored in a magnetic field?
J. How many turns of wire are needed (see eq 33.29)?
K. What is the maximum speed of the rod?
L. How large is the conductive loop?
M. How long is one turn?
N. What is the power output of the plant?
O. How large is the magnetic field?
P. How large is the loop of wire?
3
Q. What is the source of the magnetic field?
R. How much power does a typical house use?
S. What power does it carry?
T. How large is the maximum magnetic field?
U. How large is the maximum magnetic field?
V. What is the 'radius' of the car?
Key: A. Estimate the power and the potential difference to get the current. B. Between 1
and 10 times per second. C. A typical dog house is about 1 m on a side. D. Typically
about 0.75 MV. E. 220 V. F. It goes from positive maximum to negative maximum 60
times per second. G. Local power lines carry power at more than 110 V but less then 20
kV. H. The reading lamp will be between 40 and 100 W. 9 = :; and ; = 110 V. I. See
eq 33.38. J. µ0 = 4!×10"7, 0 = 0.05 m, < = 0.2 m K. There are about 3"4 m per floor. It
takes more than 1 s and less than 20 s to fall so =avg = 35 m / 5 s and =max = 2=avg.
Alternatively, you can use the equations for falling bodies. L. Your key ring is about 1-2
cm in radius so ! = !&2. Remember to use meters. M. the circumference of one turn is
2!0. N. Typically about 1 GW = 109 W. O. 0.5 G = 5×10-5 T P. About 10 cm by 30 cm.
Q. The Earth. R. Each home uses more than 1 kW (a space heater uses 1.5 kW) and less
than 100 kW. S. The power line serves between 10 and 100 homes. Each home uses
more than 1 kW (a space heater uses 1.5 kW) and less than 100 kW. T. > = µ0 : / (2!&) =
2×10-7 : / & where & ranges from 1 to 11 cm (or 0.01 to 0.1 m). U. See hint (T). How
high up are the power lines? V. About 1 m. See Eq. 33.21.
4
Worked and guided problems
?*%#%'%@/6p5%#'-"=)5=%'6/(%&-/5'1&)6'(*-#'.*/p(%&A'0,('/&%'")('/##).-/(%2'$-(*'/"B'
p/&(-.,5/&'#%.(-)"C'?Bp-./55BA'/"'%@/6p5%'(*/('-#'$)&D%2'),('-"'2%(/-5'-#'1)55)$%2'
-66%2-/(%5B'0B'/"'%@/6p5%'$*)#%'#)5,(-)"'B),'#*),52'$)&D'),('0B'1)55)$-"7'(*%'
7,-2%5-"%#'p&)=-2%2C'
WP 33.1: Faraday’s Law and Conservation of Energy. Two parallel conducting rails
of length '5 = 6.0 m are separated by a width $ = 1.0 ×101 cm . The rails are connected at
one end by a third conducting rail. The entire track is placed on a table where there is a
!
uniform magnetic field > , of magnitude > = 1.2 T , pointing downward, perpendicular to
the plane formed by the rails. A conducting bar of length $ is placed on the rails and is
pulled away from the end consisting of the third rail by an external agent with constant
velocity with magnitude = = 5.0 m/s . The resistance of the bar is < = 4.0 × 10−1 # and the
resistance of the three rails is negligibly small. What force must be applied to the bar by
the external agent to maintain its constant velocity?
!Focus problem
The moving bar and the three rails forms a rectangular closed conducting path. As the bar
moves, the area of this rectangle increases hence there is a changing magnetic flux.
Therefore there is an induced electromotive force that drives an induced current through
the moving bar. Since the induced current in the bar is in a magnetic field, there is an
!
!
induced force on the bar E0/& ,-"2,.%2 ≡ E0- . By Lenz’s Law, this induced force acts like a
resistive force so in order for the bar to move at a constant velocity, the external agent
!
!
must apply a force E0/& ,/pp5-%2 ≡ E0/ exactly equal but in an opposite direction to this
!
!
resistive force to insure that the total force on the bar is zero, E0/ + E0- = 0 , and hence will
move at a uniform velocity.
Figure 33.01: Overhead view of bar moving on rails.
5
!Plan approach
The applied force must be equal and opposite to the induced force,
!
!
E0/ = − E0- .
In order to calculate the induced force in the bar
!
! !
E0- = : -"2 # × > ,
we must calculate the induced current, : and determine the direction of the induced
current flow using Lenz’s Law. We can calculate the magnitude of the induced current
from the induced electromotive force according to
: -"2 = ε -"2 / < .
So we must first calculate the electromotive force induced in the loop
ε
-"2
=−
! !
2
>
⋅ 2! ,
2( !!
which depends on the time changing magnetic flux. So we must first calculate the
magnetic flux through the loop formed by the bar and the rails,
! !
Φ > = !! > ⋅ 2! ,
and then determine how the flux is changing in time in order to calculate the
electromotive force.
!Execute plan'
Since we will be calculating vector quantities, we should make some choices for vector
unit directions. This gives an easy reference system for any vector calculation. Choose
positive -ˆ direction in the direction of the velocity, choose the positive F̂ direction to
point from the bottom of the bar to the top of the bar, and positive D̂ out of the table
(opposite the direction of the magnetic field) to keep a consistent right handed coordinate
system, as shown in the figure.
6
! !
We should now calculate the magnetic flux Φ > = !! > ⋅ 2! through the loop formed by
!
the rails and the Choose the area element, 2! = −2! Dˆ , to point in the same direction as
!
the magnetic field. The magnetic field > = − > Dˆ is constant and in the same direction as
the area element. The dot product in the integrand of the magnetic flux integral is then
! !
> ⋅ 2! = (− > Dˆ) ⋅ (−2! Dˆ) = >2! .
Since the magnetic field is constant, it can be pulled out of the integral and the magnetic
flux is then the area times the magnetic field
Φ =
' >
!! >2! = > !! 2! = >! .
At the time '( shown in the figure the bar has moved to a position @ = =( . The area of the
loop is then '! = $@ = $=( . Therefore the magnetic flux is
Φ > = >! = >$=( .
The time derivative of the magnetic flux through the loop at time ( is
2Φ >
2(
=
2
>$=( = >=$ .
2(
So there is a non-zero electromotive force in the closed path formed by the bar and the
rails,
ε
"'
-"2
=−
r r
2
> ⋅ 2! = − >=$ = −(1.2T)(1.0 × 10−1 m)(5.0 m ⋅ s-1 ) = −6.0 × 10−1 V .
!!
2(
What is the current in the conducting loop? When viewed from overhead, does the
induced current flow clockwise (in the - ˆF direction in the bar as shown in the figure) or
counterclockwise (in the + ˆF direction in the bar as shown in the figure)?
7
The magnetic flux is downwards (into of the table) and increasing as the area increases so
by Lenz’s Law, an induced current flows in the path formed by the bar and the rails that
will produce an induced magnetic flux that must point upwards to oppose the increase
downwards in magnetic flux. This corresponds to a counterclockwise direction (as seen
from above) for the induced current. You can use your right hand to verify this. Point
your thumb upwards from the table. Then your fingers will curl counterclockwise. So the
induced current flows from the bottom of the bar to the top of the bar in the figure which
is in the + F̂ direction.
If the loop has resistance'< , then the electromotive force will produce an induced current
in the loop according to
ε
-"2
= : -"2 <.
The electromotive force is equal to the change in magnetic flux so
: -"2 < = − >=$ .
We have already determined the direction of the induced current so the magnitude of the
induced current is
: -"2
>=$ 6.0 ×10−1 V
=−
=
= 1.5 A .
4.0 ×10−1 #
<
Since there is an induced current in the bar, there is an induced force on the bar. The
direction of the induced force is opposite the direction of the motion of the bar. The
current flows from the bottom of the bar to the top (in the figure) and the magnetic field
is into of the table, so the cross product between the current and the magnetic field
(usinthe right hand rule) points to the left in the −-ˆ direction. Mathematically, we can
!
!
write both the induced current element, : -"2 # = : -"2 $ ˆF , and the magnetic field, > = − > Dˆ ,
as vectors, We can calculate the induced force by taking
!
! !
E0- = : -"2 # × > = : -"2 $ ˆF × (− > Dˆ) = − : -"2 $> -ˆ .
Since we just calculated the magnitude of the induced current, : -"2 = >=$ / < , the
magnitude of the induced force is
!
> 2 $2 = (1.2T) 2 (1.0 ×10−1 m) 2 (5.0 m ⋅ s-1 )
E0- = : -"2 $> =
=
= 1.8 ×10−1 N .
(4.0 ×10−1 #)
<
So the vector expression for the induced force is
8
!
> 2 $2 = ˆ
E0- = −
- = −1.8 ×10-1 N -ˆ .
<
!
The external agent must apply a force E0% on the bar exactly equal but in an opposite
direction to the induced (resistive) force to insure that the total force on the bar is zero,
!
!
E0% + E0- = 0 , and hence will move at a uniform velocity. Therefore the external agent
must apply an equal and opposite force
!
!
E0% = − E0- = 1.8 ×10-1 N -ˆ .
!Evaluate answer'
The first check is that the direction of the induced force opposes the applied force. The
induced force is about 0.2 N . The induced force is not so commonly experienced but it is
approximately the same force as you would feel holding about eight pennies in your
hand.
The magnetic field is quite large, 24,000 times the magnetic field of the earth, and the
velocity of the bar is a fifty times smaller than the speed of a Boeing 747 airplane. The
width of a bar is 600 times smaller that the wing span of the airplane. Since the induced
electromotive force is the product of >=$ , the order of magnitude of the induced
electromotive force is the same order of magnitude as generated in the airplane wing. So
ε = 0.6 V is not unreasonable.
' -"2
There is a very special way to check your answer. The power that the external agent
applies in moving the bar is jus the product of the force times the speed of the bar
! ! > 2 $2 =
> 2 $2 = 2 (1.2T) 2 (1.0 ×10−1 m) 2 (5.0 m ⋅ s -1 ) 2
ˆ
ˆ
90% = E0% ⋅ = =
- ⋅= - =
=
= 9.0 ×10-1 W .
−1
<
<
(4.0 ×10 #)
Where does this energy per second go? The resistance to the flow of the induced current
in the bar generated heat. You will see in the next chapter that rate that the heat is
generated (this is called Joule heating) is given by the product of the induced current
squared with the resistance,
9F),5% = : -"2 2 < = (1.5 A ) ( 4.0 ×10−1 # ) = 9.0 ×10−1 W .
2
Note that this is also
9F),5% = : -"2
2
> 2 = 2 $2
.
<=
<
9
So you see that
90% = 9F),5% .
We should expect our answer to agree because the only energy loss is through Joule
heating in the wire due to the induced current. Since the external agent is pulling the bar
at a constant speed there is no increase in kinetic energy or any type of potential energy.
Thus the power put into the system leaves as heat.
WP 33.2: A Falling Loop. A rectangular loop of wire with mass 6, width $, vertical
length 5, and resistance < falls out of a magnetic field under the influence of gravity. The
magnetic field is uniform with magnitude > and out of the paper within the area shown
(see sketch) and zero outside of that area. At the time shown in the sketch, the loop is
exiting the magnetic field at speed, =(( ) > 0 (meaning the loop is moving downward, not
upward). What is the terminal velocity of the loop?
Figure 33.2 A loop falling through a magnetic field
!Questions and suggestions
10
1. In order to get started on this problem, you need to choose a coordinate system. In
particular think about how you plan to parameterize the area of the loop where there
is magnetic flux.
2. As the loop falls, the magnetic flux through the loop is changing that generates and
induced current. The induced current will slow down the falling loop (Lenz’s Law).
So you will first need to calculate the magnetic flux Φ > through your loop at time (?
Be careful, because the magnetic field is not uniform across the entire area of the
loop! (it is zero over some parts and constant and non-zero over other parts).
3. What is changing magnetic flux 2Φ > / 2( ? Is this positive or negative at time (? Be
careful here, your answer should involve =(( ) , and remember that '=(( ) > 0 .
4. Lenz’s Law says that the induced current should be such as to create a self-magnetic
field (due to the induced current alone) that tries to keep things from changing. What
is the direction of the self-magnetic field due to your induced current inside the circuit
loop, into the page or out of the page? Is it in a direction so as to keep the flux
through the loop from changing? What is the direction of your induced current, is it
clockwise or counterclockwise?
5. What is the magnitude of the current flowing in the circuit at the time shown in the
sketch (use': -"2 =| ε -"2 | / < )?
6. Draw a force diagram for all the forces acting in the vertical? Recall that there is an
induced force acting on the bar. Give the magnitude and direction of this induced
!
! !
force in terms of the quantities given (hint: E-"2 = : -"2 # × > ). Write an appropriate
component-version of Newton’s second law for vertical direction.
7. What is the condition on the acceleration in the vertical direction when the body has
reached terminal velocity (it’s speed no longer changes)? What is the magnitude of
the terminal velocity in terms of the given quantities?
8. Check the units of your answer to make sure that you have a dimensional correct
expression for the terminal velocity.
9. Check your answer by determining whether the rate at which gravity is doing work on
the loop is equal to the rate at which energy is being dissipated in the loop through
Joule heating.
WP 33.3: Rectangular loop near a wire. A very long straight wire carries a current :
that varies as a function of time according to : (( ) = / + 0( , where / = 0.5 A and
'0 = 4.0 A/s for the time interval (-) 0 < ( ≤ 2 # . For ( > 2 # the current is held constant.
The wire is placed a distance '* = 0.04 m above a rectangular loop of wire with width
'$ = 0.1 m , length '5 = 0.6 m , and resistance < = 2.8 Ω . The self-inductance of the wire is
negligible. (a) Find expressions for the induced electromotive force in the loop and the
direction of the induced current in the loop for the time interval (-) 0 < ( ≤ 2 # , and (--)
'( > 2 # ?
!Focus Problem
11
A sketch of the problem is shown in Figure 33.3. We can determine the magnetic field of
a very long straight wire by using Ampere’s Law. Recall that the magnetic field falls as
the inverse of the distance from the wire. The magnetic field lines form circles around the
wire so there is a flux of the magnetic field through the loop. The wire, loop, and
direction of the magnetic field of the wire through the loop is shown in Figure 33.3.
below.
