23
ELECTRIC POTENTIAL
23.1.
IDENTIFY: Apply Eq. (23.2) to calculate the work. The electric potential energy of a pair of point charges
is given by Eq. (23.9).
SET UP: Let the initial position of q2 be point a and the final position be point b, as shown in Figure 23.1.
ra = 0.150 m
rb = (0.250 m) 2 + (0.250 m) 2
rb = 0.3536 m
Figure 23.1
EXECUTE: Wa →b = U a − U b
Ua =
q1q2
( +2.40 × 10−6 C)(−4.30 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 /C2 )
4π ⑀ 0 ra
0.150 m
1
U a = −0.6184 J
Ub =
q1q2
( +2.40 × 10−6 C)(−4.30 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 /C2 )
4π ⑀ 0 rb
0.3536 m
1
U b = −0.2623 J
Wa →b = U a − U b = −0.6184 J − (−0.2623 J) = −0.356 J
EVALUATE: The attractive force on q2 is toward the origin, so it does negative work on q2 when q2
23.2.
moves to larger r.
IDENTIFY: Apply Wa →b = U a − U b .
SET UP: U a = +5.4 × 10−8 J. Solve for U b .
EXECUTE: Wa →b = −1.9 × 10−8 J = U a − U b . U b = U a − Wa →b = +5.4 × 10−8 J − (−1.9 × 10−8 J) = 7.3 × 10−8 J.
23.3.
EVALUATE: When the electric force does negative work the electrical potential energy increases.
IDENTIFY: The work needed to assemble the nucleus is the sum of the electrical potential energies of the
protons in the nucleus, relative to infinity.
SET UP: The total potential energy is the scalar sum of all the individual potential energies, where each
potential energy is U = (1/4π ⑀ 0 )(qq0 /r ). Each charge is e and the charges are equidistant from each other,
so the total potential energy is U =
1 ⎛ e2 e2 e2 ⎞
3e2
.
⎜⎜ + + ⎟⎟ =
4π ⑀ 0 ⎝ r
r
r ⎠ 4π ⑀0 r
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23-1
23-2
Chapter 23
EXECUTE: Adding the potential energies gives
U=
23.4.
3e2
3(1.60 × 10−19 C) 2 (9.00 × 109 N ⋅ m 2 /C2 )
=
= 3.46 × 10−13 J = 2.16 MeV
4π ⑀0 r
2.00 × 10−15 m
EVALUATE: This is a small amount of energy on a macroscopic scale, but on the scale of atoms, 2 MeV is
quite a lot of energy.
IDENTIFY: The work required is the change in electrical potential energy. The protons gain speed after
being released because their potential energy is converted into kinetic energy.
(a) SET UP: Using the potential energy of a pair of point charges relative to infinity, U = (1/4πε 0 )(qq0 /r ),
we have W = ΔU = U 2 − U1 =
1 ⎛ e2 e2 ⎞
⎜ − ⎟.
4π ⑀0 ⎜⎝ r2 r1 ⎟⎠
EXECUTE: Factoring out the e2 and substituting numbers gives
⎛
⎞
1
1
−14
−
W = (9.00 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2 ⎜⎜
J
−15
−10
⎟⎟ = 7.68 × 10
.
×
.
×
3
00
10
m
2
00
10
m
⎝
⎠
(b) SET UP: The protons have equal momentum, and since they have equal masses, they will have equal
⎛1
⎞
speeds and hence equal kinetic energy. ΔU = K1 + K 2 = 2 K = 2 ⎜ mv 2 ⎟ = mv 2 .
⎝2
⎠
EXECUTE: Solving for v gives v =
23.5.
ΔU
7.68 × 10−14 J
=
= 6.78 × 106 m/s.
m
1.67 × 10−27 kg
EVALUATE: The potential energy may seem small (compared to macroscopic energies), but it is enough
to give each proton a speed of nearly 7 million m/s.
(a) IDENTIFY: Use conservation of energy:
K a + U a + Wother = Kb + U b
U for the pair of point charges is given by Eq. (23.9).
SET UP:
Let point a be where q2 is 0.800 m from
q1 and point b be where q2 is 0.400 m
from q1, as shown in Figure 23.5a.
Figure 23.5a
EXECUTE: Only the electric force does work, so Wother = 0 and U =
1 q1q2
.
