serway1 (1)

23

Electric Fields

CHAPTER OUTLINE ANSWERS TO QUESTIONS

23.1 Properties of Electric Q23.1 A neutral atom is one that has no net charge. This
Charges means that it has the same number of electrons
orbiting the nucleus as it has protons in the nucleus.
23.2 Charging Objects by A negatively charged atom has one or more excess
Induction electrons.

23.3 Coulomb’s Law *Q23.2 (i) Suppose the positive charge has the large value 1 mC.
23.4 The Electric Field The object has lost some of its conduction electrons,
23.5 Electric Field of a in number 10−6 C (1 e/1.60 × 10−19 C) = 6.25 × 1012 and
in mass 6.25 × 1012 (9.11 × 10−31 kg) = 5.69 × 10−18 kg.
Continuous Charge Distribution This is on the order of 1014 times smaller than the
23.6 Electric Field Lines ~1g mass of the coin, so it is an immeasurably small
23.7 Motion of a Charged Particle change. Answer (d).

in a Uniform Electric Field

(ii) The coin gains extra electrons, gaining mass on
the order of 10−14 times its original mass for the charge
−1 m C. Answer (b).

Q23.3 All of the constituents of air are nonpolar except for water. The polar water molecules in the
air quite readily “steal” charge from a charged object, as any physics teacher trying to perform
electrostatics demonstrations in the summer well knows. As a result—it is difficult to accumulate
large amounts of excess charge on an object in a humid climate. During a North American
winter, the cold, dry air allows accumulation of significant excess charge, giving the potential
(pun intended) for a shocking (pun also intended) introduction to static electricity sparks.

Q23.4 Similarities: A force of gravity is proportional to the product of the intrinsic properties (masses)
of two particles, and inversely proportional to the square of the separation distance. An electrical
force exhibits the same proportionalities, with charge as the intrinsic property.

Differences: The electrical force can either attract or repel, while the gravitational force as
described by Newton’s law can only attract. The electrical force between elementary particles is
vastly stronger than the gravitational force.

Q23.5 No. The balloon induces polarization of the molecules in the wall, so that a layer of positive
charge exists near the balloon. This is just like the situation in Figure 23.4a, except that the signs
of the charges are reversed. The attraction between these charges and the negative charges on the
balloon is stronger than the repulsion between the negative charges on the balloon and the
negative charges in the polarized molecules (because they are farther from the balloon), so that
there is a net attractive force toward the wall. Ionization processes in the air surrounding the
balloon provide ions to which excess electrons in the balloon can transfer, reducing the charge
on the balloon and eventually causing the attractive force to be insufficient to support the weight
of the balloon.

1

2 Chapter 23

*Q23.6 Answer (c). Each charge produces field as if it were alone in the Universe.

*Q23.7 (i) According to the inverse square law, the field is one-fourth as large at twice the distance. The
answer is (c), 2 × 36 cm = 72 cm.

(ii) The field is four times stronger at half the distance away from the charge. Answer (b).

Q23.8 An electric field created by a positive or negative charge extends in all directions from the charge.
Thus, it exists in empty space if that is what surrounds the charge. There is no material at point A
in Figure 23.21(a), so there is no charge, nor is there a force. There would be a force if a charge
were present at point A, however. A field does exist at point A.

*Q23.9 (i) We compute qAqB/r2 in each case. In (a) it is 400/4 = 100 (nC/cm)2. In (b) and (c), 300/4 =
75 (nC/cm)2. In (d) 600/9 = 67 (nC/cm)2. In (e) 900/9 = 100 (nC/cm)2. The ranking is then a = e >

b = c > d.

(ii) We compute qA/r2 in each case. In (a) it is 20/4 = 5 nC/cm2. In (b) 30/4 = 7.5 nC/cm2. In (c)
10/4 = 2.5 nC/cm2. In (d) 30/9 = 3.3 nC/cm2. In (e) 45/9 = 5 nC/cm2. The ranking is then b > a =

e > d > c.

*Q23.10 The charge at the upper left creates at the field point electric field to the left, with magnitude we
call E1. The charge at lower right creates downward electric field with an equal magnitude E1.
These two charges together create field 2E1 downward and to the left at 45°. The positive

charge is 2 times farther from the field point so it creates field 2E1/( 2 )2 = E1 upward and

( )to the right. The net field is then 2 − 1 E1 downward and to the left. The answer to question

(i) is (d).

(ii) With the positive charge removed, the magnitude of the field becomes 2E1, larger than
before, so the answer is (a).

*Q23.11 The certain point must be on the same line as A and B, for otherwise the field components
perpendicular to this line would not add to zero. If the certain point is between A and B, it is
midway between them, and B’s charge is also +40 nC. If the certain point is 4 cm from A
and 12 cm from B, then B’s charge must be −9(40 nC) = −360 nC. These are the only two
possibilities. The answers are (a), (f), and (j).

Q23.12 The direction of the electric field is the direction in which a positive test charge would feel a force
when placed in the field. A charge will not experience two electrical forces at the same time, but
the vector sum of the two. If electric field lines crossed, then a test charge placed at the point at
which they cross would feel a force in two directions. Furthermore, the path that the test charge
would follow if released at the point where the field lines cross would be indeterminate.

Q23.13 Both figures are drawn correctly. E1 and E2 are the electric fields separately created by the point
charges q1 and q2 in Figure 23.12 or q and –q in Figure 23.13, respectively. The net electric field

is the vector sum of E1 and E2, shown as E. Figure 23.19 shows only one electric field line at
each point away from the charge. At the point location of an object modeled as a point charge, the

direction of the field is undefined, and so is its magnitude.

*Q23.14 Answer (a). The equal-magnitude radially directed field contributions add to zero.

*Q23.15 Answer (c). Contributions to the total field from bits of charge in the disk lie closer together in
direction than for the ring.

Electric Fields 3

*Q23.16 (i) Answer (c). Electron and proton have equal-magnitude charges.
(ii) Answer (b). The proton’s mass is 1836 times larger than the electron’s.

*Q23.17 Answer (b).

Q23.18 Linear charge density, l, is charge per unit length. It is used when trying to determine the
electric field created by a charged rod.

Surface charge density, s, is charge per unit area. It is used when determining the electric field
above a charged sheet or disk.

Volume charge density, r, is charge per unit volume. It is used when determining the electric field
due to a uniformly charged sphere made of insulating material.

Q23.19 No. Life would be no different if electrons were + charged and protons were – charged. Opposite
charges would still attract, and like charges would repel. The naming of + and – charge is merely
a convention.

Q23.20 In special orientations the force between two dipoles can be zero or a force of repulsion. In
general each dipole will exert a torque on the other, tending to align its axis with the field created
by the first dipole. After this alignment, each dipole exerts a force of attraction on the other.

SOLUTIONS TO PROBLEMS

Section 23.1 Properties of Electric Charges

P23.1 (a) The mass of an average neutral hydrogen atom is 1.007 9u. Losing one electron reduces its
mass by a negligible amount, to

( )1.007 9 1.660 × 10−27 kg − 9.11 × 10−31 kg = 1.67 × 10−27 kg .

Its charge, due to loss of one electron, is

( )0 − 1 −1.60 × 10−19 C = +1.60 × 10−19 C .

(b) By similar logic, charge = +1.60 × 10−19 C

( )mass = 22.99 1.66 × 10−27 kg − 9.11 × 10−31 kg = 3.82 × 10−26 kg

(c) charge of Cl− = −1.60 × 10−19 C

( )mass = 35.453 1.66 × 10−27 kg + 9.11 × 10−31 kg = 5.89 × 10−26 kg

( )(d) charge of Ca++ = −2 −1.60 × 10−19 C = +3.20 × 10−19 C
( ) ( )mass = 40.078 1.66 × 10−27 kg − 2 9.11 × 10−31 kg = 6.65 × 10−26 kg

( )(e) charge of N3− = 3 −1.60 × 10−19 C = −4.80 × 10−19 C
( ) ( )mass = 14.007 1.66 × 10−27 kg + 3 9.11 × 10−31 kg = 2.33 × 10−26 kg

continued on next page

4 Chapter 23

( )(f ) charge of N4+ = 4 1.60 × 10−19 C = +6.40 × 10−19 C
( ) ( )mass = 14.007 1.66 × 10−27 kg − 4 9.11 × 10−31 kg = 2.32 × 10−26 kg

(g) We think of a nitrogen nucleus as a seven-times ionized nitrogen atom.

( )charge = 7 1.60 × 10−19 C = 1.12 × 10−18 C
( ) ( )mass = 14.007 1.66 × 10−27 kg − 7 9.11 × 10−31 kg = 2.32 × 10−26 kg

(h) charge = −1.60 × 10−19 C

mass = ⎡⎣2(1.007 9) + 15.999⎤⎦1.66 × 10−27 kg + 9.11 × 10−31 kg = 2.99 × 10−26 kg

P23.2 (a) N = ⎛ 10.0 grams ⎞ ⎛ 6.02 × 1023 atoms ⎞ ⎛ 47 electrons ⎞ = 2.62 × 1024
⎝⎜ 107.87 grams mol⎠⎟ ⎝⎜ mol ⎟⎠ ⎝⎜ atom ⎟⎠

(b) # electrons added = Q = 1.00 × 10−3 C = 6.25 × 1015
e 1.60 × 10−19 C electron

or 2.38 electrons for every109 already present .

Section 23.2 Charging Objects by Induction

Section 23.3 Coulomb’s Law

P23.3 If each person has a mass of ≈70 kg and is (almost) composed of water, then each person
contains
ª
N ≅ ⎛ 70 000 grams ⎞ ⎛ 6.02 × 1023 molecules ⎞ ⎝⎛10 protons ⎞ ≅ 2.3 × 1028 protons
⎜⎝ 18 grams mol ⎠⎟ ⎝ mol ⎠ molecule ⎠

With an excess of 1% electrons over protons, each person has a charge

( )( )q = 0.01 1.6 × 10−19 C 2.3 × 1028 = 3.7 × 107 C

( )( )So 2
q1q2 3.7 × 107
F = ke r2 = 9 × 109 0.62 N = 4 × 1025 N ~1026 N

This force is almost enough to lift a weight equal to that of the Earth:

( )Mg = 6 × 1024 kg 9.8 m s2 = 6 × 1025 N ~ 1026 N

Electric Fields 5

*P23.4 In the first situation, FA on B,1 = ke qA qB ˆi. In the second situation, qA and qB are the same.
r12

ke qA qB
( )FB on A,2 r22 −ˆi
= −FA on B =

F2 = ke qA qB r12
F1 r22 ke qA qB

F2 = F1 r12 = 2.62 µN ⎛ 13.7 mm ⎞ 2 = 1.57 µN
r22 ⎜⎝ 17.7 mm ⎠⎟

Then FB on A,2 = 1.57 µN to the left .

( ( )( ) )(a)
P23.5 Fe = keq1q2 = 8.99 × 109 N ⋅ m2 C2 1.60 × 10−19 C 2 = 1.59 × 10−9 N (repulsion)
r2
3.80 × 10−10 m 2

( ( )( ) )(b)
Fg = Gm1m2 = 6.67 × 10−11 N ⋅ m2 C2 1.67 × 10−27 kg 2 = 1.29 × 10−45 N
r2
3.80 × 10−10 m 2

The electric force is larger by 1.24 × 1036 times .

(c) If ke q1q2 = G m1m2 with q1 = q2 = q and m1 = m2 = m, then
r2 r2

q= G= 6.67 × 10−11 N ⋅ m2 kg2 = 8.61 × 10−11 C kg
m ke 8.99 × 109 N ⋅ m2 C2

P23.6 We find the equal-magnitude charges on both spheres:

F = ke q1q2 = ke q2 so q=r F = (1.00 m) 1.00 × 104 N C2 = 1.05 × 10−3 C
r2 r2 8.99 × 109 N ⋅ m2
ke

The number of electron transferred is then

N xfer = 1.05 × 10−3 C = 6.59 × 1015 electrons
1.60 × 10−19 C e−

The whole number of electrons in each sphere is

⎛ 10.0 g ⎞
⎜⎝ 107.87 g mol ⎟⎠
( )( )Ntot
= 6.02 × 1023 atoms mol 47 e− atom = 2.62 × 1024 e−

The fraction transferred is then

f = N xfer = ⎛ 6.59 × 1015 ⎞ = 2.51 × 10−9 = 2.51 charges in every billion
N tot ⎝⎜ 2.62 × 1024 ⎠⎟

6 Chapter 23

( )( )( )F1
P23.7 = ke q1q2 = 8.99 × 109 N ⋅ m2 C2 7.00 × 10−6 C 2.00 × 10−6 C = 0.503 N
r2
(0.500 m)2

( )( )( )F2
= ke q1q2 = 8.99 × 109 N ⋅ m2 C2 7.00 × 10−6 C 4.00 × 10−6 C = 1.01 N
r2
(0.500 m)2

Fx = 0.503cos 60.0° + 1.01cos 60.0° = 0.755 N
Fy = 0.503sin 60.0° − 1.01sin 60.0° = −0.436 N

F = (0.755 N) ˆi − (0.436 N) ˆj = 0.872 N at an angle of 330°

FIG. P23.7

P23.8 Let the third bead have charge Q and be located distance x from the left end of the rod. This bead
will experience a net force given by

ke (3q)Q ke (q)Q

x2 (d − x)2
( )F= ˆi + −ˆi

The net force will be zero if 3 = 1 , or d−x= x.
x2 3
(d − x)2

This gives an equilibrium position of the third bead of x = 0.634d .

The equilibrium is stable if the third bead has positive charge .