Figure 33.3 Current in a wire and associated magnetic flux through a loop
Since the current in the wire is changing in time, the magnetic field of the wire is also
changing in time. Therefore the magnetic flux through the rectangular loop is changing in
time. Once we calculate the changing magnetic flux we can use Faraday’s Law as we did
in WP 33.1 to calculate the induced electromotive force and the induced current.
!Plan approach
We can determine the magnetic field of a very long straight wire by using Ampere’s Law.
!
!
#! > ⋅ 2 # = µ :
0 %".
.
The magnetic flux through the loop is given by
! !
Φ > = !! > ⋅ 2! .
The magnetic field falls off like the inverse of the distance from the wire so it is not
uniform over the area of the loop. We will therefore need to integrate the field over the
loop in order to calculate the flux.
12
Electromotive force depends on a time changing magnetic flux. The current in the wire
and hence the magnitude flux is only changing during the interval 0 < ( ≤ 2 # . For the
interval '( > 2 # , the magnetic field no longer varies as a function of time, the magnetic
flux through the loop is constant and therefore both the electromotive force and the
induced current is zero.
Hence we will only need to analyze the time derivative of the magnetic flux with respect
for the interval '0 < ( ≤ 2 # in order to calculate the electromotive force induced in the
loop
ε
-"2
=−
! !
2
>
⋅ 2! .
2( !!
We can calculate the magnitude of the induced current from the induced electromotive
force according to
: -"2 = ε -"2 / < .
We first calculate the electromotive force induced in the loop
ε
-"2
=−
! !
2
> ⋅ 2! .
!!
2(
which depends on the time changing magnetic flux. So we must first calculate the
magnetic flux through the loop formed by the bar and the rails,
! !
Φ > = !! > ⋅ 2! ,
and then determine how the flux is changing in time in order to calculate the
electromotive force.
For the interval '( > 2 # , the magnetic field is no longer varies as a function of time, the
magnetic flux through the loop is constant and therefore both the electromotive force and
the induced current is zero.
!Execute Plan
For the time interval '0 < ( ≤ 2 # :
First use Ampere’s law
!
!
#! > ⋅ 2 # = µ :
0 %".
13
,
to find that the magnitude of the magnetic field due to a current-carrying wire at a
distance & away is given by
>=
µ0 : µ0 (/ + 0()
=
.
2π &
2π &
Since the magnetic field varies as 1 / & , the field is essentially constant over a thin strip of
thickness ' 2& , length '5 , and area 2! = 5 2& , 2! = 5 2& as shown in the sketch below.
The differential magnetic flux '2Φ > through the this differential area element is then
2Φ > = >2! = >52& =
µ0 (/ + 0()
52& .
2π &
The total magnetic flux is integrating from & = * to & = * + $
! ! µ (/ + 0( )
Φ > = !! B ⋅ 2A = 0
2π
!
*+ $
*
5 2& µ0 (/ + 0( ) 5 " * + $ #
ln $
=
%.
2π
&
& * '
According to Faraday’s law, the induced electromotive force is the negative of the
derivative of the magnetic flux, so
ε-"2 = − 2Φ2(
'
>
=−
+
µ 05
2 ( µ0 ( /( + 0)5
ln(1 + $ / *) - = − 0 ln(1 + $ / *) .
*
2( )
2π
2π
,
Note that the electromotive force is independent of time for the interval '0 < ( ≤ 2 # . The
numerical value of the magnitude of the electromotive force is
14
ε -"2
=
µ005
ln(1 + $ / *) = (2 × 10−7 T ⋅ m/A)(4.0 A/s)(0.6 m)ln [1+(0.1 m/0.04 m) ]
.
2π
= 6.0 × 10−7 T ⋅ m 2 /s
Recall that one tesla is T = N/(A ⋅ m) . So the units of electromotive force are volts:
T ⋅ m 2 /s = N ⋅ m 2 /(A ⋅ m ⋅ s) = N ⋅ m/(A ⋅ s) = N ⋅ m/C = J/C = V .
Thus magnitude of the electromotive force is
ε -"2
= 6.0 × 10−7 V .
The straight wire carrying a current : produces a magnetic flux into the page through the
rectangular loop. The magnetic field increases as time increases, and hence the magnetic
flux also increases into the plane of the loop. By Lenz’s law, the induced current in the
loop must be flowing counterclockwise in order to produce an induced magnetic field out
of the plane of the loop in Figure 33.3 to counteract the increase in flux due to the time
changing magnetic field. The magnitude of the induced current is then
: -"2 =
ε
-"2
=
µ005
ln(1 + $ / *)
2π <
<
.
6.0 × 10−7 V
−7
=
= 2.1×10 A
2.8 Ω
!Evaluate answer'
Moving a magnet through many turns of a coil will produce an induced current on the
order of microamps so you should not be surprised that the current through one turn of
wire is about a 1/10 of a microamp even though a fairly large current is flowing through
the wire and the loop is close to the wire. Inductive effects are small.
We can use Ampere’s Law to find the magnetic field at the center of the loop at '( = 0 ,
µ/
>(( = 0 s) = 0 = (2 × 10−7 T ⋅ m/A)(0.5 A)/(2π )(0.09 m) = 1.8 × 10−7 T .
2π &
'
The magnetic field at '( = 2 s is 17 times stronger (this factor is just the ratio of the
currents), '>(( = 2 s) = 3.0 × 10−6 T . The average change in the magnetic field at the
center is then "'∆> / ∆( ; (3.0 × 10−6 T)/(2.0 s) = 1.5 × 10−6 T/s . We can estimate the
average time changing flux, by multiplying by the area of the loop and find that
−6
−7
'∆φ / ∆( ≈ (∆> / ∆()(5$) = (1.5 × 10 T/s)(0.6 m)(0.1 m) = 1.8 × 10 V .
15
This is the same order of magnitude as the induced electromotive force. We expect it to
be smaller because the magnetic field gets stronger when we are closer to the wire so
using the midpoint underestimates the induced electromotive force.
WP 33.4: Moving loop. A rectangular loop of dimensions 5 and $ moves with a constant
velocity v away from an infinitely long straight wire carrying a current : in the plane of
the loop. Let the total resistance of the loop be <. What is the direction and magnitude of
the induced current in the loop at the instant the near side is a distance & from the wire?
Figure 33.4 Loop moving away from a current flowing through a long straight wire
!Questions and suggestions
1. Begin by calculating the magnetic field at a distance # from the straight wire using
Ampere’s law.
2. Since the magnetic field is not uniform in space, you will need to integrate the
magnetic field over the loop in order to calculate the magnetic flux. Choose a
differential area element 2!G52# of the loop and then integrate to find the total flux.
You should be integrating from # = & to # = & + $ .
3. Differentiate the magnetic flux with respect to ( to find the time changing magnetic
flux. Remember that the loop is moving so that the distance from the wire '& is a
1 2&
2
ln(& + $) =
and
function of time. You may want to get in mind that
& + $ 2(
2(
1 2&
2
ln(&) =
.
& 2(
'2(
4. Now express the changing magnetic flux in terms of the velocity of the loop. What is
the electromotive force?
5. Determine using Lenz’s Law which way the induced current is flowing. Check that
your choice provides opposes the change of magnetic flux. You can also check by
determining which way the induced force points according to your determination of
the induced current. The induced force on the loop must oppose the motion.
16
6. Check that your answer for the induced current has units that are equal to amps.
WP 33.5: Solenoid. A long solenoid with length 5 = 0.6 m and a radius '< = 0.1 m
consists of 'H = 2760 turns of wire. A current : = 4.0 A is passed through the coil. Find
the energy stored in the system.
!Focus Problem
There are several approaches to finding the stored magnetic energy in the system. In the
first approach, we can think of the magnetic energy as stored in the magnetic field so we
need to find the magnetic field inside the solenoid. Alternatively, when we start to run a
current through the solenoid, there is a back emf that opposes the change. We therefore
need to do work to overcome that back emf and bring the current up to its final value.
This work is transformed into stored magnetic energy in the solenoid. The back emf is
determined by the self-induction of the solenoid so we need to calculate the selfinductance. In order to calculate the self-inductance we still need to calculate the
magnetic field in the solenoid. So either approach requires knowledge of the magnetic
field. The field inside a finite solenoid has edge effects which complicate the problem.
The best way to simplify the problem is to neglect the edge effects and assume that the
magnetic field is the same as the field inside an infinite solenoid in which case we can
use Ampere’s Law to find the magnetic field.
Figure 33.5 Very long Solenoid
!Plan approach
First use Ampere’s law,
! !
>
#! ⋅ 2 # = µ0 : %".
17
to find the magnetic field inside an infinite solenoid.
E-&#('!pp&)/.*
Calculate the magnetic field energy density
,> =
>2
.
2 µ0
Since the magnetic field is constant inside the solenoid, the total energy is just the energy
density times the volume of the solenoid
I > = ,> (=)5,6%) .
J%.)"2'!pp&)/.*
In order to calculate the self-induction of the solenoid
K=
Φ>
,
:
we need to calculate the total magnetic flux through the solenoid which is the number of
turns times the magnetic flux through one turn
Φ > = H (Φ > )(,&" = H
!!
! !
> ⋅ 2! .
)"% (,&"
Since the magnetic field is constant inside the solenoid, this integral is just the magnetic
field times the area ! so the total magnetic flux is
Φ > = H>! .
So the self induction is
K=
H>!
.
:
The work done to over come the back emf is then
L=
1 2
K: ,
2
and this is equal to the energy stored in the magnetic field.
18
!Execute Plan
E-&#('!pp&)/.*
Using Ampere’s law, the magnetic field inside a solenoid is
'> = µ0 " : = µ0 H : / 5 ,
where '" = H / 5 is the number of turns per unit length.
The magnetic energy density is then
> 2 1 µ0 H 2 : 2
=
.
,> =
2 µ0 2 5 2
The total energy is the energy density times the volume π < 2 5 of the solenoid,
1
1
I > = ,> (=)5,6%) = ( µ0 H 2 : 2 / 5 2 )(π < 25 ) = µ0 H 2 : 2π < 2 / 5
2
2
.
1
2
2
2
2
−7
= (4π ×10 T ⋅ m/A)(2760) (4.0 A) π (0.1 m) /(0.6 m) = 4.0 (T ⋅ A ⋅ m )
2
J%.)"2'!pp&)/.*
The area of each turn is '! = π < 2 , so the magnetic flux through each turn is
( Φ ) = >! = µ0 H : ⋅ (π < 2 ) / 5 = µ0 H : π < 2 / 5 .
' > (,&"
Thus, the self-inductance is
H Φ>
= µ0 H 2π < 2 / 5
.
:
−7
2
2
'= (4π × 10 T ⋅ m/A)(2760) π (0.1 m) / (0.6 m) = 0.5 H
K=
The energy of the system is given by
I> =
1 2 1
K: = ( µ0 H 2π < 2 / 5 ) : 2 .
2
2
!Evaluate answer
19
The best check is that the two approaches agree. The units are complicated but simplify
to joules T ⋅ A ⋅ m 2 = (N/A ⋅ m) ⋅ A ⋅ m 2 = N ⋅ m = J . Four joules is the change in potential
energy if you raise one kilogram about 2/5 of a meter. The self-inductance of 0.5 H is
somewhat large but that shouldn’t be surprising since this is a very big solenoid with
many turns.
WP 33.6: Self-Induction and magnetic energy in a co-axial cable. A coaxial cable
consists of two concentric long hollow cylinders of zero resistance; the inner has
radius'/ , the outer has radius'0 , and the length of both is 5 , with 5 >> 0 . A current :
flows down the inner conductor and back up the outer one.
a) What is the magnetic field everywhere?
b) What is the total energy stored in the magnetic field?
c) What is the self-inductance of the cable?
!Questions and suggestions
1. Apply Ampere’s Law for each region, outside the cable, & > 0 , inside the inner
cylinder at distance, '& < / , and between the cylinders, / < & < 0 , to find the
magnitude of the magnetic field everywhere. Be careful to indicate your amperian
loop and how much current cuts through your loop for each region.
2. For the region where there is non-zero magnetic field, determine the magnetic energy
density (energy per unit volume).
3. Now calculate the energy stored by integrating over the magnetic energy density over
volume of space where there is non-zero magnetic field. Is your magnetic field
uniform in this region of space or not? Since between the cylinders, the magnetic
field is like the field of a wire, the field should not be uniform. Therefore you will
need to choose an volume element for your integral. What symmetry is there in the
problem. If you choose for a volume element a cylindrical shell of radius &'and
thickness 2& , and length '5 , then the volume element is 2; = 2π &5 2& . You can now
integrate the magnetic field energy density between the shells & = / and '& = 0 to find
the total energy stored in the system.
20
4. Remember that the total stored magnetic energy is also equal to I > = 12 K: 2 . So you
can use your result from for the stored energy to calculate the selfinduction'K = 2I > / : 2 .
5. Check your units for both stored energy and self-induction.
WP 33.7: Induced Electric Fields, and Faraday’s Law. A wire is wrapped 'H = 100
times around a cylinder of non-magnetized material of radius / = 3.0× 10−2 m and length
5 = 2.5 × 10−1 m . A small wire loop of radius & = 1.0 × 10−2 m is placed on the solenoid’s
axis at the mid point of the solenoid. The loop has resistance < = 5.0 Ω . At '( = 0 , the
current through the wire is increased according to
': (( ) = 0( ,
0 < ( < 2#,
where '0 = 2.0 × 10−1 A s . After ( ≥ 2# the current remains constant. What is the (-)
electromotive force, (--) magnitude and direction of the induced electric field, and (---)
magnitude and direction of the induced current, in the loop as a function of time?
!Focus Problem
During time interval that the current is changing, there is a changing magnetic flux inside
the solenoid. Therefore by Faraday’s Law there must be a changing induced electric field.
The induced electric field is responsible for the electromotive force that drives the
induced current around the loop placed inside the solenoid. We can determine the
direction of the induced current by Lenz’s Law or by determining the direction for the
induced electric field. The direction of the current flowing into the solenoid is shown in
the figure. (For '( ≥ 2# noting is changing so the electromotive force is zero, induced
electric field and induced current are zero.)