4π ⑀ 0 r
K a = 12 mva2 = 12 (1.50 × 10−3 kg)(22.0 m/s) 2 = 0.3630 J
Ua =
1 q1q2
(−2.80 × 10−6 C)( −7.80 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 /C2 )
= +0.2454 J
4π ⑀ 0 ra
0.800 m
Kb = 12 mvb2
Ub =
1 q1q2
(−2.80 × 10−6 C)( −7.80 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 /C2 )
= +0.4907 J
4π ⑀ 0 rb
0.400 m
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Electric Potential
23-3
The conservation of energy equation then gives Kb = K a + (U a − U b )
1 mv 2 = +0.3630 J + (0.2454 J − 0.4907 J) = 0.1177 J
b
2
vb =
2(0.1177 J)
1.50 × 10−3 kg
= 12.5 m/s
EVALUATE: The potential energy increases when the two positively charged spheres get closer together,
so the kinetic energy and speed decrease.
(b) IDENTIFY: Let point c be where q2 has its speed momentarily reduced to zero. Apply conservation of
energy to points a and c: K a + U a + Wother = K c + U c .
SET UP: Points a and c are shown in Figure 23.5b.
EXECUTE: K a = +0.3630 J (from part (a))
U a = +0.2454 J (from part (a))
Figure 23.5b
K c = 0 (at distance of closest approach the speed is zero)
Uc =
1 q1q2
4π ⑀0 rc
1 q1q 2
= +0.3630 J + 0.2454 J = 0.6084 J
4π ⑀0 rc
Thus conservation of energy K a + U a = U c gives
rc =
q1q2
( −2.80 × 10−6 C)(−7.80 × 10−6 C)
= (8.988 × 109 N ⋅ m 2 /C2 )
= 0.323 m.
4π ⑀ 0 0.6084 J
+0.6084 J
1
EVALUATE: U → ∞ as r → 0 so q2 will stop no matter what its initial speed is.
23.6.
IDENTIFY: The total potential energy is the scalar sum of the individual potential energies of each pair of
charges.
qq′
SET UP: For a pair of point charges the electrical potential energy is U = k
. In the O-H-N
r
combination the O− is 0.170 nm from the H + and 0.280 nm from the N − . In the N-H-N combination the
N − is 0.190 nm from the H + and 0.300 nm from the other N − . U is positive for like charges and
negative for unlike charges.
EXECUTE: (a) O-H-N:
O − − H + :U = −(8.99 × 109 N ⋅ m 2 /C2 )
O − − N − : U = (8.99 × 109 N ⋅ m 2 /C2 )
(1.60 × 10−19 C) 2
= −1.35 × 10−18 J.
0.170 × 10−9 m
(1.60 × 10−19 C) 2
0.280 × 10−9 m
= +8.22 × 10−19 J.
N-H-N:
N − − H + : U = −(8.99 × 109 N ⋅ m 2 /C2 )
N − − N − : U = (8.99 × 109 N ⋅ m 2 /C2 )
(1.60 × 10−19 C) 2
0.190 × 10−9 m
(1.60 × 10−19 C) 2
0.300 × 10−9 m
= −1.21 × 10−18 J.
= +7.67 × 10−19 J.
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23-4
Chapter 23
The total potential energy is
U tot = −1.35 × 10−18 J + 8.22 × 10−19 J − 1.21 × 10−18 J + 7.67 × 10−19 J = −9.71 × 10−19 J.
23.7.
(b) In the hydrogen atom the electron is 0.0529 nm from the proton.
(1.60 × 10−19 C)2
U = −(8.99 × 109 N ⋅ m 2 /C 2 )
= −4.35 × 10−18 J.
0.0529 × 10−9 m
EVALUATE: The magnitude of the potential energy in the hydrogen atom is about a factor of 4 larger
than what it is for the adenine-thymine bond.
IDENTIFY: The total potential energy is the scalar sum of the individual potential energies of each pair of
charges.
qq′
SET UP: For a pair of point charges the electrical potential energy is U = k
. In the O-H-O
r
combination the O− is 0.180 nm from the H + and 0.290 nm from the other O−. In the N-H-N
combination the N − is 0.190 nm from the H + and 0.300 nm from the other N − . In the O-H-N
combination the O− is 0.180 nm from the H + and 0.290 nm from the N − . U is positive for like charges
and negative for unlike charges.
EXECUTE: O-H-O O− − H + , U = −1.28 × 10−18 J; O− − O − , U = +7.93 × 10−19 J.
N-H-N N − − H + , U = −1.21 × 10−18 J; N − − N − , U = +7.67 × 10−19 J.
O-H-N O− − H + , U = −1.28 × 10−18 J; O− − N − , U = +7.93 × 10−19 J.
23.8.
The total potential energy is −3.77 × 10−18 J + 2.35 × 10−18 J = −1.42 × 10−18 J.
EVALUATE: For pairs of opposite sign the potential energy is negative and for pairs of the same sign the
potential energy is positive. The net electrical potential energy is the algebraic sum of the potential energy
of each pair.