P23.9 (a) The force is one of attraction . The distance r in Coulomb’s law is the distance between

centers. The magnitude of the force is

( )( )( )F
= keq1q2 = 8.99 × 109 N ⋅ m2 C2 12.0 × 10−9 C 18.0 × 10−9 C = 2.16 × 10−5 N
r2
(0.300 m)2

(b) The net charge of −6.00 × 10−9 C will be equally split between the two spheres, or
−3.00 × 10−9 C on each. The force is one of repulsion , and its magnitude is

( )( )( )F
= keq1q2 = 8.99 × 109 N ⋅ m2 C2 3.00 × 10−9 C 3.00 × 10−9 C = 8.99 × 10−7 N
r2
(0.300 m)2

P23.10 The top charge exerts a force on the negative charge keqQ which is directed upward and to

(d /2)2 + x2

the left, at an angle of tan −1 ⎛ d⎞ to the x-axis. The two positive charges together exert force
⎜⎝ 2x ⎠⎟

⎞⎛ (−x) ˆi ⎞
⎟⎠⎟
d2 4 + x2
( ) ( )⎛ d −2keqQ
2 md 3 8
⎜
⎝
2keqQ ⎟ ⎜⎜⎝ 1 2 = ma or for x << , a ≈ x
d2 4 + x2 ⎠

(a) The acceleration is equal to a negative constant times the excursion from equilibrium, as in

a = −ω 2x, so we have Simple Harmonic Motion with ω2 = 16keqQ .
md 3

T = 2π = π md3 , where m is the mass of the object with charge −Q.
ω 2 keqQ

(b) vmax = ω A = 4a keqQ
md 3

Electric Fields 7

( )(( ))F
P23.11 (a) = kee2 = 8.99 × 109 N ⋅ m2 C2 1.60 × 10−19 C 2 8.22 × 10−8 N toward the other
r2 0.529 × 10−10 m 2 =

particle

(b) We have F = mv2 from which
r

( )v =
Fr = 8.22 × 10−8 N 0.529 × 10−10 m = 2.19 × 106 m s
m
9.11 × 10−31 kg

Section 23.4 The Electric Field

P23.12 The point is designated in the sketch. The magnitudes of the

electric fields, E1 (due to the −2.50 × 10−6 C charge) and E2
(due to the 6.00 × 10−6 C charge), are

( )( )E1
= keq = 8.99 × 109 N ⋅ m2 C2 2.50 × 10−6 C FIG. P23.12
r2 d2
(1)

( )( )E2
= keq = 8.99 × 109 N ⋅ m2 C2 6.00 × 10−6 C (2)
r2
(d + 1.00 m)2

Equate the right sides of (1) and (2)

to get (d + 1.00 m)2 = 2.40d2

or d + 1.00 m = ±1.55d

which yields d = 1.82 m

or d = −0.392 m

The negative value for d is unsatisfactory because that locates a point between the charges where
both fields are in the same direction.

Thus, d = 1.82 m to the left of the − 2.50 µC charge

8 Chapter 23

P23.13 For equilibrium, Fe = −Fg
or
( )qE = −mg −ˆj

Thus, E = mg ˆj
q

( ( )( ) ) ( )(a)
E = mg ˆj = 9.11 × 10−31 kg 9.80 m s2 ˆj = − 5.58 × 10−11 N C ˆj
q −1.60 × 10−19 C

( ( )( ) ) ( )(b)
E = mg ˆj = 1.67 × 10−27 kg 9.80 m s2 ˆj = 1.02 × 10−7 N C ˆj
q 1.60 × 10−19 C

( ( )( ) ) ( )F
P23.14 = ke q1q2 = 8.99 × 109 N ⋅ m2 C2 1.60 × 10−19 C 2 6.02 × 1023 2 514 kN
r2 ⎡⎣2 6.37 × 106 m ⎤⎦2 =

*P23.15 The first charge creates at the origin field keQ to the right. +Q x = 0 q
a2

Both charges are on the x axis, so the total field cannot have a x

vertical comonent, but it can be either to the right or to the left. FIG. P23.15
If the total field at the origin is to the right, then q must be negative:

( )keQˆi + ( keq −ˆi = 2keQ ˆi q = −9Q
a2
a2 3a)2

In the alternative, if the total field at the origin is to the left,

( ) ( )keQˆi+keq −ˆi = 2keQ −ˆi q = +27Q
9a2 a2
a2

( )( )P23.16
(a) E = keq = 8.99 × 109 2.00 × 10−6 = 14 400 N C
r2
(1.12)2

Ex = 0 and Ey = 2(14 400)sin 26.6° = 1.29 × 104 N C FIG. P23.16

so E = 1.29 × 104 ˆj N C

( )( )(b) F = qE = −3.00 × 10−6 1.29 × 104 ˆj = −3.86 × 10−2 ˆj N

keq1 keq2 keq3 ke (2q) ke (3q) ke (4q)
r12 r22 r32
a2 2a2 a2
( )E = ˆi
P23.17 (a) rˆ1 + rˆ2 + rˆ3 = + ˆi cos 45.0° + ˆjsin 45.0° + ˆj

E = 3.06 keq ˆi + 5.06 keq ˆj = 5.91 keq at 58.8°
a2 a2 a2

keq 2
a2
(b) F = qE = 5.91 at 58.8°

Electric Fields 9

P23.18 The electric field at any point x has the x-component

= − ke q + keq = − ke q ( 4 ax )
− a
(x −(−a) x2 − a2 2
( ))E ( x )2 2

When x is much, much greater than a, we find E ≈ − 4a (keq) .

x3

P23.19 (a) One of the charges creates at P a field E= keQ n at an angle q
R2 + x2
to the x-axis as shown.

When all the charges produce field, for n > 1, the components
perpendicular to the x-axis add to zero.

nke (Q n) ˆi cosθ = keQxˆi .
R2 + x2 3 2
R2 + x2
( )The total field is FIG. P23.19

(b) A circle of charge corresponds to letting n grow beyond all bounds, but the result does
not depend on n. Smearing the charge around the circle does not change its amount or its

distance from the field point, so it does not change the field .

Section 23.5 Electric Field of a Continuous Charge Distribution

P23.20 ∫E = kedq , where dq = λ0 dx
P23.21 x2

∞ dx ⎛ 1⎞ ∞ keλ0 The direction is − ˆi or left for λ0 > 0.
x2 ⎝⎜ x ⎟⎠ x0 x0
keλ0

x0
=∫E = keλ0 − =

( )( )E )
= d keλ d) = ke (Q d) = d keQ ) = 8.99 × 109 22.0 × 10−6
d( +
(+ ( +d (0.290)(0.140 + 0.290)

E = 1.59 × 106 N C, directed toward the rod . FIG. P23.21

E= keQx
x2 + a2 3 2
( )P23.22

dE ⎡ 1 3x2 ⎤
dx ⎢ x2 + a2 x2 + a2
( ) ( )For a maximum, = Qke ⎣⎢ 32 − 5 2 ⎥ = 0
⎦⎥

x2 + a2 − 3x2 = 0 or x = a
2

Substituting into the expression for E gives

keQa keQ = 2keQ Q
( )E = = 3 3a2 = 3π ⑀0a2
2 3 a2 32 3 3 a2 6
2 2

10 Chapter 23

( ) ( ( )( ) ) ( )E =
P23.23 ke xQ 32 = 8.99 × 109 75.0 × 10−6 x 6.74 × 105 x We choose the x axis along
x2 + a2 x2 + 0.1002 3 2 = x2 + 0.010 0 3 2

the axis of the ring. E = 6.64 × 106 ˆi N C = 6.64ˆi MN C
(a) At x = 0.010 0 m ,

(b) At x = 0.050 0 m , E = 2.41 × 107 ˆi N C = 24.1ˆi MN C

(c) At x = 0.300 m, E = 6.40 × 106 ˆi N C = 6.40ˆi MN C

(d) At x = 1.00 m , E = 6.64 × 105 ˆi N C = 0.664ˆi MN C

P23.24 ⎛ x⎞
E = 2π keσ ⎜⎝1 − x2 + R2 ⎠⎟

⎛ x ⎞ ⎛ x⎞
7.90 × 10−3 ⎜1 − ⎟ ⎝⎜ 1 − x2 + 0.123 ⎠⎟
(0.350)2 ⎟⎠
⎜⎝
( )( )E = 2π 8.99 × 109 = 4.46 × 108

x2 +

(a) At x = 0.050 0 m , E = 3.83 × 108 N C = 383 MN C

(b) At x = 0.100 m , E = 3.24 × 108 N C = 324 MN C

(c) At x = 0.500 m, E = 8.07 × 107 N C = 80.7 MN C

(d) At x = 2.000 m, E = 6.68 × 108 N C = 6.68 MN C

P23.25 (a) ⎛ x⎞
From the Example in the chapter text, E = 2π keσ ⎜⎝1 − x2 + R2 ⎠⎟

σ = Q = 1.84 × 10−3 C m2
π R2

( )E = 1.04 × 108 N C (0.900) = 9.36 × 107 N C = 93.6 MN C

approximation: E = 2π keσ = 104 MN C (about 11% high)

⎛ 30.0 cm ⎞
⎜⎝ 1 − 30.02 + 3.002 cm ⎠⎟ =
( ) ( )(b)
E= 1.04 × 108 N C 1.04 ϫ108 N C (0.004 96) = 0.516 MN C

( )approximation: Q 5.20 × 10−6 0.519 MN C (about 0.6% high)
E = ke r2 = 8.99 × 109 =
(0.30)2

Electric Fields 11

P23.26 The electric field at a distance x is ⎡ x⎤
Ex = 2π keσ ⎢1 − ⎥
x2 + R2 ⎦
⎣

2π keσ ⎡ 1 ⎤
= ⎢1 − ⎥
This is equivalent to Ex ⎢⎣ 1 + R2 x2 ⎥⎦

For large x, R2 << 1 and 1+ R2 ≈ 1 + R2
x2 x2 2x2

= 2π keσ ⎛ 1 ⎞
so ( )Ex ⎜⎝⎜ 1 − R2 2x2 ⎤⎦ ⎠⎟⎟
⎡⎣1 +

( )( )1+ R2 2x2 −1

( )= 2π keσ ⎣⎡1 + R2 2x2 ⎦⎤

( ( ) )Ex x2
Substitute σ = Q , = keQ 1 2x2 ⎤⎦ = keQ ⎛ x2 + R2 ⎞
π R2 ⎣⎡1 + R2 ⎜⎝ 2 ⎟⎠

But for x >> R, x2 1 ≈ 1 , so Ex ≈ keQ for a disk at large distances
+ R2 x2 x2
2

∫ ∫ ∫Ey = dq sinθ
P23.27 Due to symmetry dEy = 0, and Ex = dE sinθ = ke r2
where
so that dq = λds = λrdθ,

where ∫Ex = keλ π sin θ dθ = keλ (− cosθ ) π = 2keλ
r 0 r
Thus, r0 FIG. P23.27

λ = q and r = L
Lπ

( )( )Ex
= 2keqπ 2 8.99 × 109 Nиm2 C2 −7.50 × 10−6 C π
L2 =
(0.140 m)2

Solving, Ex = −2.16 × 107 N C

( )Since the rod has a negative charge, E = −2.16 × 107 ˆi N C = −21.6ˆi MN C .

12 Chapter 23

P23.28 (a) We define x = 0 at the point where we are to find the field. One ring, with thickness dx, has

charge Qdx and produces, at the chosen point, a field
h

( )dE = x2 ke x 32 Qdx ˆi
+ R2 h

The total field is

∫ ∫ ( ) ∫ ( )d+h keQˆi d+h
E = dE = keQxdx 32 ˆi = 2h x=d x2 + R2 −3 2
x2 + R2
2 xdx

all charge dh

x2 + R2 −1 2 d+h ⎡ ⎤
= ⎢ ⎥
( )( )E = keQˆi ( ) ( )keQˆi 12 ⎥
2h −1 2 h⎢ 1 12 − 1 ⎥⎦
d2 + R2
x=d ⎢⎣ (d + h)2 + R2

(b) Think of the cylinder as a stack of disks, each with thickness dx, charge Qdx , and charge-
h
Qdx
per-area σ = π R2h . One disk produces a field

= 2π keQdx ⎛ x ⎞
π R2h ⎜⎜⎝1 − + R2 1 2 ⎟⎟⎠ ˆi
( )dE x2

d+h 2keQdx ⎛ x ⎞ ˆi
x=d R2h ⎜⎝⎜1 + R2 1 2 ⎠⎟⎟
∫ ∫ ( )So, E = dE = − x2

all charge

∫ ∫ ( ) ( )E2keQˆi 2keQˆi ⎡ x2 + R2 12 d+h ⎤
R2h ⎡d+h d+h ⎤ R2h ⎥
= ⎢ dx − 1 x=d x2 + R2 −1 2 2 xdx ⎥ = ⎢ x d+h −1
⎣d 2 ⎢ d 2
⎦ 12 ⎥
⎢⎣ d ⎦⎥

2keQˆi
R2h
( ) ( )E ⎡⎣⎢d 12 12⎤
= + h − d − (d + h)2 + R2 d2 + R2 ⎦⎥
+

2keQˆi
R2h
( ) ( )E = ⎣⎢⎡h 12 − 12⎤
+ d2 + R2 (d + h)2 + R2 ⎦⎥

P23.29 (a) The electric field at point P due to each element of length dx is

dE = kedq and is directed along the line joining the element to
x2 + y2

point P. By symmetry,

∫Ex = dEx = 0 and since dq = λdx,

E = Ey = ∫ dEy = ∫ dE cosθ where cosθ = y
x2 + y2

FIG. P23.29

Therefore, ∫ ( )2 dx 32 = 2keλ sinθ0
x2 + y2 y
E = 2keλy

0

(b) For a bar of infinite length, θ0 = 90° and Ey = 2keλ
y

Electric Fields 13

P23.30 (a) The whole surface area of the cylinder is A = 2π r2 + 2π rL = 2π r (r + L).