Figure 33.7 Induced electric field inside a solenoid
!Plan approach
We must first calculate the magnetic field inside the solenoid using Ampere’s law that we
found in WP 33.5,
21
> = µ 0 " : = µ0 H : / 5 .
The changing magnetic flux inside the loop
! !
2Φ> 2
= !! > ⋅ 2! .
2(
2(
The induced electric field is related to the electromotive force according to
ε
-"2
#!
=
!
!
M-"2 ⋅ 2& .
5))p
Then by Faraday’s Law, the electromotive force in the loop is
ε
-"2
=−
2Φ >
.
2(
which we can determine by calculating the changing magnetic flux.
In order to solve for the induced electric field, we need to be able to calculate the line
integral of the electric field around the loop and set that equal to the (negative of)
changing magnetic flux.
!
! !
2
!
M
⋅
2&
=
−
>
⋅ 2! .
-"2
#!
2( !!
5))p
The induced current in the loop is found as in WP 33.1
: -"2 =
ε -"2
<
=
1 2>
−
!.
< 2(
The direction of the induced current is determined by Lenz’s Law.
!Execute Plan
The magnitude magnetic field inside the solenoid can be found using Ampere’s Law,
! µ H:
µ H0( ˆ
> = 0 Dˆ = 0
D.
5
5
Choose the orientation of the area element in the magnetic flux integral to point in the
!
positive D̂ direction. Then the area element is 2! = 2! Dˆ .
22
The magnetic flux integral through the loop is then
! !
Φ > = !! > ⋅ 2! = !! > Dˆ ⋅ 2! Dˆ = !! > 2! .
Since the magnetic field is constant inside the solenoid, and the area of the loop '! = π & 2 ,
the magnetic flux inside the loop is given by
! !
µ H0( 2
Φ > = !! > ⋅ 2! = !! > Dˆ ⋅ 2! Dˆ = !! > 2! = 0
π& .
5
The changing magnetic flux is then
µ0 H0π & 2
2 µ0 H0(
2
= (
)π & =
.
2(
5
5
' 2(
2Φ >
The induced electric field is
ε
-"2
=
#!
!
!
M-"2 ⋅ 2& = M-"2 2π & ,
5))p
since the ring is centered on the symmetry axis of the solenoid, and the induced electric
field has the same magnitude everywhere on a circle of radius & defined by the loop.
Since the magnetic field is constant inside the solenoid, and the area of the loop is
2
'! = π & , the changing magnetic flux inside the loop is given by
! ! 2>
µ H0π & 2
2Φ> 2
2 µ H0(
)π & 2 = 0
.
= !! > ⋅ 2! =
!= ( 0
2(
2(
2(
2(
5
5
So the electromotive force in the loop is
ε
'
-"2
µ H0π & 2
2Φ >
=− 0
2(
5
−7
(4π × 10 T ⋅ m/A)(100)(2.0 × 10−1 A s)(1.0 × 10−2 m)2
=−
.
(2.5 × 10−1 m)
=−
= −1.0 × 10−8 T ⋅ m 2 /s = −1.0 × 10−8 V
Using these two separate calculations for each side of Faraday’s Law, we find that
M-"2 2π & = −
23
µ0 H0π & 2
5
.
So the induced electric field is
M-"2
'
2
µ H0&
1 µ0 H0π &
=−
=− 0
2π &
5
25
.
−7
(4π × 10 T ⋅ m/A)(100)(2.0 × 10−1 A s)(1.0 × 10−2 m)
−7
=−
= 5.0 × 10 T ⋅ m/s
2(2.5 × 10−1 m)
The sign of the induced electric field is with respect to the choice of positive magnetic
flux. Once a choice has been made for the unit normal for the area element then the right
hand rule defines a direction for circulating around the loop in the electromotive force
integral. This direction is the same circular direction that the current flows in the
solenoid. Since the induced electric field is negative, it points in the opposite direction
that we circulated around the loop. The direction of the induced current is in the same
direction as the induced electric field. So the induced current flows in the loop in the
opposite direction that the current flows in the solenoid. The induced current in the loop
is found as in WP 33.1
: -"2 =
ε -"2
<
=
1 µ0 H0π & 2 µ0 H0π & 2 1.0 ×10−8 V
−
=
=
= 2.0 × 10−9 A .
<
5
<5
5.0 Ω
!Evaluate answer
Then units for the induced electric field are T ⋅ m/s = (N/A ⋅ m) ⋅ m/s = N/C which are the
units for electric field. The direction of the induced current can also be determined by
Lenz’ Law. The magnetic flux through the loop is positive (the magnetic field points in
the positive D̂ direction), and is increasing as the current in the solenoid increases. So the
induced current flows in the opposite direction of the current in the solenoid in order to
produce an induce d magnetic field that generated flux in the loop that points in the
negative D̂ direction to oppose the change. The order of magnitude for the electromotive
force and induced current are consistent for the values found in WP 33.1,3,5 for single
turn loops.
WP 33.8: Betatron Induced Electric Fields, and Faraday’s Law. An electron of the
mass '6% , and charge '+ = − % is constrained to move in a circle of radius '< by a nonuniform magnetic field with a magnitude >( & ) . Assume that the field varies as a function
of distance from an axis passing through the center of the circle but that it is symmetric
about this axis. The magnetic field is perpendicular to the plane of the orbit of the
electron. The magnitude of the average magnetic field over the disk with with the
electron’s circular orbit has the circumference is
24
>/=% =
1
π <2
!! >2! .
2-#D
Find a condition relating the rate of change in time of the average magnetic field
2>/=% / 2( , to the rate of change in time of the magnetic field, 2> / 2( , in order for the
electron to stay in circular orbit.
!Questions and suggestions
1. Assume the electron is undergoing uniform circular motion. There is magnetic force
on the moving electron. Is that force pointing radial inward or outward? Find an
expression for this force in terms of the + = − % , the electron’s speed '= , the radius <
of the circular orbit, and the magnetic field > along the electron’s circular orbit.
2. What is the acceleration of a particle that undergoes uniform circular motion?
3. Use Newton’s Second Law, to find an expression for the speed of the electron in
terms of the charge '+ = − % , the mass 6% , and the radius < of the circular orbit.
4. Differentiate the speed with respect to time using your expression from suggestion 3.
to find expression for the rate of change of the electron’s speed, 2= / 2( in terms of
the rate of change in time of the magnetic field, 2> / 2( , and the radius '< of the
circular orbit, the charge '+ = − % , and the mass 6% .
5. When the magnetic field starts to increase at a rate 2> / 2( , the changing magnetic
flux produces an induced tangential electric field. Can you express the changing
magnetic flux in terms of the derivative of the average magnetic field, '2>/=% / 2( , and
the radius '< of the circular orbit?
6. Once you do this, find an expression for the induced electric field a distance '< from
the center of the electron’s orbit. Express your answer in terms of derivative of the
average magnetic field, '2>/=% / 2( , and the radius < of the circular orbit. Does the
induced electric field point in the direction of the electron’s velocity or opposite.
Explain your answer.
7. The induce electric field will cause the electron to speed up. Use Newton’s Second
Law to find an expression relating the change of the electron’s speed, '2= / 2( in terms
of the induced electric field.
8. You can use your result from suggestion 6 to find an expression for the rate of change
of the electron’s speed, '2= / 2( in terms of the rate of change in time of the magnetic
25
field, '2>/=% / 2( , and the radius < of the circular orbit, the charge + = −% , and the
mass 6% .
9. Compare your two expressions for the rate of change of the electron’s speed,
'2= / 2( in order to find a condition relating the average magnetic field, '>/=% to the rate
of the magnetic field, '> , in order for the electron to stay in circular orbit.
26
Questions and problems
Section 33.1
1.(I)EDW33.01.01.What happens if the conducting rod in Figure 33.1 moves (a) out of or
into the page, (b) up or down in the page?
2.(II)EDW33.01.02. What happens if the conducting rod in Figure 33.1 rotates clockwise
(looking back along the axis) about an axis through its center, which is pointing (a) out of
the page, (b) in the page left-to-right, (c) in the page down-to-up?
3.(I)EDW33.01.03. A square conducting loop (centered on the x-axis, with sides parallel
to the y & z-axes) moves with constant velocity (in the negative x-direction) towards a
small bar magnet (located at the origin, with its polar axis in the positive x-direction), as
shown. At the instant shown, what is the direction of the magnetic force on electrons
(negative charge carriers) at the midpoints of each side? Does the motion of the loop
induce a charge separation or a current, & if so, describe it?
4.(I)EDW33.01.04. The Air Force Thunderbirds aerial demonstration team is performing
at an air show located on the Earth’s magnetic equator. I what directions can one of the F16’s fly so that there will be no induced charge separation on its metal surfaces?
5.(II)EDW33.01.05. In the preceding question, how could an F-16 fly in order to produce
a maximum induced charge separation between its wing tips? (Consider both the
direction of motion & orientation of the wing.)
Section 33.2
Section 33.3
Section 33.4
Section 33.5
6.(I)@STE33.05.01. The magnetic flux through a conducting loop is increasing at a rate
of 3.0 Tm2/s. What is the magnitude of the induced emf in the loop?
27
7.(I)@STE33.05.02. A square conducting loop of side length 0.2 m lies in the @B plane.
The loop is contained within a uniform magnetic field that points in the negative Ndirection and is decreasing at a rate of 0.07 T/s. (a) What is the magnitude of the induced
emf in the loop? (b) What is the direction of the induced current flow in the loop?
8.(I)@STE33.05.03. Initially there is no magnetic flux through a certain conducting loop.
A magnetic field near the loop is then suddenly turned on, and five seconds later, the
magnetic flux through the loop is 1.0 Tm2. What is the average magnitude of the induced
emf in the loop during those five seconds?
9.(II)@STE33.05.04. An· infinite cylindrical solenoid of radius 0.5 m has 10 loops of
wire per centimeter along its length. A circular conducting loop of radius 1.0 m encircles
the solenoid, with the axis of the solenoid passing perpendicularly through the center of
the loop. The solenoid initially carries a steady current :, but the current is then reduced
to zero during a 0.1-s time period. If the average emf induced in the loop during that 0.1s interval is of magnitude 0.1 V, what was the magnitude of the initial current?
10.(II)@STE33.05.05. The region of space to the right of the B-axis contains a uniform
magnetic field of unknown magnitude pointing in the positive z-direction; there is no
field to the left of the B-axis. As a square conducting loop in the @B plane (parallel to the
axes) with sides of length 0.3 m is moving to the right across the y-axis with constant
velocity 2.0 m/s, an emf of magnitude 0.24 V is induced in the loop. (a) What is the
magnitude of the magnetic field? (b) In what direction does the induced current flow in
the loop?
28
11.(II)@STE33.05.06. A region of space contains a changing magnetic field given by
!
> (( ) = >0 %-( /τ Nˆ , and a circular conducting loop of radius < lies within this region in the @B
plane. (a) Calculate the magnetic flux through the wire as a function of time. Is it
increasing or decreasing as time passes? (b) Calculate the induced emf in the loop as a
function of time. (c) Determine the direction of current flow in the loop.
12.(II)@STE33.05.07. As in Example 33.7, a coil of H=100 conducting loops and area
!=0.01m2 is rotated at a constant angular speed. If the magnetic field driving the
generator is that of the earth (>=0.5×10-4 T), how many 360 rotations must the coil
complete in each second to generate a maximum induced emf of 1 V? Based on this
calculation, does it seem practical to use the Earth’s magnetic field to drive an electric
generator?
13.(II)@STE33.05.08. A uniform magnetic field of magnitude B fills all space and points
in the positive N-direction. A circular conducting loop in the @B plane is growing larger,
with its radius as a function of time given by &(() = =(, where = is a positive constant. (a)
Calculate the magnitude of the emf induced in the loop as a function of time. (b) What is
the direction of the induced current flow?
29
Section 33.6
?*%'"%@('1),&'p&)05%6#'/&%'0/#%2')"'(*%'1)55)$-"7'p*B#-./5'#-(,/(-)"8'/',"-1)&6'6/7"%(-.'
1-%52'p)-"(-"7'-"'(*%'p)#-(-=%'NO2-&%.(-)"'1-55#'/'.B5-"2&-./5'&%7-)"')1'#p/.%')1'&/2-,#'<'
$*)#%'.%"(&/5'/@-#'-#'(*%'NO/@-#P'),(#-2%'(*-#'&%7-)"A'(*%&%'-#'")'6/7"%(-.'1-%52C''?*%'
.))&2-"/(%'&'6%/#,&%#'(*%'2-#(/".%'1&)6'(*%'NO/@-#C'
'
'
14.(I)@STE33.06.09. If < = 0.25 m and the magnitude of the magnetic field is decreasing
at a rate of 0.3 T/s, what is the magnitude of the induced electric field (a) at & = 0.2 m,
and (b) at & = 0.5 m? (c) Do the induced electric field lines encircle the z-axis in a
clockwise or counterclockwise manner, when looking from the positive N-direction?
15.(I)@STE33.06.10. If < = 0.12 m, and the magnitude of the induced electric field at r =
0.06 m is 10 V/m, what is the instantaneous rate of change of the magnitude of the
magnetic field with respect to time?
16.(II)@STE33.06.11. If the magnitude of the magnetic field is changing with time
according to > = >Q sin(!(), calculate the magnitude of the induced electric field as a
function of t and r, both (a) for & < <, and (b) for & > <. (c) At the instant ( = 0, what is
the direction of the induced electric field?
17.(II)@STE33.06.12. If the magnitude of the induced electric field -"#-2%'the region
containing an magnetic field is given by E(r,t) = 34&(2 , where 4 is a positive constant
and & is the distance from an axis, and there is no magnetic field at time ( = 0, calculate
the magnitude of the magnetic field as a function of time for ( > 0.
30
Section 33.7
18.(I)@STE33.07.13. An inductor of 2-H inductance carries a current that is increasing at
a rate of 0.4 A/s. What is the magnitude of the emf induced in the inductor? Is this emf
encouraging or resisting the current flow?