IDENTIFY: Call the three charges 1, 2 and 3. U = U12 + U13 + U 23
SET UP: U12 = U 23 = U13 because the charges are equal and each pair of charges has the same separation,
0.500 m.
3kq 2
3k (1.2 × 10−6 C) 2
=
= 0.078 J.
0.500 m
0.500 m
EVALUATE: When the three charges are brought in from infinity to the corners of the triangle, the
repulsive electrical forces between each pair of charges do negative work and electrical potential energy is
stored.
IDENTIFY: The protons repel each other and therefore accelerate away from one another. As they get
farther and farther away, their kinetic energy gets greater and greater but their acceleration keeps
decreasing. Conservation of energy and Newton’s laws apply to these protons.
SET UP: Let a be the point when they are 0.750 nm apart and b be the point when they are very far apart.
A proton has charge +e and mass 1.67 × 10−27 kg. As they move apart the protons have equal kinetic
EXECUTE: U =
23.9.
energies and speeds. Their potential energy is U = ke2 /r and K = 12 mv 2 . K a + U a = Kb + U b .
EXECUTE: (a) They have maximum speed when they are far apart and all their initial electrical potential
energy has been converted to kinetic energy. K a + U a = Kb + U b .
K a = 0 and U b = 0, so
Kb = U a = k
e2
(1.60 × 10−19 C)2
= (8.99 × 109 N ⋅ m 2 /C2 )
= 3.07 × 10−19 J.
ra
0.750 × 10−9 m
Kb = 12 mvb2 + 12 mvb2 , so Kb = mvb2 and vb =
Kb
3.07 × 10−19 J
=
= 1.36 × 104 m/s.
−27
m
1.67 × 10
kg
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Electric Potential
23-5
(b) Their acceleration is largest when the force between them is largest and this occurs at r = 0.750 nm,
when they are closest.
a=
23.10.
9
2
2 ⎛ 1.60 × 10
2
−19
C⎞
−10
N.
F = k 2 = (8.99 × 10 N ⋅ m /C ) ⎜
⎟⎟ = 4.09 × 10
−9
⎜
.
×
0
750
10
m
r
⎝
⎠
e2
F 4.09 × 10−10 N
=
= 2.45 × 1017 m/s 2 .
m 1.67 × 10−27 kg
EVALUATE: The acceleration of the protons decreases as they move farther apart, but the force between
them is repulsive so they continue to increase their speeds and hence their kinetic energies.
IDENTIFY: The work done on the alpha particle is equal to the difference in its potential energy when it is
moved from the midpoint of the square to the midpoint of one of the sides.
SET UP: We apply the formula Wa →b = U a − U b . In this case, a is the center of the square and b is the
midpoint of one of the sides. Therefore Wcenter →side = U center − U side is the work done by the Coulomb force.
There are 4 electrons, so the potential energy at the center of the square is 4 times the potential energy of a
single alpha-electron pair. At the center of the square, the alpha particle is a distance r1 = 50 nm from each
electron. At the midpoint of the side, the alpha is a distance r2 = 5.00 nm from the two nearest electrons and
a distance r3 = 125 nm from the two most distant electrons. Using the formula for the potential energy
(relative to infinity) of two point charges, U = (1/4π ⑀ 0 )( qq0 /r ), the total work done by the Coulomb force is
Wcenter →side = U center − U side = 4
qα qe ⎛
1 qα qe
1 qα qe ⎞
−⎜2
+2
⎟
4π ⑀ 0 r1
4π ⑀ 0 r3 ⎠
⎝ 4π ⑀ 0 r2
1
Substituting qe = −e and qα = 2e and simplifying gives
Wcenter →side = −4e2
1 ⎡ 2 ⎛ 1 1 ⎞⎤
⎢ − ⎜ + ⎟⎥
4π ⑀ 0 ⎣⎢ r1 ⎝ r2 r3 ⎠ ⎦⎥
EXECUTE: Substituting the numerical values into the equation for the work gives
23.11.
⎡
2
1
1
⎛
⎞⎤
−21
W = −4(1.60 × 10−19 C) 2 (9.00 × 109 N ⋅ m 2 /C2 ) ⎢
−⎜
+
⎟ ⎥ = 6.08 × 10 J
5
00
nm
.
50
nm
125
nm
⎝
⎠
⎣
⎦
EVALUATE: Since the work done by the Coulomb force is positive, the system has more potential energy with
the alpha particle at the center of the square than it does with it at the midpoint of a side. To move the alpha
particle to the midpoint of a side and leave it there at rest an external force must do −6.08 × 10−21 J of work.
IDENTIFY: Apply Eq. (23.2). The net work to bring the charges in from infinity is equal to the change in
potential energy. The total potential energy is the sum of the potential energies of each pair of charges,
calculated from Eq. (23.9).