( )Q = σ A = 15.0 × 10−9 C m2 2π (0.025 0 m)[0.025 0 m + 0.060 0 m] = 2.00 × 10−10 C

(b) For the curved lateral surface only, A = 2π rL.

( )Q = σ A = 15.0 × 10−9 C m2 ⎣⎡2π (0.025 0 m)(0.060 0 m)⎤⎦ = 1.41 × 10−10 C
( )(c) Q = ρV = ρπ r2 L = 500 × 10−9 C m3 ⎣⎡π (0.025 0 m)2 (0.060 0 m)⎤⎦ = 5.89 × 10−11 C

P23.31 (a) Every object has the same volume, V = 8(0.030 0 m)3 = 2.16 × 10−4 m3.

( )( )For each, Q = ρV = 400 × 10−9 C m3 2.16 × 10−4 m3 = 8.64 × 10−11 C

(b) We must count the 9.00 cm2 squares painted with charge:

(i) 6 × 4 = 24 squares

( ) ( )Q = σ A = 15.0 × 10−9 C m2 24.0 9.00 × 10−4 m2 = 3.24 × 10−10 C

(ii) 34 squares exposed

( ) ( )Q = σ A = 15.0 × 10−9 C m2 24.0 9.00 × 10−4 m2 = 3.24 × 10−10 C

(iii) 34 squares

( ) ( )Q = σ A = 15.0 × 10−9 C m2 34.0 9.00 × 10−4 m2 = 4.59 × 10−10 C

(iv) 32 squares

( ) ( )Q = σ A = 15.0 × 10−9 C m2 32.0 9.00 × 10−4 m2 = 4.32 × 10−10 C

(c) (i) total edge length: = 24 × (0.030 0 m)

( )Q = λ = 80.0 × 10−12 C m 24 × (0.030 0 m) = 5.76 × 10−11 C

( )(ii) Q = λ = 80.0 × 10−12 C m 44 × (0.030 0 m) = 1.06 × 10−10 C

( )(iii) Q = λ = 80.0 × 10−12 C m 64 × (0.030 0 m) = 1.54 × 10−10 C

( )(iv) Q = λ = 80.0 × 10−12 C m 40 × (0.030 0 m) = 0.960 × 10−10 C

Section 23.6 Electric Field Lines P23.33
P23.32

FIG. P23.32 FIG. P23.33

14 Chapter 23

P23.34 (a) q1 = −6 = − 1
q2 18 3

(b) q1 is negative, q2 is positive

P23.35 (a) The electric field has the general appearance shown. It is
zero at the center , where (by symmetry) one can see that
the three charges individually produce fields that cancel out.

In addition to the center of the triangle, the electric field lines in
the second figure to the right indicate three other points near the
middle of each leg of the triangle where E = 0, but they are
more difficult to find mathematically.

(b) You may need to review vector addition in Chapter Three.
The electric field at point P can be found by adding the
electric field vectors due to each of the two lower point
charges: E = E1 + E2 .

The electric field from a point charge is E = ke q rˆ .
r2

As shown in the solution figure at right,

E1 = ke q to the right and upward at 60°
a2

E2 = ke q to the left and upward at 60° FIG. P23.35
a2

( ) ( ) ( )E q q
= E1 + E2 = ke a2 ⎣⎡ cos60°ˆi + sin 60°ˆj + − cos60°ˆi + sin 60°ˆj ⎦⎤ = ke a2 ⎣⎡2 sin 60°ˆj ⎦⎤

= 1.73ke q ˆj
a2

Section 23.7 Motion of a Charged Particle in a Uniform Electric Field

( )( )a = qE =
( )m
P23.36 (a) 1.602 × 10−19 6.00 × 105 = 5.76 × 1013 m s so a = −5.76 × 1013 ˆi m s2
1.67 × 10−27

( )(b) 2 2
v f = v i + 2a xf − xi

( )0 = vi2 + 2 −5.76 × 1013 (0.070 0) vi = 2.84 × 106 ˆi m s

(c) v f = vi + at t = 4.93 × 10−8 s

( )0 = 2.84 × 106 + −5.76 × 1013 t

Electric Fields 15

P23.37 (a) a = qE = 1.602 × 10−19 (640) = 6.14 × 1010 m s2
(b) m
1.67 × 10−27

v f = vi + at ( )1.20 × 106 = 6.14 × 1010 t t = 1.95 × 10−5 s

( )(c) 1 ) )( (xf 1
xf − xi = 2 vi + v f t = 2 1.20 × 106 1.95 × 10−5 = 11.7 m

( )( )(d) K = 1 mv2 = 1 1.67 × 10−27 kg 1.20 × 106 m s 2 = 1.20 × 10−15 J
22

( ) ( )( )P23.38 The particle feels a constant force: F = qE = 1 × 10−6 C (2 000 N C) −ˆj = 2 × 10−3 N −ˆj

( ) ( )∑ ( ) ( )a =
F 2 × 10−3 kg ⋅ m s2 − ˆj − ˆj
and moves with acceleration: = = 1 × 1013 m s2
m 2 × 10−16 kg

Note that the gravitational force is on the order of a trillion times smaller than the electrical force
exerted on the particle. The particle’s x-component of velocity is constant at

)(1.00 × 105 m s cos37° = 7.99 × 104 m s. Thus it moves in a parabola opening downward. The

maximum height it attains above the bottom plate is described by

( )vy2f = vy2i + 2ay yf − yi : ( ) ( )( )0 = 6.02 × 104 m s 2 − 2 × 1013 m s2 yf − 0

yf = 1.81 × 10−4 m

Since this is less than 10 mm, the particle does not strike the top plate, but moves in a symmetric
parabola and strikes the bottom plate after a time given by

yf = yi + vyit + 1 ayt 2 ( ) ( )0 = 0 + 6.02 × 104 m s t + 1 −1 × 1013 m s2 t 2
2 2

since t > 0, t = 1.20 × 10−8 s

The particle’s range is ( )( )x f = xi + vxt = 0 + 7.99 × 104 m s 1.20 × 10−8 s = 9.61 × 10−4 m.

In sum,

The particle strikes the negative plate after moving in a parabola with a height of 0.181 mm
and a width of 0.961 mm.

P23.39 The required electric field will be in the direction of motion .

Work done = ∆K −Fd = − 1 mvi2 (since the final velocity = 0)
so, 2
which becomes
eEd = K

and E = K
ed

16 Chapter 23

P23.40 vi = 9.55 × 103 m s

( ( ) )(a)
ay = eE = 1.60 × 10−19 (720) = 6.90 × 1010 m s2
m
1.67 × 10−27

From the large magnitude of this vertical acceleration, FIG. P23.40
we can note that the gravitational force on the particle is
negligible by comparison to the electrical force.

R = vi2 sin 2θ = 1.27 × 10−3 m so that
ay

( )9.55 × 103 2 sin 2θ = 1.27 × 10−3

6.90 × 1010

sin 2θ = 0.961 θ = 36.9° 90.0° − θ = 53.1°

(b) t = R = R If θ = 36.9°, t = 167 ns . If θ = 53.1° , t = 221 ns .
vix vi cosθ

P23.41 (a) t = x = 0.050 0 = 1.11 × 10−7 s= 111 ns
(b) vx 4.50 × 105

( ( )( ) )ay
= qE = 1.602 × 10−19 9.60 × 103 = 9.21 × 1011 m s2
m 1.67 × 10−27

1 ( )( )yf1
yf − yi = vyit + 2 ay t 2 : = 2 9.21 × 1011 1.11 × 10−7 2 = 5.68 × 10−3 m = 5.68 mm

(c) vx = 4.50 × 105 m s ( )( )vyf = vyi + ayt = 9.21 × 1011 1.11 × 10−7 = 1.02 × 105 m s

Additional Problems

P23.42 The two given charges exert equal-size forces of attraction d x=0 15 cm
on each other. If a third charge, positive or negative, were –
placed between them they could not be in equilibrium. If the q +x
third charge were at a point x > 15 cm, it would exert a –12 m C 45 m C
stronger force on the 45 m C than on the −12 m C, and could
not produce equilibrium for both. Thus the third charge must FIG. P23.42
be at x = −d < 0. Its equilibrium requires

ke q (12 µ C) = keq (45 µ C) ⎛ 15 cm + d⎞2 = 45 = 3.75
⎝ d ⎠ 12
d2 (15 cm + d )2

15 cm + d = 1.94d d = 16.0 cm

The third charge is at x = −16.0 cm . The equilibrium of the −12 m C requires

keq (12 µ C) = ke (45 µ C)12 µC q = 51.3 µ C

(16.0 cm)2 (15 cm)2

All six individual forces are now equal in magnitude, so we have equilibrium as required, and this
is the only solution.

Electric Fields 17

( )P23.43 C)
The proton moves with acceleration ap = qE = 1.60 × 10−19 C (640 N = 6.13 × 1010 m s2
m
1.673 × 10−27 kg

( )ae C)
while the e− has acceleration = 1.60 × 10−19 C (640 N = 1.12 × 1014 m s2 = 1 836ap

9.110 × 10−31 kg

(a) We want to find the distance traveled by the proton (i.e., d = 1 a t 2 ), knowing:
2
p

4.00 cm = 1 apt 2 + 1 aet 2 = 1 837 ⎛ 1 a t 2 ⎞
2 2 ⎝ 2 ⎠
p

Thus, d = 1 apt 2 = 4.00 cm = 21.8 µm
2 1 837

(b) The distance from the positive plate to where the meeting occurs equals the distance the

sodium ion travels (i.e., dNa = 1 aNa t 2 ). This is found from:
2

4.00 cm = 1 aNat 2 + 1 aClt 2: 4.00 cm = 1 ⎛ eE ⎞ t2 + 1 ⎛ eE ⎞ t2
2 2 2 ⎝ 22.99 u⎠ 2 ⎝ 35.45 u⎠

)(4.00 1 1 t2 1.65 ⎝⎛ 1 2 ⎞
This may be written as cm = 2 aNat 2 + 2 0.649aNa = 2 aNa t ⎠
so
dNa = 1 aNat 2 = 4.00 cm = 2.43 cm
2 1.65

P23.44 (a) The field, E1, due to the 4.00 × 10−9 C charge is in the
−x direction.

( )( )E1
= keq rˆ = 8.99 × 109 N ⋅ m2 C2 −4.00 × 10−9 C ˆi
r2
(2.50 m)2

= −5.75ˆi N C FIG. P23.44 (a)

Likewise, E2 and E3 , due to the 5.00 × 10−9 C charge and the 3.00 × 10−9 C charge, are

( )( )E2
= keq rˆ = 8.99 × 109 N ⋅ m2 C2 5.00 × 10−9 C ˆi = 11.2 N C ˆi
r2
(2.00 m)2

( )( )E3 =
8.99 × 109 N ⋅ m2 C2 3.00 × 10−9 C ˆi = 18.7 N C ˆi

(1.20 m)2

ER = E1 + E2 + E3 = 24.2 N C in +x direction.

continued on next page

18 Chapter 23

( )(b) keq
E1 = r2 rˆ = (−8.46 N C) 0.243ˆi + 0.970ˆj

( )E2 keq C) + ˆj
= r2 rˆ = (11.2 N

( )E3 keq
= r2 rˆ = (5.81 N C) −0.371ˆi + 0.928ˆj

Ex = E1x + E3x = −4.21ˆi N C Ey = E1y + E2y + E3y = 8.43ˆj N C

ER = 9.42 N C θ = 63.4º above − x axis FIG. P23.44 (b)

P23.45 (a) Let us sum force components to find

∑ ∑Fx = qEx − T sinθ = 0 , and Fy = qEy + T cosθ − mg = 0

Combining these two equations, we get

( ) ( )q =
mg = 1.00 × 10−3 (9.80) = 1.09 × 10−8 C
Ex cotθ + Ey (3.00 cot 37.0° + 5.00) × 105

= 10.9 nC

(b) From the two equations for ∑ Fx and ∑ Fy we also find FIG. P23.45

T = qEx = 5.44 × 10−3 N = 5.44 mN
sin 37.0°

P23.46 This is the general version of the preceding problem. The known quantities are A, B, m, g, and θ.
The unknowns are q and T.

The approach to this problem should be the same as for the last problem, but without numbers to
substitute for the variables. Likewise, we can use the free body diagram given in the solution to
problem 45.

Again, Newton’s second law: ∑ Fx = −T sinθ + qA = 0 (1)
and ∑ Fy = +T cosθ + qB − mg = 0 (2)
(a) Substituting T = qA into Eq. (2),
qA cosθ + qB = mg
sin θ sin θ

Isolating q on the left, q = ( A mg + B)
cot θ

(b) Substituting this value into Eq. (1), T = mgA

(A cosθ + B sinθ )

If we had solved this general problem first, we would only need to substitute the appropriate
values in the equations for q and T to find the numerical results needed for problem 45. If you
find this problem more difficult than problem 45, the little list at the first step is useful. It shows
what symbols to think of as known data, and what to consider unknown. The list is a guide for
deciding what to solve for in the analysis step, and for recognizing when we have an answer.