19.(I)@STE33.07.14. When the current flowing across an inductor is decreasing at a rate
of 2.0 A/s, the magnitude of the emf across the inductor is measured to be 6 V. What is
the inductance of the inductor?
20.(I)@STE33.07.15. For a toroidal coil like that discussed in Example 33.27, the radius
of the ring is < = 0.1 m, the radius of the loops is 0 = 0.01 m, and the total number of
loops is 400. Calculate the inductance of the coil.
21.(II)@STE33.07.16. The flow of current through an inductor of inductance L is given
by the function :(()= :Q sin(!(). (a) Calculate the induced emf in the inductor as a
function of time. (b) At the instant ( = 0, is the current through the inductor increasing or
decreasing? (c) At the same instant, is the induced emf pushing with or against the
current flow? (Remember that a positive induced emf is directed in the same direction as
the current flow, and a negative induced emf is directed against it.) (d) How are the
answers to parts (b) and (c) consistent with the behavior of inductors discussed in the
text?
22.(III)@STE33.07.17. Show that the inductance of a cylindrical solenoid of length D
and radius R with n coils per unit length is given by K = ! "Q'"R'<R'S. (When calculating
the magnetic field inside the solenoid, use the infinite solenoid approximation: >= "Q":.)
23.(II)@STE33.07.18. Use the result of the previous problem to calculate the inductance
of a cylindrical solenoid with length 0.2 m, radius 0.03 m, and a total of 2000 coils.
24.(II)@STE33.07.19. The induced emf in an inductor of inductance K varies with time
according to #-"2(()= –24(, where 4 is a positive constant. (a) If no current flows through
the inductor at time t = 0, calculate the current flowing through the inductor as a function
of time for ( > 0. (b) Is the current increasing or decreasing for times ( > 0? (c) Discuss
how your answer to part (b) is consistent with the sign of #-"2 and what you know about
the behavior of inductors.
Section 33.8
25.(I)@STE33.08.20. Calculate the amount of magnetic energy stored in a 0.6-H inductor
when a 6.0-A current flows through it.
26.(I)@STE33.08.21. If 10 J of magnetic energy is stored in a 5.0-H inductor, how much
current is passing through it?
31
27.(II)@STE33.08.22. How much magnetic energy is stored in the toroidal coil described
in Problem 33.07.15 when a 12-A current is passing through it?
28.(II)@STE33.08.23. A cylindrical region of space, with radius 0.04 m and length 0.06
m, contains a uniform magnetic field of magnitude 0.12 T. How much magnetic energy
is stored in this region of space?
29.(II)@STE33.08.24. A cubical region of space with side lengths 0.05 m contains a
uniform magnetic field. If the total magnetic energy stored in this region is 12 J, what is
the magnitude of the magnetic field?
30.(III)@STE33.08.25. As a continuation of the previous problem, use two different
methods to calculate the energy stored in a cylindrical solenoid of length S and radius <
with n coils per unit length which carries a current :. (a) Use the inductance L found in
the previous problem and the formula relating stored energy to inductance and current.
(b) Use the formula relating stored energy to magnetic energy density and the volume of
the solenoid. (You should get the same answer both ways.)
32
Answers to review questions
1. The magnetic force on the charge carriers in the conducting object can cause either
the separation of charge carriers on the surfaces of the object, or the circulation of an
induced current.
2. An induced current is one that is caused by the motion of charge carriers relative to a
magnetic field (or more generally, by any time-varying magnetic flux), whereas a regular
current is caused by a source of potential difference, like a battery. (This distinction is
sometimes moot, since ordinary house current is caused by a potential difference
produced by induction in an electric generator.) Of course, there is no physical difference
in the currents themselves.
3. Yes, the induced effects are reversed when the direction of the relative velocity is
reversed.
4. An induced current is generated if the magnetic flux through the loop changes with
time.
5. With respect to conducting loops, Faraday’s law states that a current is induced
whenever the magnetic flux through the loop changes.
6. Electromagnetic induction is the process by which a changing magnetic flux causes
the separation of charge carriers or the flow of an induced current.
7. No, in the reference frame where the loop is at rest, there is no magnetic force.
Rather, there is an induced electric field which causes the current to flow.
8. The nature of the force that causes electromagnetic induction does depends on the
choice of reference frame (as well as on the situation). In a frame where a conducting
loop is at rest, induced electric fields, caused by time-varying magnetic flux, can account
for the effects of electromagnetic induction; in a frame where conductors are moving
through a magnetic field, the magnetic force on charge carriers can be the account. In
some situations, like that pictured in Fig. 33.19, either point of view can account for the
electromagnetic induction.
9. Induced electric field lines form closed loops, whereas electrostatic field lines, which
originate on positively charged particles and terminate on negatively charged ones, never
do.
10. When converted from a mathematical statement into a qualitative statement, Lenz’s
law says that a circuit experiencing a changing magnetic field will develop a current that
produces a magnetic field that opposes (i.e., lessens) the change in magnetic flux through
the circuit.
33
11. The conservation of energy underlies Lenz’s law. If there were positive feedback in
electromagnetic induction, energy could increase without any work being done.
12. Eddy currents are induced currents in a conducting object. They are not confined to a
single path, as in a wire, but circulate as tiny whirlpools.
13. Faraday’s law states that the induced emf (around a closed path) is equal to the
negative of the rate of change of magnetic flux (through an area bounded by the path).
The relative negative sign reflects Lenz’s law, that induced effects oppose the changes
that generate them.
14. In general, the agency causing the change in magnetic flux supplies the energy for the
resulting induced effects, either from external mechanical work or energy stored in fields,
etc. In a situation like that in Fig. 33.19, the person pulling the rod does the work which is
transferred to the charge carriers by induction (see paragraph and footnote following
Equation 33.2).
15. Potential difference is path-independent, i.e, the work done by electrostatic forces on
a charged particle only depends on the starting and ending points, not on the path taken
by the charge between them. In contrast, an emf induced by a changing magnetic field
does depend upon the path taken by the charge, i.e. the work done by the nonelectrostatic forces depends upon the shape of the circuit, which affects the magnetic flux
enclosed by it.
16. The emf induced under such circumstances (as shown in Fig. 33.22) is an alternating
sinusoidal emf (see Equation 33.13).
17. The induced emf is the line integral of the induced electric field, around the path
bounding the area through which the magnetic flux is changing.
18. The direction of the induced electric field is tangent to circles around the center of the
magnetic field region, in planes perpendicular to the magnetic field, circulating in the
sense dictated by Lenz’s law. The magnitude of the induced electric is proportional to the
derivative of the magnetic field strength, and varies directly with the distance from the
center, inside the magnetic field region, and inversely with the distance outside.
19. Yes; if the magnetic flux through a circuit changes for any reason, Faraday’s law
requires that an emf will be induced.
20. The inductance of a circuit describes how much the magnetic flux through it changes
when the current changes. The inductance depends only on the geometrical properties of
the circuit, that is, its size and shape.
21. The unit of inductance is the henry, which equals a volt per ampere. (It honors the
American Joseph Henry (1797-1879), who independently of Faraday, discovered the
phenomenon of induction, and was the first to investigate self-induction.)
34
22. The work done changing the current in an inductor goes into, or comes out of, the
energy stored in the magnetic field of the inductor.
23. The energy in an inductor depends of the square of the current, and on the inductance.
(It is one half the product of these; see Equation 33.34.)
24. The magnetic energy in an inductor is also equal to the volume integral of the energy
density in its magnetic field (see the first parts of Equation 33.39). The latter is
proportional to the square of the field strength, which is a general relationship for any
magnetic field (see Equation 33.38). (If the field strength is approximately uniform over
the volume of the inductor, the energy is just the product of the energy density and the
volume; see Equation 33.37.)
35
Chapter 34 (version 04/16/06)
Changing electric fields
Review questions
Answers to these questions can be found at the end of this chapter.
Section 34.1
1. What electrical system can be used to demonstrate that the steady-state form of
Ampere’s law, first introduced in Chapter 32, is incomplete? Briefly describe how
you would choose a path (Amperian loop) and a surface spanned by the path to
illustrate this.
2. What else, besides electric currents (including atomic Amperian currents), can produce
a magnetic field?
3. How is the direction of the induced magnetic field related to that of the changing
electric field that gave rise to it?
Section 34.2
4. What symmetries or similarities are there in the sources that produce electric and
magnetic fields?
5. What symmetries or similarities are there in the way electric and magnetic fields are
detected?
6. What symmetries or similarities are there in the spatial appearance of electric and
magnetic field lines?
Section 34.3
7. Is the electric field from a charged particle, moving with constant velocity, spherically
symmetric? If not, how does it differ from spherical symmetry?
8. When do “kinks” appear in the field pattern of a charged particle, and what is the
collection of such kinks, generated in a brief time interval, called?
9. Describe the motion of an electromagnetic wave pulse.
Section 34.4
10. How are the electric and magnetic fields oriented in a transverse electromagnetic
wave?
11. What does it mean to say that the electric and magnetic fields in an electromagnetic
wave are “in phase”?
12. How is the polarization of an electromagnetic wave defined?
13. What is an antenna and what can it do?
Section 34.5
1
14. How is Maxwell’s displacement current defined for electric fields in free space? Does
this definition also hold in the presence of dielectric materials? If not, how should the
definition be changed?
15. Is displacement current really a current? If not, what is useful in applying this word?
16. What is the general form of Ampere’s law, which applies when electric fields are
changing?
Section 34.6
17. What individual names are applied to the four equations that comprise Maxwell’s
equations?
18. Briefly, what is the nature of the experimental evidence for each of Maxwell’s
equations?
19. Given that each of the four laws for electric and magnetic fields were discovered
before Maxwell modified one of them, why is it still fitting to refer to all of them as
his?
Section 34.7
20. What is the speed of an electromagnetic pulse or wave in free space or in a dielectric,
in terms of fundamental electric and magnetic constants?
21. What is the relation between the strengths of the electric and magnetic fields in an
electromagnetic pulse or wave in free space or in a dielectric?
22. Describe the electromagnetic fields in a harmonic plane wave?
23. What are the names of the major divisions of the electromagnetic spectrum?
24. If all electromagnetic waves are governed by the same physics as visible light, what
differentials them?
Section 34.8
25. How do the energy densities in the electric and magnetic fields of a transverse
electromagnetic wave compare? What is the total energy density, in terms of the
fields?
26. What is the Poynting vector and what does it represent?
27. What is the relation between the intensity of an electromagnetic wave and the energy
density it carries?
2
Developing a feel
Calculate or estimate the following quantities:
1.
2.
3.
4.
5.
The frequency of visible light. (F, L)
The wavelength of FM 99. (F, A, K)
The wavelength of AM 1010. (F, A, N)
The time for a radio signal to travel around the Earth. (B, G)
The time for a radio signal to travel to and from a communications satellite in
geostationary orbit. (B, P, M, H)
6. The maximum magnetic field 1 km from an FM radio station transmitter. (Q, E, I)
7. The maximum electric field 1 km from an AM radio station transmitter. (Q, E, I)
8. The maximum magnetic field 10 cm from a 125 mW cell phone. (E, I)
9. The change in the electric flux when a spark jumps from your finger to a doorknob in
the winter. (C, J, R)
10. The total power radiated by all the AM radio stations in the US (Q, D, O)
Hints:
A. What does the radio station number refer to?
B. What is the speed of the radio signal?
C. What is the electric field at which air breaks down (ie: sparks)?
D. How many AM stations are there in your area?
E. What is the relationship between power, intensity and distance?
F. What is the relationship between wavelength and frequency?
G. What is the circumference of the Earth?
H. What is the distance from the center of the Earth to geostationary orbit?
I. What is the relationship between power, distance, and electric field?
J. What is the typical distance a spark jumps?
K. What are the units of frequency for FM?
L. What is the wavelength of visible light?
M. What is the relationship between orbital period and height?
N. What are the units of frequency for AM?
O. What percent of the US population lives in your area?
P. What is the orbital period of a geostationary orbit?
Q. What is the power radiated by a typical commercial radio station?
R. What is the cross sectional area where the electric field is large?
Key: A. The frequency of the electromagnetic radiation. B. c0 ! 3×108 m/s. C. 3×106
V/m. D. More than 1, less than 100. In an area with a population of about 2×106 there are
3
about 20. E. The intensity S = P/A where A = 4 ! R2 and S = EB/µ0. F. f = c0/". G. C =
2#R where R = 6400 km. H. 4.2×104 km. I. B = E/c0. J. Between 1 and 10 mm. K. FM
frequencies are given in MHz. L. 400 < " < 700 nm. M. You find this by balancing the
centripetal force F = mv2/R needed with the gravitational force F = GMm/R2 to get: R3 =
GMT2/(4#2). N. AM Frequencies are given in kHz. O. The US population is 3×108. P. T
= 1 day = 86400 s. Q. About 10 to 50 kW, depending on the station. R. Somewhere
between the cross sections of your finger (about 1 cm2) and the doorknob (about 10 cm2).
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any
particular section. Typically, an example that is worked out in detail is followed
immediately by an example whose solution you should work out by following the
guidelines provided.
NONE SUPPLIED WITH THIS CHAPTER.
5
Questions and problems
Section 34.1
1. (I)@STE34.01.01 A uniform electric field points out of the page and is contained in a
region of space with circular cross section, as shown below. If the electric field is
suddenly turned off, a magnetic field will be induced. (a) What is the direction of the
induced magnetic field? (b) Does the induced magnetic field exist only inside the region
where the electric field exists, only outside that region, or both inside and outside that
region?
2. (I)@STE34.01.02 The capacitor shown below is being charged by a constant current
I. What are the directions of (a) the electric field between the plates, (b) the change in the
electric field, and (c) the induced magnetic field between the plates?
3. (I)@STE34.01.03 The electric field shown below points out of the page and has a
magnitude that is changing with time. As a result of this change, an upward-pointing
magnetic field is induced at point P. Is the magnitude of the electric field increasing or
decreasing with time?