SET UP: Let 1 be where all the charges are infinitely far apart. Let 2 be where the charges are at the
corners of the triangle, as shown in Figure 23.11.
Let qc be the third, unknown charge.
Figure 23.11
EXECUTE: W = −ΔU = −(U 2 − U1 ), where W is the work done by the Coulomb force.
U1 = 0
U 2 = U ab + U ac + U bc =
1
(q 2 + 2qqc )
4π ⑀ 0d
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23-6
Chapter 23
Want W = 0, so W = −(U 2 − U1) gives 0 = −U 2
0=
1
(q 2 + 2qqc )
4π ⑀ 0d
q 2 + 2qqc = 0 and qc = − q/2.
EVALUATE: The potential energy for the two charges q is positive and for each q with qc it is negative.
There are two of the q, qc terms so must have qc < q.
23.12.
IDENTIFY: Use conservation of energy U a + K a = U b + Kb to find the distance of closest approach rb .
The maximum force is at the distance of closest approach, F = k
q1q2
rb2
.
SET UP: Kb = 0. Initially the two protons are far apart, so U a = 0. A proton has mass 1.67 × 10−27 kg
and charge q = + e = +1.60 × 10−19 C.
e2
qq
and
EXECUTE: K a = U b . 2 12 mva2 = k 1 2 . mva2 = k
rb
rb
(
rb =
ke 2
mva2
2
F =k
23.13.
e
=
)
(8.99 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C) 2
(1.67 × 10−27 kg)(1.00 × 106 m/s) 2
= (8.99 × 109 N ⋅ m 2 /C2 )
2
= 1.38 × 10−13 m.
(1.60 × 10−19 C) 2
= 0.012 N.
(1.38 × 10−13 m) 2
EVALUATE: The acceleration a = F/m of each proton produced by this force is extremely large.
IDENTIFY and SET UP: Apply conservation of energy to points A and B.
EXECUTE: K A + U A = K B + U B
rb
U = qV , so K A + qVA = K B + qVB
K B = K A + q (VA − VB ) = 0.00250 J + (−5.00 × 10−6 C)(200 V − 800 V) = 0.00550 J
vB = 2 K B /m = 7.42 m/s
23.14.
EVALUATE: It is faster at B; a negative charge gains speed when it moves to higher potential.
Wa→b
IDENTIFY: The work-energy theorem says Wa →b = Kb − K a .
= Va − Vb .
q
SET UP: Point a is the starting point and point b is the ending point. Since the field is uniform,
Wa →b = Fs cos φ = E q s cos φ . The field is to the left so the force on the positive charge is to the left. The
particle moves to the left so φ = 0° and the work Wa →b is positive.
EXECUTE: (a) Wa →b = Kb − K a = 1.50 × 10−6 J − 0 = 1.50 × 10−6 J
(b) Va − Vb =
Wa →b 1.50 × 10−6 J
=
= 357 V. Point a is at higher potential than point b.
q
4.20 × 10−9 C
(c) E q s = Wa →b , so E =
23.15.
Wa →b Va − Vb
357 V
=
=
= 5.95 × 103 V/m.
−2
qs
s
6.00 × 10 m
EVALUATE: A positive charge gains kinetic energy when it moves to lower potential; Vb < Va .
G
b G
IDENTIFY: Apply the equation that precedes Eq. (23.17): Wa →b = q′∫ E ⋅ dl .
a
G
SET UP: Use coordinates where + y is upward and + x is to the right. Then E = Eˆj with
E = 4.00 × 104 N/C.
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Electric Potential
23-7
(a) The path is sketched in Figure 23.15a.
Figure 23.15a
G
G G
b G
EXECUTE: E ⋅ dl = ( Eˆj ) ⋅ (dxiˆ) = 0 so Wa →b = q′∫ E ⋅ dl = 0.
a
EVALUATE: The electric force on the positive charge is upward (in the direction of the electric field) and
does no work for a horizontal displacement of the charge.
(b) SET UP: The path is sketched in Figure 23.15b.
G
dl = dyˆj
Figure 23.15b
EXECUTE:
G G
E ⋅ dl = ( Eˆj ) ⋅ (dyˆj ) = E dy
G
b G
b
Wa →b = q′∫ E ⋅ dl = q′ E ∫ dy = q′ E ( yb − ya )
a
a
yb − ya = +0.670 m, positive since the displacement is upward and we have taken + y to be upward.
Wa →b = q′ E ( yb − ya ) = (+28.0 × 10−9 C)(4.00 × 104 N/C)( +0.670 m) = +7.50 × 10−4 J.