Electric Fields 19

P23.47 F = keq1q2 : tanθ = 15.0
r2 60.0

θ = 14.0°

) )( (8.99 × 109 10.0 × 10−6 2 FIG. P23.47

F1 = (0.150)2 = 40.0 N

) )( (8.99 × 109 10.0 × 10−6 2

F3 = (0.600)2 = 2.50 N

) )( (8.99 × 109 10.0 × 10−6 2

F2 = (0.619)2 = 2.35 N

Fx = −F3 − F2 cos14.0° = −2.50 − 2.35cos14.0° = −4.78 N
Fy = −F1 − F2 sin14.0° = −40.0 − 2.35sin14.0° = −40.6 N

Fnet = Fx2 + Fy2 = (4.78)2 + (40.6)2 = 40.9 N

tanφ = Fy = −40.6
Fx −4.78

φ = 263°

P23.48 From Figure (a) we have d cos30.0° = 15.0 cm

or d = 15.0 cm
From Figure (b) we have cos 30.0°

or θ = sin−1 ⎛ d ⎞
⎝ 50.0 cm ⎠

θ = sin−1 ⎛ 50.0 15.0 cm ⎞ = 20.3° Figure (a)
⎝⎜
cm (cos30.0°)⎠⎟ Figure (b)

Fq = tanθ Figure (c)
mg FIG. P23.48

Fq = mg tan 20.3° (1)

From Figure (c) we have Fq = 2F cos30.0°

Fq = 2 ⎡ ( keq2 )2 ⎤ cos 30.0° (2)
⎢⎣ 0.300 m ⎥⎦

Combining equations (1) and (2),

2 ⎡ ( keq2 )2 ⎤ cos 30.0° = mg tan 20.3°
⎢⎣ 0.300 m ⎦⎥

q2 = mg(0.300 m)2 tan 20.3°

2ke cos 30.0°

( ( )( ) )q2 = (0.300 m)2 tan 20.3°
2.00 × 10−3 kg 9.80 m s2
2 8.99 × 109 N ⋅ m2 C2 cos 30.0°

q = 4.20 × 10−14 C2 = 2.05 × 10−7 C = 0.205 µC

20 Chapter 23

Q
P23.49 Charge resides on each of the blocks, which repel as point charges:

2

F = ke (Q 2)(Q 2) = k(L − Li )

L2

Solving for Q, Q = 2L k (L − Li )

ke

P23.50 If we place one more charge q at the 29th vertex, the total force on the central charge will add up

to zero: F28 charges + keqQ away from vertex 29 =0 F =28 charges keqQ toward vertex 29
a2 a2

P23.51 According to the result of an Example in the chapter text, the left-
hand rod creates this field at a distance d from its right-hand end:

E = keQ d)

d (2a + FIG. P23.51

dF = keQQ dx
2a
d (d + 2a)

keQ2 b dx keQ2 ⎛ 1 2a + x⎞b
2a x=b−2a 2a ⎝ 2a x ⎠ b−2a
∫F = x (x + 2a) = − ln

F = + keQ 2 ⎛ − ln 2a + b + ln b⎞ = keQ2 ln b2 = ⎛ keQ2 ⎞ ln ⎛ b2 ⎞
4a2 ⎝ b − 2a⎠ 4a2 ⎝⎜ 4a2 ⎠⎟ ⎝⎜ − 4a2 ⎟⎠
b (b − 2a)(b + 2a) b2

*P23.52 We model the spheres as particles. They have different charges. They exert on each other forces
of equal magnitude. They have equal masses, so their strings make equal angles q with the
vertical. The distance r between them is described by sin q = (r/2)/40 cm,

so r = 80 cm sin q

Let T represent the string tension. We have

ΣFx = 0: keq1q2/r2 = T sinq

ΣFy = 0: mg = T cosq

Divide to eliminate T. keq1q2 = tanθ = r/2
r 2 mg (40 cm)2 − r2/4

Cleared of fractions, keq1q2 (80 cm)2 − r2 = mgr3

8.99 × 109(N ⋅ m2/C2) 300 × 10−9C (200 × 10−9C) (0.8 m)2 − r2 = 2.4 × 10−3 (9.8) N r3

(0.8 m)2 − r 2 = 1901 r 6

We home in on a solution by trying values.

r 0.64 − r 2 − 1901 r 6
0 +0.64
0.5 −29.3
0.2 +0.48
0.3 −0.84
0.24 +0.22
0.27 −0.17
0.258 +0.013
0.259 −0.001

Thus the distance to three digits is 0.259 m .

Electric Fields 21

90.0 º

90.0 º
−90.0 º
[ ]λ0
*P23.53 ∫ ∫Q = λd = cosθ Rdθ = λ0 R sinθ = λ0 R 1− (−1) = 2λ0R

−90.0 º

Q = 12.0 µC = (2λ0 )(0.600) m = 12.0 µC so λ0 = 10.0 µC m

)(dFy µC)
= 1 ⎛ (3.00 µC)(λd )⎞ cosθ = 1 ⎛ (3.00 λ0 cos2 θ Rdθ ⎞ cos q
4π ∈0 ⎝⎜ 4π ∈0 ⎜⎝ R2 ⎟⎠ 1
R2 ⎠⎟ 0
–1
∫ ( ) ( )( )90.0º
0°
Fy = 8.99 × 109 N ⋅ m2 C2

−90.0 º
3.00 × 10−6 C 10.0 × 10−6 Cm cos2 θdθ

(0.600 m) 360°

= 8.99(30.0) π 2 ⎛ 1 1 ⎞ cos2 q
−π 2 ⎝ 2 2 ⎠ 1
0.600 0
( ) ∫Fy
10−3 N + cos 2θ dθ 0° 360°
FIG. P23.53
Fy = (0.450 N)⎛⎝ 1 θ + 1 sin 2θ ⎞ π2 0.707 N downward
2 4 ⎠
=

−π 2

Since the leftward and rightward forces due to the two halves of the semicircle cancel out, Fx = 0.

*P23.54 (a) The two charges create fields of equal magnitude, both with outward components along the
x axis and with upward and downward y components that add to zero. The net field is then

keq x ˆi + keq x ˆi = 2(8.99 × 109 N ⋅ m2 )52 × 10−9 C x ˆi
r2 r r2 r C2 ((0.25 m)2 + x2 )3/2

= 935 N ⋅ m2 x ˆi
C(0.0625 m2 + x2 )3/2

(b) At x = 0.36 m,

E= 935 N ⋅ m2 0.36 m ˆi = 4.00 kN/C ˆi
C (0.0625 m2 + (0.36 m)2 )3/2

(c) We solve 1 000 = 935 x (0.0625 + x2)−3/2 by tabulating values for the field function:

x 935 x (0.0625 + x2)−3/2
00
0.01 597
0.02 1 185
0.1 4 789
0.2 5 698
0.36 4 000
0.9 1 032
1 854
∞0

We see that there are two points where E = 1 000. We home in on them to determine their
coordinates as (to three digits) x = 0.016 8 m and x = 0.916 m.

(d) The table in part (c) shows that the field is nowhere so large as 16 000 N/C.

(e) The field of a single charge in Question 7 takes on all values from zero to infinity, each
at just one point along the positive x axis. The vector sum of the field of two charges,
in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and then decreases
asymptotically to zero. In the question and the problem, the fields at x = 36 cm happen
to take similar values. For large x the field of the two charges in this problem shows
just the same inverse proportionality to x2 as the field in the question, being larger

/by the factor 2(52 nC) (57.7 nC) = 1.80 times.

22 Chapter 23

P23.55 (a) From the 2Q charge we have Fe − T2 sinθ2 = 0 and mg − T2 cosθ2 = 0

Combining these we find Fe = T2 sinθ2 = tan θ 2
mg T2 cosθ2

From the Q charge we have Fe = T1 sinθ1 = 0 and mg − T1 cosθ1 = 0

Combining these we find Fe = T1 sinθ1 = tanθ1 or θ2 = θ1 FIG. P23.55
mg T1 cosθ1

(b) Fe = ke 2QQ = 2keQ2
r2 r2

If we assume q is small then tanθ ≈ r 2

Substitute expressions for Fe and tanθ into either equation found in part (a) and solve for r.

Fe = tanθ, then 2keQ2 ⎛ 1⎞ ≈r and solving for r we find r ≈ ⎛ 4 keQ 2 ⎞1 3
mg r2 ⎝⎜ mg ⎟⎠ 2 ⎝⎜ mg ⎠⎟ .

P23.56 The bowl exerts a normal force on each bead, directed along
the radius line or at 60.0° above the horizontal. Consider the
free-body diagram shown for the bead on the left side of the
bowl:

∑ Fy = n sin 60.0° − mg = 0,

or n = mg n
Also, sin 60.0° Fe 60.0°

∑ Fx = −Fe + n cos 60.0° = 0,

or keq2 = n cos60.0° = mg = mg mg
R2 tan 60.0° 3 FIG. P23.56

Thus, q= ⎛ mg ⎞ 1 2
R ⎜⎝ ke 3 ⎟⎠

P23.57 (a) The total non-contact force on the cork ball is: F = qE + mg = m ⎛ g + qE ⎞ ,
⎝⎜ m ⎠⎟

which is constant and directed downward. Therefore, it behaves like a simple pendulum in
the presence of a modified uniform gravitational field with a period given by:

L = 2π 0.500 m
( )( )T = 2π
g + qE m 9.80 m s2 + ⎣⎡ 2.00 × 10−6 C 1.00 × 105 N C 1.00 × 10−3 kg⎤⎦

= 0.307 s

(b) Yes . Without gravity in part (a), we get T = 2π L
qE m

0.500 m
1.00 × 105 N C
( )( )T = 2π 1.00 × 10−3 kg = 0.314 s (a 2.28% difference).
2.00 × 10−6 C

Electric Fields 23

*P23.58 (a) At A the top charge makes the dominant contribution to the +P
field and the net field is downward. At B the total electric A
field is zero. Between B and C the main change is weakening
of the downward electric field of the top charge, so the net field B
at C is upward. At E the fields of the two bottom charges cancel C
out and the total field is downward. At F the total field is D
downward.
E
(b) The field is zero at B as it changes from downward at A to + F +
upward at C. As a continuous function, the field must pass
through the value zero near D as it changes from upward FIG. P23.58
at C to downward at E and F.

(c) Let y represent the distance from E up to the zero-field point.
The distance from P to E is (32 − 1.52)1/2 cm = 2.60 cm. Then
the requirement that the field be zero is

keq y)2 = (1.5 keq + y2 (1.5 y + y2 + ⎣⎡(1.5 keqy
(2.60 cm − cm)2 cm)2 + y2 ⎤⎦3/2
cm )2

keq y)2 = ⎣⎡(1.5 2keqy y2 ⎦⎤3/2
(2.60 cm − cm)2 +

(1.52 + y2)3/2 − 2 y (2.60 − y)2 = 0

As a check on our algebra, we note that y = (1/3)2.60 cm = 0.866 cm should be a solution,
corresponding to point B. Substituting 0.866 gives 5.20 − 5.20 = 0 as it should. We home in
on the smaller answer:

y (1.52 + y2)3/2 − 2y(2.60 − y)2

0 +3.375
0.3 +0.411
0.4 −0.124
0.37 +0.014
0.373 −0.000 6

To three digits the answer is 0.373 cm .

P23.59 (a) There are 7 terms which contribute: FIG. P23.59

3 are s away (along sides)
3 are 2s away (face diagonals) and sinθ = 1 = cosθ

2
1 is 3s away (body diagonal) and sin φ = 1

3
The component in each direction is the same by symmetry.

( ) ( )Fkeq2⎡⎣⎢1 + 2 1⎤ keq2
= s2 22 + 3 3 ⎥⎦ ˆi + ˆj + kˆ = s2 (1.90) ˆi + ˆj + kˆ

(b) F= Fx2 + Fy2 + Fz2 = 3.29 keq2 away from the origin
s2

24 Chapter 23

P23.60 (a) Zero contribution from the same face due to symmetry, opposite
face contributes

⎛ keq ⎞ ⎛ s⎞2 ⎛ s ⎞ 2
⎝⎜ r2 ⎟⎠ ⎜⎝ 2⎟⎠ ⎜⎝ 2 ⎠⎟
4 sin φ where r= + + s2 = 1.5s = 1.22s

sin φ = s E = 4 keqs = 4 keq = 2.18 keq
r r3 s2 s2
(1.22)3 FIG. P23.60

(b) The direction is the kˆ direction.

P23.61 The field on the axis of the ring is calculated in an Example in the chapter text as

ke xQ
x2 + a2 3 2
( )E = Ex =

The force experienced by a charge −q placed along the axis of the ring is

F = − keQq ⎢ ( )⎡ x ⎤ and when x << a, this becomes F = − ⎛ keQq ⎞ x.
⎢⎣ x2 + a2 ⎥ ⎝ a3 ⎠
3 2 ⎦⎥

This expression for the force is in the form of Hooke’s law, with an effective spring

constant of k= keQq .
a3

Since ω = 2π f = k, we have f = 1 keQq .
m 2π ma3

P23.62 dE = kedq ⎛ −xˆi + 0.150 mˆj ⎞
⎜ ⎟
x2 + (0.150 m )2 ⎝⎜ x2 + (0.150 m)2 ⎠⎟

( )=
keλ −xˆi + 0.150 mˆj dx

⎣⎡x2 + (0.150 m)2 ⎤⎦3 2

( )0.400 m −xˆi + 0.150 mˆj dx FIG. P23.62

∫ ∫ )(E = dE = keλ
all charge
x=0 ⎣⎡x2 + 0.150 m 2 ⎤⎦3 2

⎡ +ˆi 0.400 m (0.150 m) ˆjx ⎤0.400 m
⎢ 0 m)2 x2 + (0.150 ⎥
E = keλ ⎢ + (0.150 m )2 + m )2 ⎥

⎢⎣ x2 (0.150 0 ⎥⎦

( )( )E = 8.99 × 109 N ⋅ m2 C2 35.0 × 10−9 C m ⎣⎡ˆi (2.34 − 6.67) m−1 + ˆj(6.24 − 0) m−1 ⎦⎤

( ) ( )E = −1.36ˆi + 1.96ˆj × 103 N C = −1.36ˆi + 1.96ˆj kN C

Electric Fields 25

P23.63 The electrostatic forces exerted on the two charges result in a net torque
τ = −2Fa sinθ = −2Eqa sinθ .