4. (II)@STE34.01.04 The figure below shows an increasing magnetic field and the
electric field that is induced around it according to Faraday’s Law. If the magnetic field
6
shown is not only increasing, but increasing at an increasing rate, (a) will the magnitude
of the electric field shown be constant, increasing, or decreasing? (b) Will the behavior
of the electric field induce an additional magnetic field? If so, draw or describe its
direction, and show whether it reinforces or counteracts the original magnetic field.
Section 34.2
5. (I)@STE34.02.05 True or false: “All electric field lines must either begin or end at a
charged particle.” If false, when is this not the case?
6. (II)@STE34.02.06 For each of the following situations, indicate whether or not an
electric field is produced: (a) A wire with zero net charge carries a constant current. (b)
A wire with zero net charge carries a time-varying current. (c) A bar magnet moves
through space at constant velocity. (d) A bar magnet rotates about an axis that passes
symmetrically through its north and south poles.
7. (II)@STE34.02.07 For each of the following situations, indicate whether or not a
magnetic field is produced: (a) A uniformly charged sphere rotates with constant angular
speed about an axis that passes through its fixed center. (b) A sphere with a fixed
amount of uniformly distributed charge and a fixed center pulsates (expands and shrinks
continuously) while constantly retaining its spherical symmetry. (c) A charged particle
moves through space with constant velocity. (d) A charged particle accelerates through
space.
8. (I)@STE34.02.08 Which of the following field line patterns could represent (a)
electric field lines or (b) magnetic field lines?
Section 34.3
7
9. (I)@STE34.03.09 In order to detect an incoming electromagnetic wave pulse, we
suspend a single charged particle at the origin. If we observe the particle oscillating
along the y-axis, what can we say about the direction in which the wave pulse is
propagating?
10. (II)@STE34.03.10 In the figure below, a charged particle has been suddenly
accelerated downward, producing the electric field line pattern shown. When the kinks in
the field lines reach the rod at the left, they produce a downwardly directed current in the
rod. Is the particle that has been accelerated positively or negatively charged?
11. (I)@STE34.03.11 Do the electric field lines produced by a single charged particle
point directly at or away from the particle (a) when the particle is stationary, (b) when the
particle moves with constant velocity, (c) when the particle accelerates?
Section 34.4
12. (I)@STE34.04.12 An electric dipole located at the origin oscillates along the x-axis,
creating an electromagnetic wave. (a) To which plane (x-y, x-z, or y-z) will all the
magnetic field lines be parallel? (b) At a point on the z-axis far from the origin, what will
be the polarization of the wave? (Along which direction will the electric field lines
point?)
13. (I)@STE34.04.13 The figure below shows a magnetic field line pattern, which
radiates outward from an oscillating electric dipole at the center of the figure. (a) What is
the direction of the change (over time) in the magnetic field at the point P? (b) Use your
answer to part (a) to determine the direction of the electric field loop around point P. (c)
In what direction is the electric dipole oscillating (right-left, up-down, or in and out of the
page)?
8
14. (I)@STE34.04.14 What total antenna length (twice the rod length) should be used to
create a dipole antenna that transmits electromagnetic waves with frequency 100 MHz, a
frequency typical of FM radio broadcasts?
15. (II)@STE34.04.15 The receiving dipole antenna in checkpoint 34.9 has to be aligned
with the oscillating electric field in order to produce a measurable potential difference in
the antenna. What if, instead, we wanted to detect the potential difference caused by the
changing magnetic field? (a) What kind of antenna would we use in this case (what
would its shape be)? (b) How would this antenna need to be oriented with respect to
figure 34.15 to maximize its ability to detect the wave? (c) Referring to equation 33.13,
explain why this antenna would be more effective for either the low or high frequency
range.
Section 34.5
16. (I)@STE34.05.01 A parallel-plate capacitor has plate-size large compared to plate
separation. At the instant a current of 5.0 A is flowing through the circuit, find (a) the
time rate of change of the electric flux, (b) the displacement current. (c) How do these
answers change if there is a dielectric, κ=1.5, between the plates of the capacitor?
17. (I)@STE34.05.02 For the capacitor in the previous problem (with no dielectric), if the
plates are circular, with radius 4.0 cm centered on the wires in the circuit, what is the
magnetic field at 3 cm from the axis of the plates?
9
18. (I)@STE34.05.03 Instead of a capacitor in a circuit, we could get the same effect if
we simply had a circuit made up of very thick wires, with a smooth break in the wire,
causing two parallel flat surfaces. If the surfaces are 0.01 mm apart and the wire is 1.00
cm in diameter, find the magnetic field 0.20 from the axis at the instant the current is 2.0
A in the wire. You may assume that the current is uniformly distributed over the crosssectional area of the wire.
19. (I)@STE34.05.04 A region of space contains an electric field that is given by
E=E0cos(ωt). If E0=10 N/C and ω=107 s-1, typical for AM radio. (a) What is the
maximum displacement current through a 0.5 m2 area? (b) How would this area have to
be oriented with respect to the field to get this maximum?
20. (II)@STE34.05.05 For the instant B=1.6×10-8 T at 7.0 mm from the center of the
wires in Example 34.5, (a) find the current in the wire, and (b) Draw the current in the
left plate to show why you get the same value of B, no matter where you choose to place
the surface bounded by your Amperian loop.
21. (II)@STE34.05.06 Assume the surface charge on the plates of a capacitor is uniform
as the current flows through the circuit containing it. The capacitor starts out uncharged.
The radius of the plates of the capacitor is 0.03 m, and a steady 2.0-A current flows
through the circuit. (Note, a variable potential difference must be supplied across the
circuit to keep the current constant.) (a) Find the electric field as a function of time. (b)
Find the displacement current through a circle (your Amperian loop) of radius 0.01 m in
the plane between the plates, using the flat surface bounded by the circle as your
Amperian surface. (c) Repeat part (b) for the same Amperian loop, only now use a
cylindrical surface open between the plates, that extends through the plate and is crossed
by the wire. (d) Why are these answers different, and different from 2 A? (This probably
needs a figure)
10
22. (II)@STE34.05.09 Now let us do the previous problem in a slightly more realistic
fashion. Assume that the resistance in the circuit is 3 Ω, and that the potential difference
across the circuit is 6.0 V. The separation of the plates is 0.001 m. The radius of the
plates of the capacitor is 0.03 m. (a) Find the capacitance of the capacitor. (b) Find the
electric field between the plates as a function of time. (c) Find the current in the wire as a
function of time. (d) Find the magnetic field a 0.01 m from the axis of the plates as a
function of time. (e) Find the current flowing outward toward the edge of the plates at
this radius as a function of time.
23. (II)@STE34.05.07 If the capacitance of a circuit is C, and the resistance is negligible,
show that the displacement current between the capacitor’s plates is C dV/dt, where V is
the potential difference across the capacitor.
24. (II)@STE34.05.08 A parallel-plate capacitor has plate of area A whose length and
width are large compared to plate separation d. It is connected to a potential difference
source such that Vc(t)=ct, where c is a constant. There is an imperfect dielectric with
dielectric constant κ and resistivity ρ between the plates of the capacitor. (a) Find the
conduction current as a function of time. (b) Find the displacement current. (c) Find the
time at which the displacement current and conduction current are equal.
25. (III)@STE34.05.10 A short segment of current carrying wire runs from x=a to x=-a.
At x=a a very small sphere with charge +Q0 exists. At x=-a, a very small sphere with
charge -Q0 exists. Use both the generalized form for Ampere’s law and the Biot-Savart
law to find the magnetic field at a point a distance R up the y-axis, and show that they are
the same. Hint: to find the flux, consider a flat ring of radius r and width dr in the y-z
plane, centered at the origin. Integrate over these rings to get the total flux, since the field
is NOT constant over the surface.
11
Section 34.6
26. (I)@STE34.06.11 Rewrite Maxwell’s equations as they would be if magnetic
monopoles, qm, existed.
27. (I)@STE34.06.12 Using equation 31.10, show that the normal component of the
magnetic field is continuous across any surface (think back to how we found the change
in the normal for E using Gauss’ Law).
28. (II)@STE34.06.13 If magnetic monopoles did exist, one way to search for them
would be to look for an induced current through a loop due to changing magnetic flux, as
one zoomed through. If the loop had negligible resistance and a self-inductance L, what
would be the current induced in the loop if a magnetic monopole, qm, passed through the
loop?
29. (II)@STE34.06.14 For a perfect conductor, E=0 everywhere inside, and all the charge
resides on the surface even when a current is flowing. (a) Use Maxwell’s equations to
show the magnetic field is constant in time inside the conductor. (b) Use Maxwell’s
equations to reason or show that the magnetic flux through a perfectly conducting loop is
constant. A superconductor is more than just a perfect conductor. It also has the property
that B=0 inside the superconductor. (c) Use Maxwell’s equations to show why the current
in a superconductor is confined to the surface. Most wires we work with in this book are
not superconductors, so it is perfectly reasonable to assume the current is uniformly
distributed throughout the material.
Section 34.7
30. (I)@STE34.07.15 Starting from ε0=8.85×10-12C2/Nm2 and µ0=4π×107 T·m2, show that
equation 34.30 does in fact give the speed of light, including correct units!
12
31. (I)@STE34.07.16 Your house is just full of wires with electric currents in them
oscillating back and forth at 60Hz. What is the wavelength of the electromagnetic waves
generated by these currents?
32. (I)@STE34.07.17 The frequency for AM radio channels is typically on the order of
105 or 106 Hz. What is the wavelength for these waves? FM waves have wavelengths
between 1 and 100 m. What is the frequency for these waves?
33. (I)@STE34.07.18 Can you hear the electromagnetic waves broadcast by radio
stations? Justify your answer. A common-sense response works.
34. (I)@STE34.07.19 You may have heard of the “3-cm cosmic background radiation”.
What is the frequency of this radiation?
35. (I)@STE34.07.20 The speed of light drops to 2.6×108 m/s in water. What is the
dielectric constant of water?
36. (I)@STE34.07.21 If the electric field 500m from a radio broadcast antenna has a
maximum amplitude of 12 V/m, what is the maximum magnetic field at this distance?
37. (II)@STE34.07.22 At a distance from an antenna such that the electric field has a
maximum amplitude of 15 V/m when the air is dry, (a) what is the maximum magnetic
field at this distance? (b) what happens to the maximum electric and magnetic fields
when the air is very damp, increasing the dielectric constant of the air?
Section 34.8
38. (I)@STE34.08.23 Lasers in fact may have very small wattages compared to light
bulbs, but the light does not spread out very much. Explain why this has such an effect
over long distances.
39. (I)@STE34.08.24 A 1.0-mW laser has a beam radius of 0.6 mm. What is the intensity
of this beam?
40. (I)@STE34.08.25 Assume that half of the energy radiated by an incandescent bulb
goes into heating, or infrared radiation, instead of visible light. By how much does this
diminish the intensity of the visible light, compared to if all of the energy went into the
visible spectrum? This is one reason why fluorescent light bulbs are brighter at lower
wattage.
41. (II)@STE34.08.26 An incandescent light bulb radiates at 60 W, assume this occurs
uniformly in all directions. At a distance of 2 m from the bulb, find (a) the intensity of the
electromagnetic waves, (b) the maximum electric field, and (c) the maximum magnetic
field.
42. (II)@STE34.08.27 The intensity of sunlight striking the Earth’s upper atmosphere is
approximately 1.35×103 W/m2. (a) Find Emax and Bmax at this location. (b) Given that the
13
Earth is 1.5×1011 m from the sun, what is the total power output of the sun. (c) Assume
that all of the sunlight that strikes the upper atmosphere reached the ground. What would
be the total rate of energy striking the Earth? There is a trick here, because the sun is far
enough away, it is as if all the light is perpendicular to the cross section of the Earth.
43. (II)@STE34.08.28 For a particular electromagnetic wave, Brms is 0.30×10-6 T. For this
wave, find (a) Erms, (b) the average energy density and (c)the intensity.
44. (II)@STE34.08.29 A 100kW radio station (definitely AM) radiates equally in all
directions. (a) What are the amplitudes Emax and Bmax at a distance of 100 m from the
radio tower? (b) What are they 50 km from the radio tower? (c)Find the potential
difference caused over a 1.0-m-long car antenna for these two locations, assuming the
antenna is aligned perfectly with the electric field direction.
45. (III)@STE34.08.30 A Poynting vector for an electromagnetic wave is given by
(100 W/m 2 ) sin 2 [(1000 m -1 )x-(3 × 1011 s -1 )t ]kˆ . (a)What is the direction of propagation of
the electromagnetic wave? (b)What is the energy per unit time being radiated through a
1.0-m2 surface aligned with its normal perpendicular to the direction of propagation.
(c)At an instant when the electric field is in the +x-direction, in what direction is the
magnetic field? (d)What are the wavelength and frequency of this electromagnetic wave
(you may need to think back to chapter 18, and you can check to see if your answers
make sense)? (e) What are the electric and magnetic field vectors, as a function of
position and time?
14
Answers to review questions
Section 34.1
1. A charging capacitor demonstrates that the steady-state form of Ampere’s law must
be modified. Choose either a path encircling the wire, spanning a surface passing
between the capacitor plates which is not pierced by the wire, as in Figure 34.2(a), or a
path encircling between the capacitor plates, spanning a surface which is pierced by the
wire, as in Figure 34.2(b).
2. Besides currents, a changing electric field also produces a magnetic field.
3. They are perpendicular to one another.
Section 34.2
4. Electric fields are produced by charged particles, moving or at rest; magnetic fields
only by moving charged particles. Fields of each kind are produced by changing fields of
the other kind.
5. Electric fields can be detected by the forces they exert on charged particles, either
moving or at rest; magnetic fields only by the forces they exert on moving charged
particles (which can be distinguished from electric forces, since they depend on the
velocity of the particle and are perpendicular to it).
6. Magnetic field lines always form closed loops, around either moving charged
particles or changing electric fields. Electric field lines emanate or terminate on charged
particles, or form loops around changing magnetic fields.
Section 34.3
7. The electric field is not spherically symmetric, but weaker along the line of the
velocity, and stronger in directions transverse to the velocity, depending on the speed of
the particle relative to the speed of light.
8. When a charged particle is accelerated, changing its velocity, kinks appear that join
the field lines before the acceleration began, to the altered field lines after the velocity has
changed. A ring of kinks, generated during a brief time interval of acceleration,
represents an electromagnetic wave pulse.