EVALUATE: The electric force on the positive charge is upward so it does positive work for an upward
displacement of the charge.
(c) SET UP: The path is sketched in Figure 23.15c.
ya = 0
yb = − r sinθ = − (2.60 m) sin 45° = −1.838 m
The vertical component of the 2.60 m
displacement is 1.838 m downward.
Figure 23.15c
G
EXECUTE: dl = dxiˆ + dyˆj (The displacement has both horizontal and vertical components.)
G G
E ⋅ dl = ( Eˆj ) ⋅ (dxiˆ + dyˆj ) = E dy (Only the vertical component of the displacement contributes to the
work.)
G
b G
b
Wa →b = q′∫ E ⋅ dl = q′ E ∫ dy = q′E ( yb − ya )
a
a
−9
Wa →b = q′ E ( yb − ya ) = (+28.0 × 10 C)(4.00 × 104 N/C)(−1.838 m) = −2.06 × 10−3 J.
EVALUATE: The electric force on the positive charge is upward so it does negative work for a
displacement of the charge that has a downward component.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
23-8
Chapter 23
23.16.
IDENTIFY: Apply K a + U a = Kb + U b .
SET UP: Let q1 = +3.00 nC and q2 = +2.00 nC. At point a, r1a = r2 a = 0.250 m. At point b,
r1b = 0.100 m and r2b = 0.400 m. The electron has q = − e and me = 9.11 × 10−31 kg. K a = 0 since the
electron is released from rest.
EXECUTE: −
keq1 keq2
keq keq2 1
−
=− 1−
+ mevb2 .
r1a
r2 a
r1b
r2b
2
⎛ (3.00 × 10−9 C) (2.00 × 10−9 C) ⎞
−17
Ea = K a + U a = k (−1.60 × 10−19 C) ⎜
+
⎟⎟ = −2.88 × 10 J.
⎜
.
.
0
250
m
0
250
m
⎝
⎠
⎛ (3.00 × 10−9 C) (2.00 × 10−9 C) ⎞ 1
1
−17
2
2
Eb = Kb + U b = k (−1.60 × 10−19 C) ⎜
+
⎟⎟ + mevb = −5.04 × 10 J + mevb
⎜
.
.
0
100
m
0
400
m
2
2
⎝
⎠
Setting Ea = Eb gives vb =
2
9.11 × 10−31 kg
(5.04 × 10−17 J − 2.88 × 10−17 J) = 6.89 × 106 m/s.
EVALUATE: Va = V1a + V2 a = 180 V. Vb = V1b + V2b = 315 V. Vb > Va . The negatively charged electron
gains kinetic energy when it moves to higher potential.
23.17.
IDENTIFY: The potential at any point is the scalar sum of the potentials due to individual charges.
SET UP: V = kq /r and Wab = q (Va – Vb ).
EXECUTE: (a) ra1 = ra 2 =
⎛q
1
q ⎞
(0.0300 m) 2 + (0.0300 m) 2 = 0.0212 m. Va = k ⎜ 1 + 2 ⎟ = 0.
2
⎝ ra1 ra 2 ⎠
(b) rb1 = 0.0424 m, rb 2 = 0.0300 m.
⎛ +2.00 × 10−6 C −2.00 × 10− 6 C ⎞
⎛q
q ⎞
5
Vb = k ⎜ 1 + 2 ⎟ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜
+
⎟⎟ = −1.75 × 10 V.
⎜ 0.0424 m
.
r
r
0
0300
m
b
b
1
2
⎝
⎠
⎝
⎠
(c) Wab = q3 (Va − Vb ) = (−5.00 × 10−6 C)[0 − (−1.75 × 105 V)] = −0.875 J.
EVALUATE: Since Vb < Va , a positive charge would be pulled by the existing charges from a to b, so they
would do positive work on this charge. But they would repel a negative charge and hence do negative work
on it, as we found in part (c).
23.18.
IDENTIFY: The total potential is the scalar sum of the individual potentials, but the net electric field is the
vector sum of the two fields.
SET UP: The net potential can only be zero if one charge is positive and the other is negative, since it is a
scalar. The electric field can only be zero if the two fields point in opposite directions.
EXECUTE: (a) (i) Since both charges have the same sign, there are no points for which the potential is
zero.
(ii) The two electric fields are in opposite directions only between the two charges, and midway between
them the fields have equal magnitudes. So E = 0 midway between the charges, but V is never zero.
(b) (i) The two potentials have equal magnitude but opposite sign midway between the charges, so V = 0
midway between the charges, but E ≠ 0 there since the fields point in the same direction.