For small θ, sinθ ≈ θ and using p = 2qa, we have τ = −Epθ

The torque produces an angular acceleration given by τ = Iα = I d 2θ
Combining these two expressions for torque, we have dt 2
FIG. P23.63
d 2θ
dt 2 + ⎛ Ep ⎞ θ = 0
⎝⎜ I ⎠⎟

This equation can be written in the form d 2θ = −ω 2θ which is the standard
equation characterizing simple harmonic motion, with dt 2
ω 2 = Ep
I

Then the frequency of oscillation is f = w /2p, or f = 1 pE = 1 2qaE
2π I 2π I

P23.64 keq∑ ( ) ( ) ( )E =rˆ=keq−ˆi + ( keq −ˆi + keq −ˆi +…= − ke qˆi ⎛⎝ 1 + 1 + 1 +…⎠⎞ = − π 2keq ˆi
P23.65 r2 a2 a2 22 32 6a2
2a)2 (3a)2

∫ ∫ ( ) ∫ ( )E =⎡ −ˆi ⎤ ∞⎞
dE = ∞ ⎢ keλ0 x0dx ⎥ = ∞ x −3 dx = −keλ0 x0ˆi ⎛ − 1 x0 ⎠⎟ = keλ0 −ˆi
x0 ⎢⎣ x3 ⎦⎥ ⎝⎜ 2x2 2x0
−keλ0 x0ˆi

x0

ANSWERS TO EVEN PROBLEMS

P23.2 (a) 2.62 ϫ 1024 (b) 2.38 electrons for every 109 present

P23.4 1.57 µN to the left

P23.6 2.51 ϫ 10−9

P23.8 x = 0.634d. The equilibrium is stable if the third bead has positive charge.

P23.10 (a) period = π md 3 (b) 4a keqQ
P23.12 where m is the mass of the object with charge −Q md 3
2 keqQ

1.82 m to the left of the negative charge

P23.14 514 kN
P23.16 (a) 12.9ˆj kN C (b) −38.6ˆj mN

P23.18 See the solution.
P23.20
P23.22 ( )keλ0 −ˆi

x0
See the solution.

P23.24 (a) 383 MN/C away (b) 324 MN/C away (c) 80.7 MN/C away (d) 6.68 MN/C away

P23.26 See the solution.

26 Chapter 23

keQˆi ⎡ 2keQˆi
h ⎣⎢ R2h
( ) ( )(a) −1 2 ⎤ ( ) ( )(b)⎡⎢⎣h + 12 − 12⎤
P23.28 d2 + R2 −1 2 − (d + h)2 + R2 ⎦⎥ d2 + R2 (d + h)2 + R2 ⎦⎥
P23.30
(a) 200 pC (b) 141 pC (c) 58.9 pC

P23.32 See the solution.
P23.34
P23.36 (a) − 1 (b) q1 is negative and q2 is positive
3

(a) −57.6ˆi Tm s2 (b) 2.84ˆi Mm s (c) 49.3 ns

P23.38 The particle strikes the negative plate after moving in a parabola 0.181 mm high and 0.961 mm wide.

P23.40 (a) 36.9°, 53.1° (b) 167 ns, 221 ns

P23.42 It is possible in just one way: at x = −16.0 cm place a charge of +51.3 µC.

P23.44 (a) 24.2 N/C at 0° (b) 9.42 N/C at 117°

P23.46 (a) mg (b) mgA
P23.48 A cotθ + B A cosθ + B sinθ

0.205 µC

P23.50 keqQ toward the 29th vertex
P23.52 a2
P23.54
25.9 cm

E= 935 N ⋅ m2 x ˆi (b) 4.00 ˆi kN/C (c) At x = 0.016 8 m and at x = 0.916 m
0.0625 m2 + x2
( )(a) C 3/2

(d) Nowhere is the field so large. (e) The field of a single charge in Question 7 takes on all

values from zero to infinity, each at just one point along the positive x-axis. The vector sum of

the field of two charges, in this problem, is zero at the origin, rises to a maximum at 17.7 cm, and

then decreases asymptotically to zero. In the question and the problem, the fields at x = 36 cm

happen to take similar values. For large x the field of the two charges in this problem shows

just the same inverse proportionality to x2 as the field in the question, being larger by the

factor 2(52 nC)/(57.7 nC) = 1.8 times.

P23.56 ⎛ mg ⎞ 1 2
R ⎝⎜ ke 3 ⎠⎟

P23.58 (a) At A downward. At B zero. At C upward. At E downward. At F downward. (b) See the
solution. (c) 0.373 cm

P23.60 (a) See the solution. (b) kˆ .

( )P23.62 −1.36ˆi + 1.96ˆj kN C

P23.64 − π 2keq ˆi
6a2

24

Gauss’s Law

CHAPTER OUTLINE ANSWERS TO QUESTIONS

24.1 Electric Flux Q24.1 The luminous flux on a given area is less when the sun is
24.2 Gauss’s Law low in the sky, because the angle between the rays of the
24.3 Application of Gauss’s Law to
sun and the local area vector, d A , is greater than zero.
Various Charge Distributions The cosine of this angle is reduced. The decreased flux
24.4 Conductors in Electrostatic results, on the average, in colder weather.

Equilibrium

Q24.2 The surface must enclose a positive total charge.

Q24.3 The net flux through any gaussian surface is zero. We
can argue it two ways. Any surface contains zero charge,
so Gauss’s law says the total flux is zero. The field is
uniform, so the field lines entering one side of the closed
surface come out the other side and the net flux is zero.

*Q24.4 (i) Equal amounts of flux pass through each of the six faces of the cube. Answer (e).

(ii) Move the charge to very close below the center of one face, through which the flux is then
q/2∈0. Answer (c).
(iii) Move the charge onto one of the cube faces. Then the field has no component perpendicular
to this face and the flux is zero. Answer (a).

*Q24.5 (i) Answer (a).
(ii) the flux is zero through the two faces pierced by the filament. Answer (b).

*Q24.6 (i) Answer (a).

(ii) The flux is nonzero through the top and bottom faces, and zero through the other four faces.
Answer (c).

*Q24.7 (i) Both spheres create equal fields at exterior points, like particles at the centers of the spheres.
Answer (c).

(ii) The field within the conductor is zero. The field within the insulator is 4/5 of its surface
value. Answer (f).

Q24.8 Gauss’s law cannot tell the different values of the electric field at different points on the surface.

When E is an unknown number, then we can say ∫ E cosθdA = E ∫ cosθdA. When E (x, y, z)

is an unknown function, then there is no such simplification.

Q24.9 The electric flux through a sphere around a point charge is independent of the size of the sphere.
A sphere of larger radius has a larger area, but a smaller field at its surface, so that the product
of field strength and area is independent of radius. If the surface is not spherical, some parts are
closer to the charge than others. In this case as well, smaller projected areas go with stronger
fields, so that the net flux is unaffected.

27

28 Chapter 24

Q24.10 Inject some charge at arbitrary places within a conducting object. Every bit of the charge repels
every other bit, so each bit runs away as far as it can, stopping only when it reaches the outer
surface of the conductor.

*Q24.11 (a) Let q represent the charge of the insulating sphere. The field at A is (4/5)3q/[4p (4 cm)2∈0].
(b) The field at B is q/[4p (8 cm)2∈0]. The field at C is zero. The field at D is q/[4p (16 cm)2∈0].
The ranking is A > B > D > C.
The flux through the 4-cm sphere is (4/5)3q/∈0. The flux through the 8-cm sphere and
through the 16-cm sphere is q/∈0. The flux through the 12-cm sphere is 0. The ranking is
B = D > A > C.

*Q24.12 The outer wall of the conducting shell will become polarized to cancel out the external field. The
interior field is the same as before. Answer (c).

Q24.13 If the person is uncharged, the electric field inside the sphere is zero. The interior wall of the shell
carries no charge. The person is not harmed by touching this wall. If the person carries a (small)
charge q, the electric field inside the sphere is no longer zero. Charge –q is induced on the inner
wall of the sphere. The person will get a (small) shock when touching the sphere, as all the charge
on his body jumps to the metal.

*Q24.14 (i) The shell becomes polarized. Answer (e).
(ii) The net charge on the shell’s inner and outer surfaces is zero. Answer (a).
(iii) Answer (c).
(iv) Answer (c).
(v) Answer (a).

Q24.15 There is zero force. The huge charged sheet creates a uniform field. The field can polarize the
neutral sheet, creating in effect a film of opposite charge on the near face and a film with an equal
amount of like charge on the far face of the neutral sheet. Since the field is uniform, the films of
charge feel equal-magnitude forces of attraction and repulsion to the charged sheet. The forces
add to zero.

SOLUTIONS TO PROBLEMS

Section 24.1 Electric Flux

P24.1 ΦE = EA cosθ A = π r2 = π (0.200)2 = 0.126 m2

P24.2 5.20 × 105 = E (0.126)cos 0° E = 4.14 × 106 N C = 4.14 MN C
P24.3
) )( (ΦE = EA cosθ = 2.00 × 104 N C 18.0 m2 cos10.0° = 355 kN ⋅ m2 C

( )(a) ΦE = E ⋅ A = aˆi + bˆj ⋅ Aˆi = aA
( )(b) ΦE = aˆi + bˆj ⋅ Aˆj = bA
( )(c) ΦE = aˆi + bˆj ⋅ Akˆ = 0

Gauss’s Law 29

P24.4 (a) A′ = (10.0 cm)(30.0 cm)

A′ = 300 cm2 = 0.030 0 m2
ΦE,A′ = EA′ cosθ

( )ΦE,A′ = 7.80 × 104 (0.030 0)cos180°

ΦE,A′ = −2.34 kN ⋅ m2 C

( )(b) ΦE,A = EA cosθ = 7.80 × 104 (A)cos 60.0° FIG. P24.4

A = ( 30.0 cm)(w) = ( 30.0 cm ) ⎛ 10.0 cm ⎞ = 600 cm2 = 0.060 0 m2
⎝⎜ cos 60.0°⎠⎟

( )ΦE,A = 7.80 × 104 (0.060 0)cos 60.0° = +2.34 kN ⋅ m2 C

(c) The bottom and the two triangular sides all lie parallel to E, so ΦE = 0 for each of these.
Thus,

ΦE,total = −2.34 kN ⋅ m2 C + 2.34 kN ⋅ m2 C + 0 + 0 + 0 = 0

P24.5 ΦE = EA cosθ through the base

ΦE = (52.0)(36.0)cos180° = −1.87 kN ⋅ m2 C FIG. P24.5

Note that the same number of electric field lines go through the base as go
through the pyramid’s surface (not counting the base).

For the slanting surfaces, ΦE = +1.87 kN ⋅ m2 C .

Section 24.2 Gauss’s Law

)(8.90 × 102 =
P24.6 (a) E = keQ ; 8.99 × 109 Q
r2
(0.750)2

But Q is negative since E points inward. Q = −5.57 × 10−8 C = −55.7 nC

(b) The negative charge has a spherically symmetric charge distribution, concentric with
the spherical shell.

P24.7 (a) ΦE = qin = (+5.00 µC − 9.00 µC + 27.0 µC − 84.0 µC) = −6.89 × 106 N ⋅ m2 C2
∈0 8.85 × 10−12 C2 N ⋅ m2

ΦE = −6.89 MN ⋅ m2 C

(b) Since the net electric flux is negative, more lines enter than leave the surface.

30 Chapter 24

P24.8 (a) One-half of the total flux created by the charge q goes through the plane. Thus,

Φ E,plane = 1 Φ E,total = 1⎛ q ⎞ = q
2 2 ⎝⎜ ∈0 ⎟⎠ 2∈0

(b) The square looks like an infinite plane to a charge very close to the surface. Hence,

ΦE,square ≈ ΦE,plane = q
2∈0

(c) The plane and the square look the same to the charge.

P24.9 ΦE = qin
∈0
−2Q + Q −Q
Through S1 ΦE = ∈0 = ∈0

Through S2 ΦE = +Q − Q = 0
∈0

Through S3 ΦE = −2Q +Q −Q = − 2Q
∈0 ∈0

Through S4 ΦE = 0

P24.10 (a) Φ E, shell = qin = 12.0 × 10−6 = 1.36 × 106 N ⋅ m2 C= 1.36 MN ⋅ m2 C
(b) ∈0 8.85 × 10−12

( )ΦE,half shell 1 = 6.78 × 105 N ⋅ m2
= 2 1.36 × 106 N ⋅ m2 C C= 678 kN ⋅ m2 C

(c) No , the same number of field lines will pass through each surface, no matter how the
radius changes.

P24.11 (a) With d very small, all points on the hemisphere are Q
nearly at a distance R from the charge, so the field
δ →0
everywhere on the curved surface is keQ radially
R2

outward (normal to the surface). Therefore, the flux

is this field strength times the area of half a sphere:

∫Φcurved = E ⋅ dA = E Alocal hemisphere

Φ curved = ⎛ ke Q ⎞ ⎛ 1 4π R 2 ⎞ = 1 Q ( 2π ) = +Q
⎝ R2 ⎠ ⎝ 2 ⎠ 4π ∈0 2∈0
FIG. P24.11

(b) The closed surface encloses zero charge so Gauss’s law gives

Φcurved + Φflat = 0 or Φflat = −Φcurved = −Q
2∈0

P24.12 Consider as a gaussian surface a box with horizontal area A, lying between 500 and 600 m elevation.

∫ E ⋅ dA = q : (+120 N C) A + (−100 N C) A = ρA(100 m)
∈0
∈0

( )(20 N C) 8.85 × 10−12 C2 N ⋅ m2 = 1.77 × 10−12 C m3

ρ=
100 m

The charge is positive , to produce the net outward flux of electric field.