9. An electromagnetic wave pulse travels outward, from the location of the accelerated
charged particle that produced it, with a speed characteristic of electromagnetic
interactions.
Section 34.4
10. The magnetic field is perpendicular to the electric field, and both are perpendicular to
the direction of propagation, as determined by the (appropriate) right-hand rule.
11.In phase means that the fields oscillate with the same frequency and both
simultaneously reach their maxima (or minima).
15
12. The polarization is defined to be the direction of the electric field of the wave, as seen
looking along the direction of propagation.
13. An antenna is a conducting device in which charge carriers are accelerated
periodically, by oscillating currents, so as to generate or respond to electromagnetic wave
fields. Antennas can transmit or receive waves.
Section 34.5
14. The displacement current through a surface, with or without the presence of dielectric
materials, is defined as the rate of change of electric flux times the product of the
permittivity constant and the dielectric constant.
15. Although the displacement current is not a real current, in the sense of involving the
displacement or movement of charge carriers, the direction of the rate of change of
electric field, associated with the displacement current, determines the direction of
induced magnetic fields according to the same right-hand rule used for ordinary
(conduction) currents.
16. In the generalized form of Ampere’s law, the contribution of displacement current is
added to the contribution of ordinary current piercing the surface. (See Equation 34.15.)
Section 34.6
17. They are Gauss’s law, zero magnetic flux through any closed surface (Gauss’s law for
magnetism), Faraday’s law, and (the generalization of) Ampere’s law.
18. The experimentally determined inverse square dependence of Coulomb’s law, arising
from the precisely measured absence of a steady state net charge on the interior surface of
a hollow conductor, is the basis for Gauss’s law. The non-existence of magnetic
monopoles guarantees the zero magnetic flux through any closed surface (Gauss’s law for
magnetism). Experiments on electromagnetic induction underlie Faraday’s law.
Measurements of the force between current-carrying wires and observations of
electromagnetic waves are the basis for Ampere’s law and Maxwell’s generalization of it.
19. Maxwell first recognized their coherence and completeness (along with charge
conservation) for describing all (classical) electromagnetic phenomena and devices
(including the prediction and subsequent discovery of electromagnetic waves).
Section 34.7
20. In free space, the speed is the square root of the inverse of the product of the electric
permittivity and magnetic permeability constants (see Equation 34.30), and in a
dielectric, it is this speed divided by the square root of the dielectric constant (see
Equation 34.35).
21. The ratio of the magnitudes of the electric and magnetic fields equals the speed of
propagation, in free space or in a dielectric.
16
22. The magnitudes and directions of the fields and the velocity of propagation are
related as in any transverse electromagnetic wave, but for a harmonic plane wave, the
fields vary sinusoidally (in space and time) and are uniform in planes perpendicular to the
direction of propagation.
23. From lowest to highest frequencies, they are various radio waves, infrared, visible
light, ultraviolet, X-rays and gamma rays.
24. Although governed by the same physics, the way electromagnetic waves interact with
matter depends strongly on their frequency.
Section 34.8
25. The electric and magnetic energy densities are equal to each other, and therefore the
total is twice either one, or the square root of the product of the two. These relations are
expressed, in terms of the fields, by Equations 34.37-34.39.
26. The Poynting vector is the cross product of the electric field with the magnetic field
divided by the permittivity constant. It represents a flow of energy per unit time, or
power, crossing a unit area perpendicular to the direction of flow. For electromagnetic
waves, the magnitude of the Poynting vector is called the intensity of the wave.
27. The intensity of a wave equals the product of the energy density and the speed of
energy propagation of the wave. (See Equation 34.42.)
17
Chapter 35 (version 04/27/06)
Electric circuits
Review questions
Answers to these questions can be found at the end of this chapter.
Section 35.1
1. What is the essential topological property (i.e. the way circuit elements are
connected) of an electric circuit?
2. What essential electrical element is necessary in order for an electric circuit to
function?
3. How can the basic energy conversions in an electric circuit be represented in terms of
two generalized circuit elements?
4. Why is tungsten used for the filaments of light bulbs?
Section 35.2
5. In electrical circuits, to what situation does the term ‘steady state’ refer?
6. In a steady state, what does the continuity principle say about the accumulation or
depletion of charge carriers at any point in a circuit?
7. When a charge carrier moves between two points in a circuit, what determines the
amount of energy that can be converted to other forms than electrical energy?
8. What electrical property of a circuit element determines how great a potential
difference must be applied to it in order to maintain a given steady current?
Section 35.3
9. In a steady state, what does continuity require for (a) the current in one branch of a
circuit, and (b) the currents flowing into and out of a junction?
10. What electrical quantity is the same for two bulbs connected (a) in series, and (b) in
parallel?
11. What is a short circuit and what does it do?
12. Does the orientation or order that elements are represented in a circuit diagram make
any difference?
Section 35.4
13. What is the electric field like inside a conductor of uniform cross section carrying a
steady current?
14. How is the electric field inside a conductor, of uniform cross section carrying a steady
current, related to the potential difference across the length of the conductor?
15. What does the resistance of an object which conducts electric current depend on?
Section 35.5
16. Describe the Drude model for a metallic conductor.
1
17. What is the electric conductivity, and in general, on what dose it depend?
18. In the Drude model, what properties of the material does the conductivity depend on?
19. How is the electrical resistance of a conducting element defined and what is this
definition called?
20. What is an I-V curve, what information does it give about resistance, and what does
such a curve look like for an ohmic element?
Section 35.6
21. What two conditions apply to a single closed loop circuit in the steady state?
22. What is the equivalent resistance of a number of resistors in series?
23. What is the equivalent resistance of a number of resistors in parallel?
Section 35.7
24. What are Kirchhoff’s circuit laws?
25. Summarize the suggested strategy for analyzing multi-loop circuits.
Section 35.8
26. What is the general expression for electrical power converted in a circuit element?
27. What is the expression for the power dissipated in a resistor, in terms of the current
through it, and what kind of energy conversion does this represent?
2
Developing a feel
Calculate or estimate the following quantities:
1.
2.
3.
4.
The current flowing through a 100-W light bulb? (E)
The resistance of a 100-W light bulb? (E)
The resistance of a 4-W nightlight plugged into a bathroom outlet? (E)
The energy a 100-W light bulb uses in an hour? If this energy was used to lift you up
a building (using some device that works with 100% efficiency), how high could it
lift you? How far is that in floors? (A)
5. The resistance of your flashlight light bulb? The current flowing through it? (H, K)
6. The resistance of the wires in your house from the fusebox or circuit breaker panel
(where the electrical lines enter your house) to the 100-W lamp in your living room?
(M, D)
7. The electric field along the filament of a 100-W light bulb? (E, B)
8. The energy is stored in a C battery? (H, Q)
9. The cost of 1 MJ (106 J) of battery energy. Compare this to the $0.03 per MJ cost of
gasoline or electrical energy. (J)
10. The electrical power used by the average US home? (N, G, L, I)
11. The number of large (1 GW) power plants needed to supply electricity to all the
homes in the US. (C, F, J)
12. The money a typical American home pays for electricity each year (at $0.10 per kWhr or $0.03 per MJ)? (O, P)
13. The height to which you could lift an American family and its cars if you used all the
electrical energy they consumed in one year. (A, J)
Hints:
A. Consider gravitational potential energy.
B. How long is the filament?
C. What is the population of the US?
D. What is the maximum current in the house wires (how much power can one outlet
provide)?
E. What is the potential difference supplied by your house electrical wiring?
F. How many people live in each home?
G. How much power does your heating and cooling system use?
H. Compare the light output of a flashlight to different household light bulbs (eg: 100 W,
40 W, night light, etc)
I. Alternatively, you can estimate it from a typical monthly power bill and the cost per
kW-hr (kilowatt-hour).
J. Answer the previous question first.
K. What potential difference do the batteries supply?
L. How much of the time do we use the various lights and appliances? What fraction of
the time is your heating and cooling system actively working?
3
M.
N.
O.
P.
Q.
What potential difference can be tolerated in the house wires?
What lights and appliances do we use?
Answer question 10 first.
How many hours are there in a year?
How long can a flashlight remain on before it dims significantly?
Key: A. U = mgh. B. The filament spans about 1 cm but it is tightly coiled so it is about
3 cm long. C. The world population is about 6 billion and we have less than 10% of the
world’s population. The US population is 300 million (3 x 108). D. If you plug too many
things into one outlet, the circuit breaker or fuse will open. This is especially obvious
with 1500-W toasters or 1000-W microwave ovens. Therefore the limit is about 2000 W.
This corresponds to I = P/V = 2000 W / 100 V = 20 A. E. 110 V. F. More than 1 and less
than 10. I’ll estimate 3. G. Your heating and cooling system uses more than 100 W (ie:
more heat than one light bulb can supply) and less than 100 kW (less than 1000 light
bulbs). Say about 3 kW. H. A flashlight emits about as much light as a 4-W night light,
therefore, it should have about the same power output. I. A typical monthly electrical
power bill is more than $10 (since then your parents wouldn’t bother to nag you to turn
off the lights) and less than $1000 (since most people couldn’t afford that). I’ll estimate
$100 per month. J. No more help here. Sorry. K. A battery (AAA, AA, C or D)
maintains a potential difference of 1.5 V. Two batteries in series (head to tail) maintain a
potential difference of 3 V. L. More than 10% of the time but less than 100%. Say about
30% (or 7 hours per day). M. The potential difference between the two ends of the wire
(between the circuit breaker panel and the lamp) must be much less than the potential
difference across the lamp filament. I’ll estimate that it is 1-2% of the 110 V. N. About a
dozen or so lights, a refrigerator, stove, and heating and cooling system. O. No more
help here. Sorry. P. (days/year)*(hours/day)*(minutes/hour)*(seconds/minute). Q. A
few hours.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any
particular section. Typically, an example that is worked out in detail is followed
immediately by an example whose solution you should work out by following the
guidelines provided.
NONE SUPPLIED WITH THIS CHAPTER.
5
Questions and problems
Section 35.1
Section 35.2
Section 35.3
Section 35.4
Section 35.5
1. (I)@STE35.05.01 What is the magnitude of the applied electric field inside an
aluminum wire of radius 1.0mm that carries a 4.0A current?
2. (I)@STE35.05.02 For each of the following situations, predict whether !, the average
time between collisions of electrons with lattice ions in a conductor, would increase or
decrease. Explain your answers. (a) The spatial density of the lattice ions is increased.
(b) The size of the lattice ions is decreased. (c) The charge of the lattice ions is increased.
3. (I)@STE35.05.03 In what case could the charge on the charge carriers in a current
have a magnitude other than |e|?
4. (I)@STE35.05.04 What happens to the resistance of a wire if, for a given material at a
fixed temperature, you double the diameter of the wire and make it three times as long?
5. (I)@STE35.05.05 Two wires made of the same material are held at the same potential
difference. One is twice as long as the other. (a) How do the currents in the two wires
compare when they are first connected to the potential source? (b) A long time later, the
currents in the two wires are measured and found to be different than when they were
first connected. Which one do you expect to have changed the most, and why?
6. (II)@STE35.05.06 In a copper wire with diameter 1.63mm, the drift velocity is found
to be 7.08 × 10−4 m s . Assume that there is one free electron per copper atom. (a) What
is the current I in the wire? (b) What is the current density J in the wire? (c) Assuming it
is 3m from the switch to the light bulb, if an electron at the switch had to make it to the
bulb in order for the bulb to light up, how long would the delay be?
7. (II)@STE35.05.07 In a particle accelerator, protons are traveling at 0.1c. If the beam
has a radius of 0.1µm and carries a current of 2.0nA, how many protons strike a target
placed in the proton beam in 1.0s?
8. (II)@STE35.05.08 Gauge is a term used to describe the size of a wire; the larger the
gauge, the smaller the wire’s diameter. 4-Gauge wire has a diameter, at room
temperature, of 5.19mm. 22-Gauge wire has a diameter, at room temperature, of 0.64mm.
6
For copper, how long would a wire of each gauge need to be to have a resistance of
1.0Ω?
Section 35.6
9. (I)@STE35.06.09 (a) If a 4Ω resistor and an 8Ω resistor are connected in parallel,
which will carry the greater current, and why? (b) What if they are connected in series?
10. (I)@STE35.06.10 When a certain non-ideal battery is connected to a 2Ω resistor, a 2A
current flows through the circuit. When the same battery is connected to a 1Ω resistor, a
3A current flows through the circuit. Use these measurements to determine the emf and
internal resistance of the battery.
11. (I)@STE35.06.11 When two resistors are connected in series, their equivalent
resistance is 8Ω. When they are connected in parallel, their equivalent resistance is 1.5Ω.
What are the resistances of the two resistors?
12. (II)@STE35.06.12 The circuit shown Figure 35.00 has resistance values of
R1 = R5 = R6 = 1.0Ω , R2 = R3 = 2.0Ω , and R4 = 3.0Ω . (a) Find R4,5 , the equivalent
resistance for R4 and R5. (b) Find R2,3,4,5 , the equivalent resistance for R2, R3, R4 and R5.
(c) Find Rtot, the equivalent resistance for the entire circuit. (d) If the battery is 14V, what
is the current through each resistor?
13. (II)@STE35.06.13 For the circuit diagram shown in Figure 35.00, (a) which resistors,
if any, are in series? (b) Which resistors, if any, are in parallel? (c) Find the equivalent
resistance of the circuit, given R1 = R3 = R5 = 2.0Ω and R2 = R4 = R6 = 1.5Ω .
7
14. (II)@STE35.06.14 For the circuit diagram shown in Figure 35.00, (a) which resistors
are in series? (b) Which resistors are in parallel? (c) Find the equivalent resistance of the
circuit, given that R1 = R3 = R5 = 2.0Ω , R2 = 1.5Ω , and R4 = R6 = 1.0Ω . (d) What is the
current through the battery and each resistor if Vbatt = 12V ? (e) Using your answers to the
previous part, verify that the current entering the point a is equal to the current leaving it.