(ii) Between the two charges, the fields point in the same direction, so E cannot be zero there. In the other
two regions, the field due to the nearer charge is always greater than the field due to the more distant
charge, so they cannot cancel. Hence E is not zero anywhere.
EVALUATE: It does not follow that the electric field is zero where the potential is zero, or that the
potential is zero where the electric field is zero.
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Electric Potential
23.19.
IDENTIFY: V =
23-9
1
q
∑ i
4π ⑀ 0 i ri
SET UP: The locations of the changes and points A and B are sketched in Figure 23.19.
Figure 23.19
EXECUTE: (a) VA =
1 ⎛ q1 q2 ⎞
+
⎜
⎟
4π ⑀ 0 ⎝ rA1 rA2 ⎠
⎛ +2.40 × 10−9 C −6.50 × 10−9 C ⎞
VA = (8.988 × 109 N ⋅ m 2 /C2 ) ⎜
+
⎟ = −737 V
⎜ 0.050 m
0.050 m ⎟⎠
⎝
1 ⎛ q1 q2 ⎞
(b) VB =
+
⎜
⎟
4π ⑀ 0 ⎝ rB1 rB 2 ⎠
⎛ +2.40 × 10−9 C −6.50 × 10−9 C ⎞
VB = (8.988 × 109 N ⋅ m 2 /C2 ) ⎜
+
⎟ = −704 V
⎜ 0.080 m
0.060 m ⎟⎠
⎝
(c) IDENTIFY and SET UP: Use Eq. (23.13) and the results of parts (a) and (b) to calculate W.
EXECUTE: WB → A = q ′ (VB − VA ) = (2.50 × 10−9 C)( −704 V − ( −737 V)) = +8.2 × 10 −8 J
23.20.
EVALUATE: The electric force does positive work on the positive charge when it moves from higher
potential (point B) to lower potential (point A).
kq
IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the
r
potentials of the two charges.
SET UP: Consider the distances from the point on the y-axis to each charge for the three regions
− a ≤ y ≤ a (between the two charges), y > a (above both charges) and y < −a (below both charges).
EXECUTE: (a) y < a : V =
y < −a :V =
kq
kq
2kqy
kq
kq
−2kqa
−
=
. y > a: V =
−
=
.
(a + y ) (a − y ) y 2 − a 2
(a + y) y − a y 2 − a 2
− kq
kq
2kqa
−
=
.
(a + y ) (− y + a ) y 2 − a 2
⎛ −q
q ⎞
A general expression valid for any y is V = k ⎜
+
⎟.
⎜ y−a
y + a ⎟⎠
⎝
(b) The graph of V versus y is sketched in Figure 23.20.
−2kqa −2kqa
(c) y >> a : V = 2
.
≈
y − a2
y2
(d) If the charges are interchanged, then the potential is of the opposite sign.
EVALUATE: V = 0 at y = 0. V → +∞ as the positive charge is approached and V → −∞ as the negative
charge is approached.
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23-10
Chapter 23
Figure 23.20
23.21.
IDENTIFY: For a point charge, V =
kq
. The total potential at any point is the algebraic sum of the
r
potentials of the two charges.
SET UP: (a) The positions of the two charges are shown in Figure 23.21a.
Figure 23.21a
(b) x > a : V =
x < 0 :V =
kq 2kq − kq ( x + a )
kq 2kq
kq (3 x − a )
. 0 < x < a :V =
.
−
=
−
=
x x−a
x( x − a)
x a−x
x( x − a)
⎛q
kq ( x + a )
− kq 2kq
2q ⎞
. A general expression valid for any y is V = k ⎜ −
+
=
⎟⎟ .
⎜
x
x−a
x( x − a)
⎝ x x−a ⎠
(c) The potential is zero at x = − a and a /3.
(d) The graph of V versus x is sketched in Figure 23.21b.
Figure 23.21 b
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Electric Potential
− kq
, which is the same as the potential of a point charge – q.
x
x
Far from the two charges they appear to be a point charge with a charge that is the algebraic sum of their
two charges.
kq
IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the
r
potentials of the two charges.
SET UP: The distance of a point with coordinate y from the positive charge is y and the distance from
EVALUATE: (e) For x >> a : V ≈
23.22.
−kqx
23-11
2
=
the negative charge is r = a 2 + y 2 .
EVALUATE: (a) V =
⎛ 1
⎞
kq 2kq
2
⎟.
−
= kq ⎜ −
2
2 ⎟
⎜ y
y
r
a
+
y
⎝
⎠
a2 + y2
a
⇒ 3 y2 = a2 ⇒ y = ±
.
4
3
(c) The graph of V versus y is sketched in Figure 23.22. V → ∞ as the positive charge at the origin is
approached.