Gauss’s Law 31

P24.13 The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q , of which
∈0

one-sixth goes through each face:

( )ΦE one face = Q−6q
6∈0

( ).ΦE = Q−6q = (5.00 − 6.00) × 10−6 C ⋅ N ⋅ m2 = −18.8 kN ⋅ m2 C
6∈0
one face 6 × 8.85 × 10−12 C2

P24.14 The total charge is Q − 6 q . The total outward flux from the cube is Q − 6 q , of which
one-sixth goes through each face: ∈0

( )Φ =E one face Q−6 q
6∈0

P24.15 If R ≤ d, the sphere encloses no charge and ΦE = qin = 0 .
∈0

If R > d , the length of line falling within the sphere is 2 R2 − d2

so ΦE = 2λ R2 − d2
∈0

( ) ( )( ) ( )P24.16
Φ E,hole = E ⋅ Ahole = ⎛ keQ ⎞ πr2 ⎛ 8.99 × 109 N ⋅ m2 C2 10.0 × 10−6 C ⎞ 1.00 × 10−3 m 2
⎝ R2 ⎠ = ⎝⎜ ⎠⎟ π
(0.100 m)2

ΦE,hole = 28.2 N ⋅ m2 C

P24.17 ΦE = qin = 170 × 10−6 C = 1.92 × 107 N ⋅ m2/C
∈0 8.85 × 10−12 C2 N ⋅ m2

( )(a) = 1 ΦE 1.92 × 107 N ⋅ m2 C ( )Φ =E one face 3.20 MN ⋅ m2 C
ΦE 6 = 6
one face

(b) ΦE = 19.2 MN ⋅ m2 C

(c) The answer to (a) would change because the flux through each face of the cube would
not be equal with an asymmetric charge distribution. The sides of the cube nearer the
charge would have more flux and the ones further away would have less. The answer
to (b) would remain the same, since the overall flux would remain the same.

32 Chapter 24

Section 24.3 Application of Gauss’s Law to Various Charge Distributions

P24.18 (a) E = keQr = 0
(b) a3
(c)
(d) ) )( (E
= keQr = 8.99 × 109 26.0 × 10−6 (0.100) 365 kN C
a3
(0.400)3 =

( )( )E
= keQ = 8.99 × 109 26.0 × 10−6 = 1.46 MN C
r2
(0.400)2

( )( )E
= keQ = 8.99 × 109 26.0 × 10−6 = 649 kN C
r2
(0.600)2

The direction for each electric field is radially outward .

P24.19 The charge distributed through the nucleus creates a field at the surface equal to that of a point

charge at its center: E = keq .
r2

( )( )8.99 × 109 Nm2 C2 82 × 1.60 × 10−19 C

E = ⎣⎡(208)1 3 1.20 × 10−15 m⎦⎤2

E = 2.33 × 1021 N C away from the nucleus

P24.20 Note that the electric field in each case is directed radially inward, toward the filament.

( )( )(a)
E = 2keλ = 2 8.99 × 109 N ⋅ m2 C2 90.0 × 10−6 C m
r 0.100 m = 16.2 MN C

( )( )(b)
E = 2keλ = 2 8.99 × 109 N ⋅ m2 C2 90.0 × 10−6 C m
r 0.200 m = 8.09 MN C

( )( )(c)
E = 2keλ = 2 8.99 × 109 N ⋅ m2 C2 90.0 × 10−6 C m
r 1.00 m = 1.62 MN C

9.00 × 10−6 C m2
8.85 × 10−12 C2 N ⋅ m2
P24.21 ( )E = σ = 2 = 508 kN C, upward
P24.22 2 ∈0

(a) E = 2keλ 3.60 × 104 = 2(8.99 × 109 )(Q 2.40)
r 0.190

Q = +9.13 × 10−7 C = +913 nC

(b) E = 0

( )Q

A
P24.23 mg = qE = ⎛ σ ⎞ = ⎛ Q A⎞ = 2∈0 mg = 2 8.85 × 10−12 (0.01)(9.8)
q ⎜⎝ 2 ∈0 ⎟⎠ q ⎜⎝ 2 ∈0 ⎠⎟ q
−0.7 × 10−6

= −2.48 µC m2

*P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed
throughout its volume, with density 5.00 µC/m3. Find the magnitude of the electric field
it creates at a point P, 3.00 cm from its axis. As a gaussian surface choose a concentric
cylinder with its curved surface passing through the point P and with length 8.00 cm.

(b) We solve for

E = (0.02 m)2 0.08 m (5 × 10−6 C/m3 ) m = 3.77 kN/C
(8.85 × 10−12 C2 /N ⋅ m2 )2(0.03 m) 0.08

Gauss’s Law 33

P24.25 The volume of the spherical shell is

4π ⎡⎣(0.25 m )3 − (0.20 m)3 ⎦⎤ = 3.19 × 10−2 m3
3

Its charge is

( )( )ρV = −1.33 × 10−6 C m3 3.19 × 10−2 m3 = −4.25 × 10−8 C

The net charge inside a sphere containing the proton’s path as its equator is
−60 × 10−9 C − 4.25 × 10−8 C = −1.02 × 10−7 C

The electric field is radially inward with magnitude

( )ke q

r2
= ∈0 q = 8.99 × 109 Nm2 1.02 × 10−7 C = 1.47 × 104 N C
4π r2
C2 (0.25m)2

For the proton

∑ F = ma eE = mv2
r

( )v 0.25 m ⎞ 1 2
⎛ eEr ⎞ 1 2 ⎛ 1.60 × 10−19 C 1.47 × 104 N C ⎟⎠
⎝ m ⎠ = ⎜⎝ 1.67 × 10−27 kg
= = 5.94 × 105 m s

( )*P24.26 ⎛ 100 cm ⎞ 2
s= 8.60 × 10−6 C cm2 ⎜⎝ m ⎟⎠ = 8.60 × 10−2 C m2

8.60 × 10−2
8.85 × 10−12
( )E = σ = 2 = 4.86 × 109 N C away from the wall
2 ∈0

So long as the distance from the wall is small compared to the width and height of the wall, the
distance does not affect the field.

P24.27 If r is positive, the field must be radially outward. Choose as the

gaussian surface a cylinder of length L and radius r, contained FIG. P24.27
inside the charged rod. Its volume is π r2L and it encloses
charge ρπ r2 L. Because the charge distribution is long, no electric
flux passes through the circular end caps; E ⋅ dA = EdA cos90.0° = 0 .
The curved surface has E ⋅ dA = EdA cos 0°, and E must be the same

strength everywhere over the curved surface.

Gauss’s law, ∫ E ⋅ dA = q , becomes ∫E dA = ρπ r2 L .
∈0 ∈0
Curved

Surface

Now the lateral surface area of the cylinder is 2π rL :

E (2π r) L = ρπ r2L Thus, E = ρr radially away from the cylinder axis .

∈0 2∈0

P24.28 The distance between centers is 2 × 5.90 × 10−15 m. Each produces a field as if it were a point
charge at its center, and each feels a force as if all its charge were a point at its center.

( ) ( ( ))F
= keq1q2 = 8.99 × 109 N ⋅ m2 C2 (46)2 1.60 × 10−19 C 2 = 3.50 × 103 N= 3.50 kN
r2
2 × 5.90 × 10−15 m 2

34 Chapter 24

P24.29 (a) E = 0

( )( )(b)
E = keQ = 8.99 × 109 32.0 × 10−6 = 7.19 MN C E = 7.19 MN C radially outward
r2
(0.200)2

P24.30 Consider two balloons of diameter 0.2 m, each with mass 1 g, hanging apart with
a 0.05 m separation on the ends of strings making angles of 10° with the vertical.

(a) ∑ Fy = T cos10° − mg = 0 ⇒T = mg
cos10°

∑ Fx = T sin10° − Fe = 0 ⇒ Fe = T sin10°, so FIG. P24.30

⎛ mg ⎞
⎝ cos10°⎠
)Fe
= sin10° = mg tan10° = (0.001 kg)(9.8 m s2 tan10°

Fe ≈ 2 × 10−3 N ~10−3 N or 1 mN

(b) Fe = keq2
r2

( )2 × 10−3 N ≈ C2
8.99 × 109 N ⋅ m2 q2

(0.25 m)2

q ≈ 1.2 × 10−7 C ~10−7 C or 100 nC

( )( )(c)
E = keq ≈ 8.99 × 109 N ⋅ m2 C2 1.2 × 10−7 C ≈ 1.7 × 104 N C ~10 kN C
r2
(0.25 m)2

(d) ΦE = q ≈ 1.2 × 10−7 C = 1.4 × 104 N ⋅ m2 C ~10 kN ⋅ m2 C
∈0 8.85 × 10−12 C2 N ⋅ m2

P24.31 (a) ( ) ( )E = 2keλ = 2 8.99 × 109 N ⋅ m2 C2 ⎡⎣ 2.00 × 10−6 C 7.00 m⎦⎤
r 0.100 m
E = 51.4 kN C, radially outward

(b) ΦE = EA cosθ = E (2π r )cos 0°

)ΦE = (5.14 × 104 N C 2π (0.100 m)(0.020 0 m)(1.00) = 646 N ⋅ m2 C

Section 24.4 Conductors in Electrostatic Equilibrium

P24.32 The fields are equal. The equation E = σ conductor suggested in the chapter for the field outside the
∈0

aluminum looks different from the equation E = σ insulator for the field around glass. But its charge
2∈0
Q
will spread out to cover both sides of the aluminum plate, so the density is σ conductor = .

2A

The glass carries charge only on area A, with σ insulator = Q . The two fields are Q , the same
A 2A∈0

in magnitude, and both are perpendicular to the plates, vertically upward if Q is positive.

Gauss’s Law 35

∫P24.33 EdA = E (2π rl) = qin E = qin l = λ
2π ∈0 r 2π∈0 r
∈0
E= 0
(a) r = 3.00 cm

30.0 × 10−9 =

8.85 × 10−12 (0.100)
(b) r = 10.0 cm ( )E = 5 400 N C, outward
2π

30.0 × 10−9

8.85 × 10−12 (1.00)
(c) r = 100 cm ( )E = = 540 N C, outward
2π

*P24.34 (a) All of the charge sits on the surface of the copper sphere at radius 15 cm. The field inside
is zero .

(b) The charged sphere creates field at exterior points as if it were a point charge at the center:

( )( )E =
keq away = 8.99 × 109 N · m2 40 × 10−9 C outward = 1.24 × 104 N C outward
r2
C2 (0.17 m)2

(c) E = ) )( (8.99 × 109 N·m2 40 × 10−9 C outward = 639 N C outward

C2 (0.75 m)2

(d) All three answers would be the same. The solid copper sphere carries charge only on its
outer surface.

P24.35 (a) E= σ ( )( )σ = 8.00 × 104 8.85 × 10−12 = 7.08 × 10−7 C m2
∈0

σ = 708 nC m2 , positive on one face and negative on the other.

(b) σ = Q )(Q = σ A = 7.08 × 10−7 (0.500)2 C
A

Q = 1.77 × 10−7 C = 177 nC , positive on one face and negative on the other.

*P24.36 Let the flat box have face area A perpendicular to its thickness dx. The flux at x = 0.3 m is into
the box −EA = −(6 000 N/C ⋅ m2)(0.3 m)2 A = −(540 N/C) A

The flux out of the box at x = 0.3 m + dx
+EA = −(6 000 N/C ⋅ m2)(0.3 m + dx)2 A = +(540 N/C) A + (3 600 N/C ⋅ m) dx A
(The term in (dx)2 is negligible.)
The charge in the box is rA dx where r is the unknown. Gauss’s law is
−(540 N/C) A + (540 N/C) A + (3 600 N/C ⋅ m) dx A = rA dx/∈0
Then r = (3600 N/C ⋅ m)∈0 = (3600 N/C ⋅ m)(8.85 × 10−12 C2/N ⋅ m2) = 31.9 nC/m3

Q
P24.37 The charge divides equally between the identical spheres, with charge on each. Then they
2
repel like point charges at their centers:

( )F 2
ke (Q 2)(Q 2) keQ2 8.99 × 109 N ⋅ m2 60.0 × 10−6 C
= = = = 2.00 N
(L + R + R)2 4(L + 2R)2 4C2 (2.01 m)2

36 Chapter 24

*P24.38 The surface area is A = 4pa2. The field is then

E = keQ = 4π Q = Q =σ
a2 ∈0 a 2 A ∈0 ∈0

It is not equal to s /2∈0. At a point just outside, the uniformly charged surface looks just like a
uniform flat sheet of charge. The distance to the field point is negligible compared to the radius

of curvature of the surface.

P24.39 (a) Inside surface: consider a cylindrical surface within the metal. Since E inside the

conducting shell is zero, the total charge inside the gaussian surface must be zero, so the

inside charge/length = −λ.

0 = λ + qin so qin = −λ

Outside surface: The total charge on the metal cylinder is 2λ = qin + qout
qout = 2λ + λ so the outside charge/length is 3λ

(b) E = 2ke (3λ ) = 6keλ = 3λ radially outward

r r 2π ∈0 r

P24.40 An approximate sketch is given at the right. Note that the electric field
lines should be perpendicular to the conductor both inside and outside.