15. (II)@STE35.06.15 A piece of copper wire has a length of 10m and a diameter of
0.2mm. (a) What is the resistance of the wire? (b) The wire is cut into N identical shorter
pieces of wire, and the pieces are all placed in a bundle (connected in parallel) to form a
single new resistor. What is the minimum value of N that will give the bundle a
resistance less than 1.0mΩ?
Section 35.7
16. (I)@STE35.07.16 Given
ε
= 4V , ε 2 = 9V , R1 = R5 = 5Ω , R2 = R3 = 10Ω , and
R4 = 15Ω , find the current in the circuit shown in Figure 35.00. In what direction does it
flow (clockwise or counterclockwise)?
1
8
17. (II)@STE35.07.17 “Jumping” a car battery is something a lot of people have to do,
but it turns out that doing it wrong can be quite dangerous. A car battery can be modeled
as in Figure 35.00. (a) If you wanted to connect another battery so that you have the
smallest possible current going through the bad battery, how do you think you should
connect the two batteries (in series or in parallel)? (b) Assume the good battery has
ε1 = 12V and R1 = 0.02Ω and the bad battery has ε 2 = 11V and R2 = 0.02Ω . Calculate
the current through the batteries if they are connected in series. (c) Calculate the current
through the batteries if they are connected in parallel.
18. (II)@STE35.07.18 For the circuit shown in Figure 35.00, ε1 = ε 2 = 6V ,
and R1 = R2 = 8Ω . Calculate the magnitudes of the three currents shown.
ε
3
= 9V ,
19. (I)@STE35.07.19 In Figure 35.00, ε1 = 8V , ε 2 = 4V , and R1 = 5Ω . If the potential
difference between points a and b is measured to be 5.5V, what is the value of the
resistance R2?
9
20. (I)@STE35.07.20 If ε1 = 3.0V and ε 2 = 5.0V in Figure 35.00, (a) what value of
will cause the potential difference across R1 to be zero? (b) In that case, how much
current flows through R2 ?
ε
3
21. (II)@STE35.07.21 For the circuit shown in Figure 35.00, ε1 = 10V , ε 2 = 2V ,
R1 = 5Ω , R2 = 3Ω , and R3 = 6Ω . (a) Find the current that flows in the circuit and
indicate its direction. (b) Calculate the potential differences Vab, Vbc, Vcd, and Vda.
22. (II)@STE35.07.22 In Figure 35.00, ε1 = 5V , ε 2 = 5V , and ε 3 = 1.5V . (a) If the
resistors all have the same resistance of 50Ω, what is the current in each branch of the
circuit? (b) What are the potential differences Vab, Vbc, Vcd, Vde, Vef and Vfa?
23. (III)@STE35.07.23 For the circuit diagram shown in Figure 35.00, all batteries supply
an emf of 6V, and all resistors have a resistance of 3Ω. Find the magnitudes of the five
currents shown in the figure.
10
Section 35.8
24. (I)@STE35.08.24 If a light bulb has a resistance of 5.5Ω and is dissipating energy at a
rate of 9.0W, (a) how much current must be flowing through the light bulb? (b) What is
the potential difference across the light bulb?
25. (I)@STE35.08.25 A 2.0A current flows through a 10Ω resistor. How much energy is
dissipated by the resistor in one hour?
26. (II)@STE35.08.26 If ε1 = 10V , ε 2 = 5V and all resistances are 50Ω , (a) find the
current through and potential difference across each resistor. (b) What is the rate of Joule
heating in each resistor? (c) What is the power in, and is it used or supplied by, each
battery?
= 4.5V , ε 2 = 8.0V , and R1 = 12Ω . If the
power dissipated by R1 is 0.75W, (a) what is the value of the resistance R2 ? (b) How
much power is dissipated by R2 ? (c) Which battery supplies energy to the circuit, and at
what rate does it do so? (d) At what rate is energy used by the other battery?
27. (II)@STE35.08.27 In Figure 35.00,
ε
1
11
Answers to review questions
1. An electric circuit must contain a closed conducting path that allows charge carriers
to circulate.
2. Every (functioning) electric circuit must contain a source of potential difference (such
as a battery or a generator).
3. One general element (the source) provides electric potential energy to the other (the
rest of the circuit, or the load) where it can be converted into other forms of energy.
4. The electric current in a light bulb causes the filament to become white hot and emit
light and heat. Tungsten has a very high melting point, so it does glows and does not
melt.
5. Steady state for an electric circuit means that the current in every part of the circuit is
constant over time.
6 .In a steady state, charge carriers are not accumulating or getting “used up” at any
point in a circuit.
7. The potential difference between the two points, multiplied by the charge on the
carrier, determines the amount of energy conversion.
8. The resistance.
9. (a) A steady current is the same everywhere in a given branch. (b) The sum of the
currents going into a junction must equal the sum of the currents coming out of that
junction.
10. (a) The current is always the same for two bulbs connected in series, whereas (b) the
potential difference is always the same when the bulbs are connected in parallel.
11. A conducting path between two points, with zero (negligible) resistance, is a short
circuit. Almost (essentially) all the current flows through the short, and (essentially) none
flows through any other branch connecting those two points.
12. No, as long as the connections between elements in the actual circuit are accurately
portrayed.
13. Such an electric field has the same magnitude everywhere inside the conductor and is
parallel to the walls of the conductor (i.e. in the direction of the current).
14. The electric field strength is equal to the potential difference divided by the length.
12
15. The resistance of an object depends on its shape and the material from which it is
made.
16. In this model, a metal consists of positive ions, which can vibrate about fixed lattice
positions, and conduction electrons, which are free to move about the lattice, colliding
occasionally with the ions. The free electrons move randomly between collisions, except
in the presence of an electric field, where their average drift velocity (over many
collisions) is proportional to the field.
17. The conductivity is the proportionality factor between the current density and the
electric field in a conductor. It depends on the type of material, the temperature, and in
some materials, also on the magnitude of the current (i.e. the field).
18. The conductivity depends on the number density of the charge carriers, their charge
and inertia, and on the average time between collisions, as in Equation 35.9.
19. The resistance, defined by Ohm’s law, is the ratio of the potential difference across
the element to the current through it.
20. An I-V curve is the plot of the current vs. the potential difference for a circuit
element. Its slope at any point is the reciprocal of the resistance of the element, and for
ohmic elements, the I-V curve is a straight line (i.e. the resistance is independent of the
potential difference).
21. The current through any part of the loop (the number of charge carriers passing any
point per unit time) is the same, and the sum of the energy changes of the charge carriers
going around the entire loop (from the emf’s and potential differences) is zero.
22. The equivalent resistance of resistors in series equals the sum of the individual
resistances.
23. The reciprocal of the equivalent resistance of resistors in parallel equals the sum of
the reciprocals of the individual resistances.
24. Kirchhoffs first circuit law is the junction rule (the sum of the currents entering and
leaving any junction are equal) and the second is the loop rule (the sum of the energy
changes, per unit positive charge carrier, going around any closed loop is zero).
25. Identify the junctions and label the currents in each branch. Write down Kirchhoff’s
first law for all but one junction, and Kirchhoff’s second law for as many loops as
necessary, so that there are as many independent equations as there are unknowns to
solve for.
26. The power converted in a circuit element is equal to the product of the current
through it and the potential difference across it. (See Equation 35.54.)
13
27. The power dissipated in a resistor is equal to the current squared time the resistance,
and represents the conversion of electric potential energy to thermal energy. (See
Equation 35.55.)
14
Chapter 37 (version 05/04/06)
Optics
Review questions
Answers to these questions can be found at the end of this chapter.
Section 37.1
1. What has to happen for you to “see” an object with your eyes?
2. What is a ray of light?
3. What is a shadow?
Section 37.2
4. What do colors of light correspond to?
5. When light falls on an object, some of it can be absorbed. In general, what can
happen to the rest of the light than isn’t absorbed?
6. What is the law of reflection?
7. What is the difference between diffuse and specular reflection?
8. Explain what it means to form an image of an object, for example in a mirror.
9. Where is the image of an object in a plane mirror located?
10. How are wavefronts and rays related to the direction of propagation and to each other,
for light traveling in a vacuum?
Section 37.3
11. What is refraction, and why does it occur?
12. Does the wavelength of light change when the light passes from one transparent
medium into another of different density, and if so, how?
13. When a ray of light passes through the interface between two transparent media, one
denser than the other, how do the angles between the ray and the normal in the two
media compare?
14. What is total internal reflection, and why does it occur?
15. What is Fermat’s principle?
16. What is dispersion and why does it occur?
Section 37.4
17. What is the principle of optical reversibility (or the reciprocity principle)?
18. What are paraxial rays?
19. What are the three principle rays for converging and diverging thin lenses, and what
are their paths?
20. What is the difference between a real and a virtual image?
Section 37.5
21. What is the definition of the index of refraction of a material?
1
22. How does the wavelength of light in a medium depend on the index of refraction?
23. What is Snell’s law?
24. How does the critical angle for total internal reflection, at the boundary between to
media, depend on the indices of refraction of the media?
Section 37.6
25. What is the lens equation (for thin lenses) and what are the sign conventions for the
distances in it (that allow it to be used for converging and diverging lenses)?
26. What is the magnification of a lens and what does its sign tell you?
27. What is the meaning of the near point?
28. How is angular magnification defined?
29. How are diopters defined and what do they signify?
Section 37.7
30. For paraxial rays, how does the focal length of a spherical mirror depend on the size
of the mirror?
31. Can the (thin) lens equation be used to relate the focal length, object and image
distances for spherical mirrors, and if so, with what sign conventions?
Section 37.8
32. What restrictions apply to the lens maker’s formula?
33. What is the sign convention for the radii of curvature in Equation 37.49?
2
Developing a feel
Calculate or estimate the following quantities:
1. The thickness of glass needed to delay light by 1 µs. (H, A, L)
2. The thickness of glass needed to delay light by a µs relative to the same thickness of
air. (H, A, L, D)
3. The parallel displacement of a horizontal light ray passing through a car windshield.
(I, A, D, M)
4. The angular range through which you can see the air when looking up from
underwater. (J, V, D)
5. The critical angle at a diamond-air interface. (J, Y, D)
6. The angular displacement of the sun by the atmosphere when sunlight is tangent to
the surface of the Earth. (I, D, Q, E, K)
7. The time you can still see the sun after it is physically below the horizon. (F, R)
8. The maximum focal length of the eye. (G, S, N)
9. The minimum focal length of the eye. (G, T, N)
10. The radius of curvature of a funhouse mirror that makes you look fat. (U, B, O, W)
11. The radius of curvature of one side of a +5 diopter glass lens if the other side is flat.
(A, P, X)
12. The height at which a large passenger airliner no longer casts a dark shadow on the
ground. (C, Z)
Hints:
A. What is the index of refraction of glass?
B. What is the distance from the mirror to the object?
C. How large is an airplane?
D. What is the index of refraction of air?
E. What are the height of the atmosphere and the radius of the Earth?
F. See the answer to the previous question.
G. What is the lens equation?
H. How are speed and index of refraction related?
I. How are angle of refraction and index of refraction related?
J. What is the critical angle?
K. What is the maximum angle the sun makes with respect the atmosphere?
L. What is the speed of light in vacuum?
M. What is the thickness of the windshield?
N. What is the distance from the lens to the image?
P. What is the relationship between diopters and focal length?
O. What is the relationship between magnification and o, i and f?
Q. How do I find the maximum angle the sun makes with respect the atmosphere?
R. How fast does the Earth rotate on its axis (in degrees per hour)?
3
S.
T.
U.
V.
W.
X.
Y.
Z.
What is the minimum distance from an object to the lens?
What is the maximum distance from an object to the lens?
How much is your image magnified horizontally to make you look fat?
What is the index of refraction of water?
What is the relationship between focal length and radius of curvature?
What is the lens maker’s formula?
What is the index of refraction of diamond?
What is the angular size of the Sun?
Key (all values are reasonable estimates, but are still only “about”)
A. n = 1.4 to 1.5. B. About 2 m or so. C. About 50 m (either length or wingspan). D. n =
1.00029 E. RE = 6.4×106 m and ha = 105 m. F. No more help here. G. 1/f = 1/o + 1/i. H. c
= c0/n I. n1sin!1 = n2sin!2 J. !c = sin-1(n1/n2) K. About 10o. L. c0 = 2.998×108 m/s. M. A
few mm. N. The diameter of the eyeball is 2-3 cm. O. M = -i/o and 1/f = 1/o + 1/i. P. d
= 1 m / f. Q. Treat the atmosphere as a spherical shell surrounding the Earth. Consider
the angle the tangent to the surface makes to the outside of the atmosphere. R. " = 360o /
24 hr = 15o/hr. S. About 0.25 m. T. Infinity. U. About a factor of 2 or so. V. n = 1.33.
W. f = R/2. X. 1/f = (n#1)(1/R1 + 1/R2). Y. n = 2.42. Z. The Sun is about the same
angular size as a finger-width at arm’s length, or about 1 cm at 1 m, which is ! = 1 cm / 1
m = 0.01 radian.
4
Worked and guided problems
These examples involve material from this chapter, but are not associated with any
particular section. Typically, an example that is worked out in detail is followed
immediately by an example whose solution you should work out by following the
guidelines provided.
NONE SUPPLIED WITH THIS CHAPTER.
5
Questions and problems
Section 37.1
Section 37.2
Section 37.3
Section 37.4
Section 37.5
1.(I)@STE37.05.01. (a) What would be the wavelength of light in a medium where its
speed is 2.4 × 108 m s if its wavelength in vacuum is 550nm? (b) What is the frequency
of this light in vacuum? (c) What is the frequency of this light in the medium?
2.(I)@STE37.05.02. You shine a laser pointer at an angle of 30° from the normal to the
surface of a thick block of glass ( n = 1.5 ) . (a) What is the angle of refraction of the light?
(b) Draw a diagram showing the normal, the surface, and the incident and refracted
waves. (c) How would your answer to part (a) change if the incident angle were 45°?
3.(I)@STE37.05.03. To produce a cool visual effect, a light ray shines up from the
bottom of a tank of mineral water ( n = 1.37 ) . Assuming that the surface is smooth, at
what angle from the normal would the light ray have to strike the top surface of the water
to reflect back into the tank and not escape?