⎛1 2⎞
kq
EVALUATE: (d) y >> a : V ≈ kq ⎜ − ⎟ = − , which is the potential of a point charge − q. Far from the
y
y
y
⎝
⎠
(b) V = 0, when y 2 =
two charges they appear to be a point charge with a charge that is the algebraic sum of their two charges.
Figure 23.22
23.23.
IDENTIFY and SET UP: Apply conservation of energy, Eq. (23.3). Use Eq. (23.12) to express U in terms
of V.
(a) EXECUTE: K1 + qV1 = K 2 + qV2 , q (V2 − V1) = K1 − K 2 ; q = −1.602 × 10−19 C.
K1 − K 2
= 156 V.
q
EVALUATE: The electron gains kinetic energy when it moves to higher potential.
K − K2
(b) EXECUTE: Now K1 = 2.915 × 10−17 J, K 2 = 0. V2 − V1 = 1
= −182 V.
q
EVALUATE: The electron loses kinetic energy when it moves to lower potential.
kq
kq
IDENTIFY: For a point charge, E = 2 and V = .
r
r
SET UP: The electric field is directed toward a negative charge and away from a positive charge.
2
V
kq/r
4.98 V
⎛ kq ⎞ ⎛ r ⎞
EXECUTE: (a) V > 0 so q > 0.
=
=
= 0.415 m.
⎜
⎟⎟ = r. r =
⎜
⎟
2
⎜
12.0 V/m
E k q /r
⎝ r ⎠ ⎝ kq ⎠
K1 = 12 mev12 = 4.099 × 10−18 J; K 2 = 12 mev22 = 2.915 × 10−17 J. ΔV = V2 − V1 =
23.24.
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23-12
Chapter 23
(b) q =
rV
(0.415 m)(4.98 V)
=
= 2.30 × 10−10 C
k 8.99 × 109 N ⋅ m 2 /C2
(c) q > 0, so the electric field is directed away from the charge.
23.25.
EVALUATE: The ratio of V to E due to a point charge increases as the distance r from the charge
increases, because E falls off as 1/r 2 and V falls off as 1/r.
G
(a) IDENTIFY and SET UP: The direction of E is always from high potential to low potential so point b is
at higher potential.
(b) Apply Eq. (23.17) to relate Vb − Va to E.
G
b G
b
EXECUTE: Vb − Va = − ∫ E ⋅ dl = ∫ E dx = E ( xb − xa ).
a
a
V − Va
+240 V
E= b
=
= 800 V/m
xb − xa 0.90 m − 0.60 m
(c) Wb →a = q (Vb − Va ) = ( −0.200 × 10−6 C)(+240 V) = −4.80 × 10−5 J.
23.26.
EVALUATE: The electric force does negative work on a negative charge when the negative charge moves
from high potential (point b) to low potential (point a).
kq
IDENTIFY: For a point charge, V = . The total potential at any point is the algebraic sum of the
r
kq
potentials of the two charges. For a point charge, E = 2 . The net electric field is the vector sum of the
r
electric fields of the two charges.
G
SET UP: E produced by a point charge is directed away from the point charge if it is positive and toward
the charge if it is negative.
EXECUTE: (a) V = VQ + V2Q > 0, so V is zero nowhere except for infinitely far from the charges. The
fields can cancel only between the charges, because only there are the fields of the two charges in opposite
directions. Consider a point a distance x from Q and d − x from 2Q, as shown in Figure 23.26a.
d
d
kQ
k (2Q)
EQ = E2Q → 2 =
→ (d − x) 2 = 2 x 2 . x =
. The other root, x =
, does not lie
1+ 2
1− 2
x
(d − x)2
between the charges.
(b) V can be zero in 2 places, A and B, as shown in Figure 23.26b. Point A is a distance x from −Q and
d − x from 2Q. B is a distance y from − Q and d + y from 2Q. At A :
k (−Q) k (2Q)
+
= 0 → x = d/3.
x
d−x
k ( − Q) k (2Q)
+
= 0 → y = d.
y
d+y
The two electric fields are in opposite directions to the left of − Q or to the right of 2Q in Figure 23.26c.
But for the magnitudes to be equal, the point must be closer to the charge with smaller magnitude of
kQ
k (2Q)
charge. This can be the case only in the region to the left of − Q. EQ = E2Q gives 2 =
and
x
(d + x)2
At B:
d
.
2 −1
G
EVALUATE: (d) E and V are not zero at the same places. E is a vector and V is a scalar. E is proportional
G
to 1/r 2 and V is proportional to 1/r. E is related to the force on a test charge and ΔV is related to the
work done on a test charge when it moves from one point to another.
x=
Figure 23.26
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Electric Potential
23.27.