FIG. P24.40

P24.41 (a) The charge density on each of the surfaces (upper and lower) of the plate is:

( )σ
= 1 ⎛ q⎞ = 1 4.00 × 10−8 C = 8.00 × 10−8 C m2 = 80.0 nC m2
2 ⎝ A⎠ 2
(0.500 m)2

(b) E = ⎛ σ ⎞ kˆ = ⎛ 8.00 × 10−8 C m2 2 ⎞ kˆ = (9.04 kN C)kˆ
⎜⎝ ∈0 ⎠⎟ ⎝⎜ 8.85 × 10−12 C2 N⋅m ⎟⎠

(c) E = (−9.04 kN C)kˆ

Additional Problems

P24.42 In general, E = ayˆi + bzˆj + cxkˆ z
x=0
y=0 y=h
y
In the xy plane, z = 0 and E = ayˆi + cxkˆ

( )ΦE = ∫ E ⋅ dA = ∫ ayˆi + cxkˆ ⋅ kˆdA x=w dA = hdx
x FIG. P24.42
∫ΦE = w = ch x2 w = chw2
2 x=0 2
ch xdx

x=0

Gauss’s Law 37

P24.43 (a) Uniform E, pointing radially outward, so ΦE = EA. The arc length
is ds = Rdθ , and the circumference is 2π r = 2π R sinθ.

θθ FIG. P24.43

A = ∫ 2π rds = ∫ (2π Rsinθ ) Rdθ = 2π R2 ∫ sinθdθ
00

= 2π R2 ( − cosθ ) θ = 2π R2 (1− cosθ )
0

ΦE =1 Q ⋅ 2π R2 (1 − cosθ ) = Q (1− cosθ ) [independent of R!]
4π ∈0 R2
2 ∈0

(b) For θ = 90.0° (hemisphere): ΦE = Q (1− cos 90°) = Q .
2 ∈0 2 ∈0

(c) For θ = 180° (entire sphere): ΦE = Q (1− cos180°) = Q [Gauss’s Law].
2 ∈0 ∈0

*P24.44 (a) qin = +3 µC − 1 µC = +2.00 µC

(b) The charge distribution is spherically symmetric and qin > 0. Thus, the field is directed
radially outward or to the right at point D.

(c) E = ke qin = 8.99 × 109 2.00 × 10−6 N/C = 702 kN/C
r2 (0.16)2

(d) Since all points within this region are located inside conducting material, E = 0 .

∫(e) ΦE = E ⋅ dA = 0 ⇒ qin = ∈0 ΦE = 0

(f ) qin = +3.00 µC

(g) E = ke qin = 8.99 × 109 3.00 × 10−6 = 4.21 MN/C to the right (radially outward).
r2 (0.08)2

(h) qin = ρV = ⎛ +3 µC⎞ ⎛ 4π 43 ⎞ = +1.54 µC
⎜⎝ ⎠⎟ ⎝ 3 ⎠
4 π 53
3

(i) E = ke qin = 8.99 × 109 1.54 × 10−6 = 8.63 MN/C to the right (radially outward)
r2 (0.04)2

(j) As in part (d), E = 0 for 10 cm < r < 15 cm . a bc r
Thus, for a spherical gaussian surface with E FIG. P24.44(l)
10 cm < r < 15 cm, qin = +3 µC + qinner = 0
where qinner is the charge on the inner surface
of the conducting shell. This yields

qinner = −3.00 µC .

(k) Since the total charge on the conducting
shell is qnet = qouter + qinner = −1 µC, we have

)(qouter = −1 µC − qinner = −1 µC − −3 µC = +2.00 µC

(l) This is shown in the figure to the right.

38 Chapter 24

*P24.45 (a) The field is zero within the metal of the shell. The exterior electric field lines end at equally
spaced points on the outer surface. The charge on the outer surface is distributed uniformly.
Its amount is given by

EA = Q/∈0
Q = −(890 N/C) 4p (0.75 m)2 8.85 × 10−12 C2/N ⋅ m2 = −55.7 nC

(b) and (c) For the net charge of the shell to be zero, the shell must carry +55.7 nC on its inner
surface, induced there by −55.7 nC in the cavity within the shell. The charge in the
cavity could have any distribution and give any corresponding distribution to the
charge on the inner surface of the shell. For example, a large positive charge might
be within the cavity close to its topmost point, and a slightly larger negative charge
near its easternmost point. The inner surface of the shell would then have plenty
of negative charge near the top and even more positive charge centered on the
eastern side.

P24.46 The sphere with large charge creates a strong field to polarize the other sphere. That means it
pushes the excess charge over to the far side, leaving charge of the opposite sign on the near side.
This patch of opposite charge is smaller in amount but located in a stronger external field, so it
can feel a force of attraction that is larger than the repelling force felt by the larger charge in the
weaker field on the other side.

E ⋅ dA = E 4π r2 = qin
∈0
( )∫P24.47 (a)

For r < a, qin = ρ ⎛ 4 π r 3 ⎞
⎝ 3 ⎠

so E = ρr
3 ∈0

For a < r < b and c < r, qin = Q FIG. P24.47
So
E= Q
4π r2 ∈0

For b ≤ r ≤ c, E = 0, since E = 0 inside a conductor.

(b) Let q1 = induced charge on the inner surface of the hollow sphere. Since E = 0 inside the
conductor, the total charge enclosed by a spherical surface of radius b ≤ r ≤ c must be zero.

Therefore, q1 + Q = 0 and σ1 = q1 = −Q
4π b 2 4π b 2

Let q2 = induced charge on the outside surface of the hollow sphere. Since the hollow
sphere is uncharged, we require

q1 + q2 = 0 and σ2 = q1 = Q
4π c2 4π c2

Gauss’s Law 39

P24.48 First, consider the field at distance r < R from the center of a uniform sphere of positive charge

(Q = +e) with radius R.

( )4πr2 qin ρV ⎛ +e ⎞ 4 π r 3 ⎛ e ⎞
∈0 ∈0 ⎝⎜ 3 ⎜⎝ ∈0 ⎟⎠
E= = = ⎠⎟ so E = r directed outward
4 π R3 ∈0 4π R3
3

(a) The force exerted on a point charge q = −e located at distance r from the center is then

F = qE = −e ⎛ 4π e R3 ⎞ r = − ⎛ 4π e2 R3 ⎞ r = −Kr
⎝⎜ ∈0 ⎟⎠ ⎜⎝ ∈0 ⎠⎟

(b) K = e2 = kee2
4π ∈0 R3 R3

(c) Fr = mear = − ⎛ kee2 ⎞ so ar = − ⎛ kee2 ⎞ r = −ω 2r
⎜ R3 ⎟ r, ⎜⎝ me R3 ⎠⎟
⎝ ⎠

Thus, the motion is simple harmonic with frequency f=ω = 1 kee2
2π 2π me R3

( ( )( ) )(d)
f = 2.47 × 1015 Hz = 1 8.99 × 109 N ⋅ m2 C2 1.60 × 10−19 C 2
2π 9.11 × 10−31 kg R3

which yields R3 = 1.05 × 10−30 m3, or R = 1.02 × 10−10 m = 102 pm

P24.49 The vertical velocity component of the moving charge y vy
increases according to v vx x

m dvy = Fy m dvy dx = qEy 0
dt dx dt qd

Now dx = vx has the nearly constant value v. So Q
dt
FIG. P24.49

dvy = q Ey dx ∫ ∫vy = vy dvy = q ϱ
mv 0 mv
Ey dx

−ϱ

The radially outward component of the electric field varies along the x axis, but is described by

∫ ∫ )(ϱ ϱ 2π d dx = Q
∈0
EydA = Ey

−ϱ −ϱ

∫So ϱ = Q and vy = qQ . The angle of deflection is described by
2π d ∈0 mv2π d ∈0
Ey dx

−ϱ

tanθ = vy = qQ θ = tan−1 2π qQ
v 2π ∈0 dmv2 ∈0 dmv2

40 Chapter 24

P24.50 Consider the field due to a single sheet and let E+ and E− represent
the fields due to the positive and negative sheets. The field at any

distance from each sheet has a magnitude given by the textbook equation

E+ = E− = σ
2 ∈0

(a) To the left of the positive sheet, E+ is directed toward the
left and E− toward the right and the net field over this

region is E = 0 .

(b) In the region between the sheets, E+ and E− are both directed FIG. P24.50
toward the right and the net field is

E = σ to the right
∈0

(c) To the right of the negative sheet, E+ and E− are again
oppositely directed and E = 0 .

P24.51 The magnitude of the field due to each sheet given by
Equation 24.8 is

E = σ directed perpendicular to the sheet
2 ∈0

(a) In the region to the left of the pair of sheets, both fields are
directed toward the left and the net field is

E = σ to the left FIG. P24.51
∈0

(b) In the region between the sheets, the fields due to the individual sheets are oppositely
directed and the net field is

E= 0

(c) In the region to the right of the pair of sheets, both fields are directed toward the right and
the net field is

E = σ to the right
∈0

P24.52 The resultant field within the cavity is the superposition of

two fields, one E+ due to a uniform sphere of positive charge
of radius 2a, and the other E− due to a sphere of negative
charge of radius a centered within the cavity.

4⎛ πr3ρ⎞ = 4π r2E+ so E+ = ρr rˆ = ρr
3 ⎜⎝ ∈0 ⎟⎠ 3∈0 3∈0

4 ⎛ π r13ρ ⎞ )(E− ρ r1 −ρ
− 3 ⎜⎝ ∈0 ⎠⎟ = 4π r12 E− so = 3∈0 −rˆ1 = 3∈0 r1

FIG. P24.52

continued on next page

Gauss’s Law 41

Since r = a + r1, E− = −ρ(r − a)

Thus, 3∈0
and
E = E+ + E− = ρr − ρr + ρa = ρa = 0ˆi + ρ a ˆj
3∈0 3∈0 3∈0 3∈0 3∈0

Ex = 0

Ey = ρa at all points within the cavity
3∈0

P24.53 Consider the charge distribution to be an unbroken charged spherical shell with uniform charge
density s and a circular disk with charge per area −σ . The total field is that due to the whole

sphere, Q = 4π R2σ =σ outward plus the field of the disk − σ = σ radially
4π ∈0 R2 4π ∈0 R2 ∈0 2 ∈0 2 ∈0

inward. The total field is σ − σ = σ outward .
∈0 2 ∈0 2 ∈0

P24.54 The electric field throughout the region is directed along x; therefore,
E will be perpendicular to dA over the four faces of the surface which are
perpendicular to the yz plane, and E will be parallel to dA over the two faces FIG. P24.54
which are parallel to the yz plane. Therefore,

( ) ( ) ( ) ( )ΦE = − Ex x=a A + Ex x=a+c A = − 3 + 2a2 ab + 3 + 2(a + c)2 ab

= 2abc(2a + c)

Substituting the given values for a, b, and c, we find ΦE = 0.269 N ⋅ m2 C .
Q = ∈0 ΦE = 2.38 × 10−12 C = 2.38 pC

P24.55 ( )∫ E ⋅ dA = E 4π r2 = qin
∈0

(a) For r > R, ∫ ( )R 4π r2 dr = 4π AR5
5
qin = Ar2

0

and E= AR5
5 ∈0 r2

(b) For r < R, ∫ ( )r 4π r2 dr = 4π Ar5
5
qin = Ar2

0

and E = Ar3
5 ∈0

42 Chapter 24

Q
P24.56 The total flux through a surface enclosing the charge Q is .
∈0
The flux through the disk is

∫Φdisk = E ⋅ dA

where the integration covers the area of the disk. We must evaluate this

integral and set it equal to 1 Q to find how b and R are related. In the figure,
4

∈0 FIG. P24.56

take d A to be the area of an annular ring of radius s and width ds. The flux

through d A is E ⋅ dA = EdA cosθ = E (2π sds)cosθ.

The magnitude of the electric field has the same value at all points within the annular ring,

E = 1 Q = 1 Q cosθ = b = b .
4π ∈0 r2 4π ∈0 s2 + b2 r s2 + b2
( )and 12

Integrate from s = 0 to s = R to get the flux through the entire disk.

Qb R sds Qb R Q ⎡ b ⎤
2 ∈0 0 s2 + b2 2 ∈0 2 ∈0 ⎢1 − + b2
∫ ( ) ( ) ( )ΦE, disk= 32 = ⎣⎡− s2 + b2 12⎤ = ⎢⎣ R2 1 2 ⎥
⎦ ⎦⎥

0

The flux through the disk equals ( )Q b 1
4 ∈0 provided that R2 + b2 12 = 2 .

This is satisfied if R = 3b .

E ⋅ dA = qin = 1 r a 4π r2dr
∈0 ∈0 0 r
∫ ∫P24.57

∫E4π r2 = 4π a r rdr = 4π a r2
∈0 0 ∈0 2

E = a = constant magnitude
2 ∈0

(The direction is radially outward from center for positive a; radially inward for negative a.)

P24.58 In this case the charge density is not uniform, and Gauss’s law is written as ∫ E ⋅ dA = 1 ∫ ρdV .
∈0
We use a gaussian surface which is a cylinder of radius r, length , and is coaxial with the charge

distribution.

E (2π r ) = ρ0 r ⎛ r⎞
0 ⎝ b⎠
∈0
∫(a) a − dV .
When r < R, this becomes The element of volume is a

cylindrical shell of radius r, length , and thickness dr so that dV = 2π r dr.

E (2π r ) = ⎛ 2π r2 ρ0 ⎞ ⎛ a − r⎞ so inside the cylinder, E = ρ0r ⎛ a − 2r ⎞
⎝⎜ ∈0 ⎠⎟ ⎝⎜ 2 3b ⎟⎠ 2 ∈0 ⎜⎝ 3b ⎠⎟

(b) When r > R , Gauss’s law becomes

) = ρ0 R ⎛ r⎞ dr ) ρ0 R2 ⎛ 2R⎞
0 ⎜⎝ b ⎠⎟ 2∈0 r ⎝ 3b ⎠
∈0
∫E (2πr a − ( 2π r or outside the cylinder, E = a −

Gauss’s Law 43

P24.59 (a) Consider a cylindrical shaped gaussian surface y
perpendicular to the yz plane with one end in the yz plane
and the other end containing the point x:

∫Use Gauss’s law: E ⋅ dA = qin gaussian
surface
∈0
x
By symmetry, the electric field is zero in the yz plane z x
and is perpendicular to d A over the wall of the gaussian
cylinder. Therefore, the only contribution to the integral FIG. P24.59
is over the end cap containing the point x:

∫ E ⋅ dA = qin or EA = ρ (Ax)

∈0 ∈0

so that at distance x from the mid-line of the slab, E = ρx .
∈0

(b) a = F = (−e) E = − ⎛ ρe ⎞ x
me ⎝⎜ me ∈0 ⎠⎟
me

The acceleration of the electron is of the form a = −ω 2 x with ω = ρe
me ∈0

Thus, the motion is simple harmonic with frequency f = ω = 1 ρe
2π 2π me ∈0

P24.60 Consider the gaussian surface described in the solution to problem 59.