4.(II)@STE37.05.04. You shine a laser pointer ( λvacuum = 538nm ) at an angle of 60° from
the normal to the surface of a calm pond ( n = 1.33) . How do each of the following
properties of the light change: (a) the wavelength, (b) the frequency, (c) the speed and
(d) the direction of propagation? (e) If instead, you shine the light directly onto the
surface (along the normal), do any of these answers change, and if so, how?
5.(II)@STE37.05.05. A ray of light is incident on a layer of oil floating on water. The
oil, water and air all have different indices of refraction. Call the angle of incidence in the
air θa, the angle of refraction in oil θo, and the angle of refraction in water θw. (a) Draw a
diagram including all three layers, showing all angles. (b) Show that na sin θ a = nw sin θ w .
(c) In part (b) you showed that you could neglect the oil in finding the final direction of
the transmitted ray. Is this also true of the final location of the transmitted ray? Justify
your answer.
6.(II)@STE37.05.06. As we saw earlier in the chapter, some media have a property
called dispersion, which causes the index of refraction of light to vary with its
wavelength. Flint glass is one such material, with an index of refraction of 1.66 for blue
light, and one of 1.61 for red light. Other colors fall somewhere in between. Consider a
6
ray of white light that is traveling through a slab of flint glass. The ray strikes the surface
of the slab at an angle of 30° from the normal and exits into the air. (a) Calculate the
angle of refraction for the blue light. (b) Calculate the angle of refraction for the red light.
(c) A flat screen is placed in the path of the refracted ray such that the ray is
perpendicular to the screen. If the distance along the refracted ray from the slab to the
screen is 0.5m, what is the distance between the point of red light and the point of blue
light in the rainbow produced on the screen? (Use a small angle approximation.)
7.(III)@STE37.05.07. Figure 37.00 depicts a light ray entering an optical fiber from air
and being transmitted through the fiber with very little loss of light by means of total
internal reflection. (a) If the index of refraction of the fiber is n, what is the maximum
value of !, the angle that the entering ray makes with the normal, for which this will
occur? (b) When n is greater than a certain number no, the light ray can enter the fiber at
any angle ! (between 0° and 90°, of course), and total internal reflection will still occur
inside the fiber. What is no?
Section 37.6
8.(I)@STE37.06.08. (a) Draw a simplified ray diagram for an object outside the focal
length of a converging lens, including all three principal rays. State whether the image in
such a case is (b) real or virtual, and (c) upright or inverted. If the focal length of the lens
is 50cm, and the object distance is 80cm, what are (d) the image distance and (e) the
magnification of the image?
9.(I)@STE37.06.09. (a) Draw a simplified ray diagram for an object inside the focal
length of a converging lens, including all three principal rays. State whether the image in
such a case is (b) real or virtual, and (c) upright or inverted. If the focal length of the lens
is 40cm, and the object distance is 15cm, what are (d) the image distance and (e) the
magnification of the image?
10.(I)@STE37.06.10. (a) Draw a simplified ray diagram for an object outside the focal
length of a diverging lens, including all three principal rays. State whether the image in
such a case is (b) real or virtual, and (c) upright or inverted. If the focal length of the lens
is -30cm, and the object distance is 60cm, what are (d) the image distance and (e) the
magnification of the image?
11.(I)@STE37.06.11. (a) Draw a simplified ray diagram for an object inside the focal
length of a diverging lens, including all three principal rays. State whether the image in
such a case is (b) real or virtual, and (c) upright or inverted. If the focal length of the lens
7
is -50 cm, and the object distance is 20cm, what are (d) the image distance and (e) the
magnification of the image?
12.(II)@STE37.06.12. An astronomical telescope of angular magnification M! consists of
two lenses, an objective and an eyepiece, separated by a distance d, such that the focal
points of the two lenses coincide inside the telescope. In terms of M! and d, what are (a)
the focal length of the objective, and (b) the focal length of the eyepiece?
13.(II)@STE37.06.13. A schematic diagram of a refractive microscope is shown in
Figure 37.00. The focal length of the objective is 2.5cm, the focal length of the eyepiece
is 6.3cm, and they are separated by a distance of 20cm. If the distance of the object from
the objective is 3cm, (a) what is the distance of the image produced by the eyepiece from
the eyepiece? (b) What is the overall magnification of this image?
14.(II)@STE37.06.14. Figure 37.00 shows a schematic diagram of a projection device,
which consists of two convex lenses. The focal length of lens 1 is 10cm, that of lens 2 is
18cm, and the two lenses are separated by a distance of 15cm. If an object is placed 5cm
to the left of lens 1, (a) at what distance from lens 2 should a screen be placed if the
object’s image is to be focused on the screen? (b) What will be the overall magnification
of this image? (c) Will this image be upright or inverted?
15.(II)@STE37.06.15. A convex lens of focal length f is used to produce an image of an
object that is placed a distance o from the lens. (a) For what value of o will the size of
the image be the same as that of the object? (b) Will such an image be real or virtual?
(c) Will it be upright or inverted?
16.(II)@STE37.06.16. An object of height 4.0cm is placed 6.0cm in front of a convex
lens. If the resultant image is inverted and has a height of 8.0cm, (a) what is the distance
of the image from the lens? (b) What is the focal length of the lens?
8
17.(III)@STE37.06.17. In Figure 37.00, a convex lens of focal length 10cm and a
concave lens of focal length -8cm are separated by a distance of 16cm. An object is
placed 18cm to the left of the convex lens. (a) What is the distance of the image
produced by the concave lens from the concave lens? (b) What is the overall
magnification of this image?
Section 37.7
18.(I)@STE37.07.18. An object is placed 6.0cm in front of a concave mirror with a
radius of curvature of 20cm. What are (a) the resultant image distance and (b) the
magnification of the image? State whether the image is (c) real or virtual and (d) upright
or inverted. (e) Draw a ray diagram illustrating this situation, including the principal
rays.
19.(I)@STE37.07.19. A concave mirror with a radius of curvature of 7.0cm is used to
form an image of an object that is placed 15cm in front of the mirror. (a) If the height of
the object is 2.0cm, what is the height of the image? State whether the image is (b) real
or virtual and (c) upright or inverted. (d) Draw a ray diagram illustrating this situation,
including the principal rays.
20.(I)@STE37.07.20. You stand 0.5m in front of a concave mirror with a radius of
curvature of -3.5m. (a) Is your image real or virtual? (b) Is it upright or inverted? (c)
Where is it located? (d) What is its magnification?
21.(II)@STE37.07.21. When you look at your self in a concave mirror, you appear to be
one fourth your normal size. If you are standing 1.0m in front of the mirror, what is the
radius of curvature of the mirror?
22.(II)@STE37.07.22. When an object is placed 1.2m in front of a convex mirror, an
image of the object is formed at a distance of 0.75m from the mirror. (a) Is the image
real or virtual? (b) Is it upright or inverted? (c) What is the radius of curvature of the
mirror?
23.(II)@STE37.07.23. You wish to use a concave mirror to form an image that is N times
larger than the object. In terms of N and R, the radius of curvature of the mirror, at what
distance o from the mirror should you place the object?
9
24.(III)@STE37.07.24. The kind of image a spherical mirror can produce for a real object
has to do with where the object is in relation to its focal point. There are actually three
types of spherical mirrors, if you use a mathematical trick: a plane mirror is a part of a
curved surface, if the radius of curvature of that surface is infinitely large. Use this
information to summarize what kinds of images are possible for the three types of
spherical mirrors for real objects.
Section 37.8
25.(I)@STE37.08.25. A lens has an index of refraction of 1.5. Its left side curves
outward with a radius of curvature of 0.15m, and its right side curves inward with a
radius of curvature of 0.25m. What is its focal length (a) if you look through it from the
right side, and (b) if you look through it from the left side?
26.(I)@STE37.08.26. The two surfaces of a double convex lens each have a radius of
curvature of 0.18cm. If the focal length of the lens is 0.45cm, what is the lens’ index of
refraction?
27.(I)@STE37.08.27. A plano-convex lens has an index of refraction of 1.6 and a focal
length of 17cm. What is the radius of curvature of the lens’ convex face?
28.(II)@STE37.08.28. A thin lens has an index of refraction of 1.5. When an object 1.0
cm in height is placed 50 cm from the lens, an upright image 2.15 cm tall is formed. The
lens has one side that is concave with a radius of curvature of 35 cm. Find the radius of
curvature of the other side of the lens.
29.(II)@STE37.08.29. The focal length of a spherical lens can be different for different
colors of light. How can this happen?
30.(III)@STE37.08.30. For a lens in air, (a) for what conditions on its radii of curvature
will the lens be converging? (b) For what conditions on its radii of curvature will the lens
be diverging? (c) What happens to these conditions if the lens is placed in a medium
with index of refraction greater than that of the lens?
10
Answers to review questions
1. Light coming from the object has to enter your eye and form an image, in order for
you to “see” it. The light may be emitted by the object and/or redirected by it, from
another light source.
2. A ray is a line representing the direction in which light waves are traveling. (Because
a line has no thickness, a ray does not represent the intensity of the light waves, however,
to do so, you can think of a ray as corresponding to a very thin beam of light.)
3. When light rays fall on an object, the region behind the object, where rays are
blocked, is called the shadow.
4. The colors of light correspond to different frequencies of visible electromagnetic
waves.
5. Light that is not absorbed can be transmitted or reflected.
6. For a ray striking a smooth surface, the angle of reflection is equal to the angle of
incidence.
7. Light reflected from very smooth surfaces, over which the direction of the surface
normal doesn’t change (on the scale of wavelengths of the light), is reflected specularly.
If the surface irregularities are comparable in size to the wavelength of the light, the
surface normal varies randomly, and the light is diffusely reflected in all directions from
the surface.
8. If light rays coming from a point on an object appear to come from a different point
in space (for example, if the rays are reflected by a mirror), the latter is an image of the
former.
9. An object’s image in a plane mirror is located along the normal to the mirror through
the object, the same distance behind the mirror as the object is in front of it.
10. Wavefronts are surfaces perpendicular to the direction of propagation, while rays are
in the direction of propagation, so rays are also perpendicular to wavefronts.
11. Refraction is the bending of light entering a medium in which the speed of
propagation changes. The light bends because different parts of a wavefront, on different
sides of the interface, travel at different speeds. (For normal incidence, when all parts of
the wavefront pass the interface simultaneously, there is no bending, but the spacing
between successive wavefronts still changes.)
12. The frequency of light does not change when wavefronts enter a medium, but if the
speed of propagation does change, then so must the spacing between successive
wavefronts, which determines the wavelength in each medium. The denser the medium,
the closer the wavefronts and the smaller the wavelength.
11
13. The angle between the ray and the normal is smaller in the denser medium, where the
speed of propagation is slower.
14. When light traveling in one medium enters a second, less dense medium, the angle of
refraction is greater than the angle of incidence. When the angle of incidence is greater
than the critical angle, the angle of refraction would be greater than 90°, which is
impossible (because the ray would no longer be in the second medium), so the light is
totally reflected.
15. Fermat’s principle states that the path actually taken by a ray of light is that path for
which the travel time is the least.
16. If the speed of light in a material depends on the frequency, rays representing light of
different colors separate on refraction. This is called dispersion.
17. This principle states that, the path followed by a ray of light is not changed by just
reversing its direction.
18. Paraxial rays are those rays from an object that make only small angles with the axis
of a lens, or optical system. (Parallel paraxial rays are focused in the focal plane of a
lens.)
19. Paraxial rays that strike the lens parallel to the axis pass through the focus on the
other side, for a converging lens (real focus), and appear to diverge from the focus on the
same side, for a diverging lens (virtual focus). Paraxial rays pass straight through the
center of either type of thin lens. Paraxial rays that strike a converging lens, after passing
through, or appearing to diverge from, the focus on the same side as the object, emerge
parallel to the axis of the lens. Paraxial rays that strike a diverging lens converging
towards the focus on the other side emerge parallel to the axis.
20. Light rays actually pass through points of a real image, before diverging into your
eye. Rays do not pass through the points of a virtual image, but only appear to diverge
from such points when entering your eye. (A real image can seen on a screen, whereas a
virtual image cannot.)
21. The index of refraction is the ratio of the speed of light in vacuum to the speed of
light in the material.
22. The wavelength in a medium is equal to the wavelength in vacuum divided by the
index of refraction.
23. Snell’s law (or the law of refraction) states that the sine of the angle of refraction
divided by the sine of the angle of incidence equals the index of refraction of the incident
medium divided by the index of refraction of the refractive medium.
12
24. The sine of the critical angle (in the denser medium) is equal to the index of refraction
of the less dense medium divided by that of the denser one.
25. The reciprocal of the focal length equals the sum of the reciprocals of the object and
image distances (see Equation 37.30). The focal length of a converging lens is positive,
and that of a diverging lens is negative. The image distance is positive if the image is on
the other side of the lens as the object (real image), and negative if it is on the same side
as the object (virtual image). The object distance is positive if the object is on the same
side of the lens as the illumination is coming from, and negative otherwise.
26. The magnification is the ratio of the image height to the object height. Positive
magnification indicates an upright, negative an inverted image.
27. The near point is the closest distance that a person’s eye can see an object in sharp
focus. For young adults, this is nominally about 25 cm.
28. The angular magnification of a lens is the ratio of the angular size of the image to the
angular size of the object (see the left side of Equation 37.34).
29. The strength of a lens in diopters is the reciprocal of the focal length in meters, and
(besides the focal length) signifies the power of lenses for magnifying glasses or
corrective lenses for eye glasses.
30. The focal length is (approximately) one half of the radius of curvature (see Equation
37.38).
31. Yes, where the focal length is positive, negative or infinity for concave, convex or
planar mirrors, respectively. Object and image distances are positive in front of the mirror
(the same side as the illumination), and negative behind the mirror.
32. The lens maker’s formula given by Equation 37.49 applies to paraxial rays and thin
lenses only.
33. The radii of curvature are positive for concave, negative for convex and infinite for
planar surfaces. (Note: concave and convex are relative to the side of the surface outside
the lens material.)
13