(a) For x > d , dq = ρ dV = ρAdx = CAx2dx
2

∫ E ⋅ dA = 1 ∫ dq
∈0

∫EA = CA d2 x 2 dx = 1 ⎛ CA ⎞ ⎛ d3 ⎞
∈0 0 3 ⎜⎝ ∈0 ⎟⎠ ⎝⎜ 8 ⎠⎟

E = Cd 3 or E = Cd 3 ˆi for x > d ; E = − Cd 3 ˆi for x < − d
24∈0 24 ∈0 2 24 ∈0 2

(b) For − d < x < d dq = CA x CAx 3
22 ∫ ∫ ∫E ⋅ dA = 1
∈0 ∈0 0 x2dx = 3∈0
E = Cx3 ˆi for x > 0;
3∈0 E = − Cx3 ˆi for x < 0
3∈0

44 Chapter 24

P24.61 (a) A point mass m creates a gravitational acceleration g = − Gm rˆ at a distance r
The flux of this field through a sphere is r2

( )∫ g Gm
⋅ dA = − r2 4π r2 = −4π Gm

Since the r has divided out, we can visualize the field as unbroken field lines. The same
flux would go through any other closed surface around the mass. If there are several or no
masses inside a closed surface, each creates field to make its own contribution to the net
flux according to

∫ g ⋅ dA = −4π Gmin

(b) Take a spherical gaussian surface of radius r. The field is inward so

∫ g ⋅ dA = g4π r2 cos180° = −g4π r2

and −4π Gmin = −4π G 4 πr3ρ
3

Then, −g4π r2 = −4π G 4 π r3ρ and g = 4 π rρG
33

Or, since ρ= ME , g= ME Gr or g= ME Gr inward
RE3 RE3
4 π RE3
3

P24.62 The charge density is determined by Q = 4 π a3ρ ρ = 3Q
3 4π a3

(a) The flux is that created by the enclosed charge within radius r:

ΦE = qin = 4π r3ρ = 4π r3 3Q = Qr 3
∈0 3∈0 3∈0 4π a3 ∈0a3

(b) ΦE = Q . Note that the answers to parts (a) and (b) agree at r = a .
∈0

(c) ΦE

Q
∈0

0 r
0a

FIG. P24.62(c)

Gauss’s Law 45

P24.63 ( )∫ E ⋅ dA = E 4π r2 = qin
∈0

( )(a) (0.100 m)2 Q (a < r < b)
−3.60 × 103 N C 4π = 8.85 × 10−12 C2 N ⋅ m2

Q = −4.00 × 10−9 C = −4.00 nC

(b) We take Q′ to be the net charge on the hollow sphere. Outside c,

( )+2.00 × 102N C 4π (0.500 m )2 = 8.85 Q +Q′ N ⋅ m2 (r > c)
× 10−12 C2

Q + Q′ = +5.56 × 10−9 C , so Q′ = +9.56 × 10−9 C = +9.56 nC

(c) For b < r < c : E = 0 and qin = Q + Q1 = 0 where Q1 is the total charge on the inner

surface of the hollow sphere. Thus, Q1 = −Q = +4.00 nC .
Then, if Q2 is the total charge on the outer surface of the hollow sphere,
Q2 = Q′ − Q1 = 9.56 nC − 4.0 nC = +5.56 nC .

P24.64 The field direction is radially outward perpendicular to the axis. The field strength depends on
r but not on the other cylindrical coordinates q or z. Choose a gaussian cylinder of radius r and
length L. If r < a,

ΦE = qin and E (2π rL) = λL
∈0
∈0

E= λ or E = λ rˆ (r < a)
2π r∈0 2π r∈0

If a < r < b, ( )E (2π rL) = λL + ρπ r2 − a2 L
∈0

)(λ + ρπ r2 − a2 (a < r < b)

E = rˆ
2π r∈0

If r > b, ( )E (2π rL) = λL + ρπ b2 − a2 L
∈0

)(λ + ρπ b2 − a2 (r > b)

E = rˆ
2π r∈0

*P24.65 (a) Consider a gaussian surface in the shape of a rectangular
box with two faces perpendicular to the direction of the
field. It encloses some charge, so the net flux out of the ++
box is nonzero. The field must be stronger on one side +
than on the other. The field cannot be uniform in magnitude.

(b) Now the volume contains no charge. The net flux out of the + +
box is zero. The flux entering is equal to the flux exiting. + +
The field magnitude is uniform at points along one field
line. The field magnitude can vary over the faces of the
box perpendicular to the field.

FIG. P24.65(a)

46 Chapter 24

ANSWERS TO EVEN PROBLEMS

P24.2 355 kN ⋅ m2 C

P24.4 (a) −2.34 kN ⋅ m2 C (b) +2.34 kN ⋅ m2 C (c) 0

P24.6 (a) −55.7 nC (b) The negative charge has a spherically symmetric distribution concentric with
P24.8 the shell.

q q (c) Plane and square both subtend a solid angle of a hemisphere at the
(a) (b)

2∈0 2∈0
charge.

P24.10 (a) 1.36 MN ⋅ m2 C (b) 678 kN ⋅ m2 C (c) No; see the solution.

P24.12 1.77 pC m3 positive
P24.14
P24.16 Q−6 q
6∈0

28.2 N ⋅ m2 C

P24.18 (a) 0 (b) 365 kN C (c) 1.46 MN C (d) 649 kN C

P24.20 (a) 16.2 MN C toward the filament (b) 8.09 MN C toward the filament (c) 1.62 MN C
toward the filament

P24.22 (a) 913 nC (b) 0

P24.24 (a) A long cylindrical plastic rod 2.00 cm in radius carries charge uniformly distributed through-
out its volume, with density 5.00 m C/m3. Find the magnitude of the electric field it creates at a
point P, 3.00 cm from its axis. As a gaussian surface choose a concentric cylinder with its curved
surface passing through the point P and with length 8.00 cm. (b) 3.77 kN/C

P24.26 4.86 GN C away from the wall. It is constant close to the wall.

P24.28 3.50 kN

P24.30 (a) ~1 mN (b) ~100 nC (c) ~10 kN/C (d) ~10 kN · m2/C

P24.32 E = Q/2∈0 A vertically upward in each case if Q > 0

P24.34 (a) 0 (b) 12.4 kN C radially outward (c) 639 N C radially outward (d) No answer changes.
The solid copper sphere carries charge only on its outer surface.

P24.36 31.9 nC/m3

P24.38 The electric field just outside the surface is given by s /∈0. At this point the uniformly charged
surface of the sphere looks just like a uniform flat sheet of charge.

P24.40 See the solution.
P24.42 chw2

2

Gauss’s Law 47

P24.44 (a) 2.00 m C (b) to the right (c) 702 kN/C (d) 0 (e) 0 (f) 3.00 m C (g) 4.21 MN/C
radially outward (h) 1.54 m C (i) 8.63 MN/C radially outward (j) −3.00 m C
(k) 2.00 m C (l) See the solution.

P24.46 See the solution.

P24.48 (a, b) See the solution. (c) 1 kee2 (d) 102 pm
2π me R3

P24.50 σ (c) 0
P24.52 (a) 0 (b) to the right

∈0
See the solution.

P24.54 0.269 N ⋅ m2 C; 2.38 pC

P24.56 See the solution.
P24.58
(a) ρ0r ⎛ a − 2r ⎞ (b) ρ0 R2 ⎛ a − 2R ⎞
2∈0 ⎝⎜ 3b ⎟⎠ 2∈0 r ⎝⎜ 3b ⎟⎠

P24.60 (a) E = Cd3 ˆi for x > d ; E = − Cd3 ˆi for x < − d (b) E = Cx3 ˆi for x > 0;
24∈0 2 24∈0 2 3∈0

E = − Cx3 ˆi for x < 0
3∈0

P24.62 (a) Qr3 (b) Q (c) See the solution
∈0 a3 ∈0

P24.64 For r < a, E = l/2p ∈0r radially outward.
For a < r < b, E = [l + rp(r2−a2)]/2p ∈0r radially outward.
For r > b, E = [l + rp(b2−a2)]/2p ∈0r radially outward.



25

Electric Potential

CHAPTER OUTLINE ANSWERS TO QUESTIONS

25.1 Electric Potential and Potential Q25.1 When one object B with electric charge is immersed
Difference in the electric field of another charge or charges A, the
system possesses electric potential energy. The energy
25.2 Potential Difference in a Uniform can be measured by seeing how much work the field does
Electric Field on the charge B as it moves to a reference location. We
choose not to visualize A’s effect on B as an action-at-a-
25.3 Electric Potential and Potential distance, but as the result of a two-step process: Charge
Energy Due to Point Charges A creates electric potential throughout the surrounding
space. Then the potential acts on B to inject the system
25.4 Obtaining the Value of the with energy.
Electric Field from the Electric
Potential

25.5 Electric Potential Due to
Continuous Charge Distributions

25.6 Electric Potential Due to a
Charged Conductor

25.7 The Millikan Oil Drop Experiment
25.8 Application of Electrostatics

*Q25.2 (i) The particle feels an electric force in the negative x
direction. An outside agent pushes it uphill against this
force, increasing the potential energy. Answer (a).
(ii) The potential decreases in the direction of the electric
field. Answer (c).

*Q25.3 The potential is decreasing toward the bottom of the page, so the electric field is downward.
Answer (f).

*Q25.4 (i) At points off the x axis the electric field has a nonzero y component. At points on the negative
x axis the field is to the right and positive. At points to the right of x = 500 mm the field is to the
left and nonzero. The field is zero at one point between x = 250 mm and x = 500 mm. Answer (b).
(ii) The electric potential is negative at this and at all points. Answer (c). (iii) Answer (d).
(iv) Answer (d).

Q25.5 To move like charges together from an infinite separation, at which the potential energy of the
system of two charges is zero, requires work to be done on the system by an outside agent. Hence
energy is stored, and potential energy is positive. As charges with opposite signs move together
from an infinite separation, energy is released, and the potential energy of the set of charges
becomes negative.

Q25.6 (a) The equipotential surfaces are nesting coaxial cylinders around an infinite line of charge.
(b) The equipotential surfaces are nesting concentric spheres around a uniformly charged sphere.

*Q25.7 Answer (b). The potential could have any value.

*Q25.8 The same charges at the same distance away create the same contribution to the total potential.
Answer (b).

49

50 Chapter 25

*Q25.9 The change in kinetic energy is the negative of the change in electric potential energy, so we

work out −q∆V = −q(Vf − Vi) in each case.

(a) −(−e)(60 V − 40 V) = +20 eV (b) −(−e)(20 V − 40 V) = −20 eV

(c) −(e)(20 V − 40 V) = +20 eV (d) −(e)(10 V − 40 V) = +30 eV

(e) −(−2e)(50 V − 40 V) = +20 eV (f) −(−2e)(60 V − 40 V) = +40 eV

With also (g) 0 and (h) +10 eV, the ranking is f > d > c = e = a > h > g > b.

Q25.10 The main factor is the radius of the dome. One often overlooked aspect is also the humidity of the
air—drier air has a larger dielectric breakdown strength, resulting in a higher attainable electric
potential. If other grounded objects are nearby, the maximum potential might be reduced.

*Q25.11 (i) The two spheres come to the same potential, so q/R is the same for both. With twice the
radius, B has twice the charge. Answer (d).
(ii) All the charge runs away from itself to the outer surface of B. Answer (a).

Q25.12 The grounding wire can be touched equally well to any point on the sphere. Electrons will drain
away into the ground and the sphere will be left positively charged. The ground, wire, and sphere
are all conducting. They together form an equipotential volume at zero volts during the con-
tact. However close the grounding wire is to the negative charge, electrons have no difficulty in
moving within the metal through the grounding wire to ground. The ground can act as an infinite
source or sink of electrons. In this case, it is an electron sink.

SOLUTIONS TO PROBLEMS
Section 25.1 Electric Potential and Potential Difference

P25.1 (a) Energy of the proton-field system is conserved as the proton moves from high to low poten-
tial, which can be defined for this problem as moving from 120 V down to 0 V.

Ki + Ui + ∆Emech = K f + U f 0 + qV + 0 = 1 mv 2 + 0
2 p

( ) ( )1.60 × 10−19 V)⎛⎝ 1J ⎞ 1
C (120 1 V⋅ C⎠ = 2 1.67 × 10−27 kg v 2
p

vp = 1.52 × 105 m s

(b) The electron will gain speed in moving the other way,

from Vi = 0 to Vf = 120 V: Ki + Ui + ∆Emech = K f + U f

0 + 0 + 0 = 1 mve2 + qV
2

( ) ( )0 = 1 (120 C)
2
9.11 × 10−31 kg ve2 + −1.60 × 10−19 C J

ve = 6.49 × 106 m s

( )( )P25.2 ∆V = −14.0 V and Q = −NAe = − 6.02 × 1023 1.60 × 10−19 = −9.63 × 104 C

∆V = W , so W = Q∆V = (−9.63 × 104 C)(−14.0 J C) = 1.35 MJ
Q