23-13
IDENTIFY: The potential at any point is the scalar sum of the potential due to each shell.
kq
kq
SET UP: V =
for r ≤ R and V =
for r > R.
R
r
EXECUTE: (a) (i) r = 0. This point is inside both shells so
⎛ 6.00 × 10−9 C −9.00 × 10−9 C ⎞
⎛q
q ⎞
+
V = k ⎜ 1 + 2 ⎟ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜
⎟.
⎜ 0.0300 m
0.0500 m ⎟⎠
⎝ R1 R2 ⎠
⎝
V = +1.798 × 103 V + (−1.618 × 103 V) = 180 V.
(ii) r = 4.00 cm. This point is outside shell 1 and inside shell 2.
⎛ 6.00 × 10−9 C −9.00 × 10−9 C ⎞
⎛q q ⎞
+
V = k ⎜ 1 + 2 ⎟ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜
⎟.
⎜ 0.0400 m
0.0500 m ⎟⎠
⎝ r R2 ⎠
⎝
V = +1.348 × 103 V + ( −1.618 × 103 V) = −270 V.
(iii) r = 6.00 cm. This point is outside both shells.
8.99 × 109 N ⋅ m 2 /C2
⎛q q ⎞ k
(6.00 × 10−9 C + (−9.00 × 10−9 C)). V = − 450 V.
V = k ⎜ 1 + 2 ⎟ = ( q1 + q2 ) =
0.0600 m
r ⎠ r
⎝ r
(b) At the surface of the inner shell, r = R1 = 3.00 cm. This point is inside the larger shell,
⎛q
q ⎞
so V1 = k ⎜ 1 + 2 ⎟ = 180 V. At the surface of the outer shell, r = R2 = 5.00 cm. This point is outside the
R
R
2⎠
⎝ 1
smaller shell, so
⎛ 6.00 × 10−9 C −9.00 × 10−9 C ⎞
⎛q q ⎞
V = k ⎜ 1 + 2 ⎟ = (8.99 × 109 N ⋅ m 2 /C2 ) ⎜
+
⎟.
⎜
0.0500 m ⎠⎟
⎝ r R2 ⎠
⎝ 0.0500 m
23.28.
V2 = +1.079 × 103 V + (−1.618 × 103 V) = −539 V. The potential difference is V1 − V2 = 719 V. The inner
shell is at higher potential. The potential difference is due entirely to the charge on the inner shell.
EVALUATE: Inside a uniform spherical shell, the electric field is zero so the potential is constant (but not
necessarily zero).
IDENTIFY and SET UP: Expressions for the electric potential inside and outside a solid conducting sphere
are derived in Example 23.8.
kq k (3.50 × 10−9 C)
EXECUTE: (a) This is outside the sphere, so V =
=
= 65.6 V.
0.480 m
r
k (3.50 × 10−9 C)
= 131 V.
0.240 m
(c) This is inside the sphere. The potential has the same value as at the surface, 131 V.
EVALUATE: All points of a conductor are at the same potential.
(a) IDENTIFY and SET UP: The electric field on the ring’s axis is calculated in Example 21.9. The force on
the electron exerted by this field is given by Eq. (21.3).
EXECUTE: When the electron is on either side of the center of the ring, the ring exerts an attractive force
directed toward the center of the ring. This restoring force produces oscillatory motion of the electron
along the axis of the ring, with amplitude 30.0 cm. The force on the electron is not of the form F = –kx so
the oscillatory motion is not simple harmonic motion.
(b) IDENTIFY: Apply conservation of energy to the motion of the electron.
SET UP: K a + U a = Kb + U b with a at the initial position of the electron and b at the center of the ring.
(b) This is at the surface of the sphere, so V =
23.29.
From Example 23.11, V =
1
Q
4π ⑀ 0
2
x + R2
, where R is the radius of the ring.
EXECUTE: xa = 30.0 cm, xb = 0.
K a = 0 (released from rest), Kb = 12 mv 2
Thus 12 mv 2 = U a − U b
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23-14
Chapter 23
2e(Vb − Va )
.
m
And U = qV = −eV so v =
Va =
1
Q
4π ⑀ 0
xa2 + R 2
= (8.988 × 109 N ⋅ m 2 /C2 )
24.0 × 10−9 C
(0.300 m) 2 + (0.150 m)2
Va = 643 V
Vb =
v=
1
Q
4π ⑀ 0
xb2 + R 2
= (8.988 × 109 N ⋅ m 2 /C2 )
24.0 × 10−9 C
= 1438 V
0.150 m
2e(Vb − Va )
2(1.602 × 10−19 C)(1438 V − 643 V)
=
= 1.67 × 107 m/s
m
9.109 × 10−31 kg
EVALUATE: The positively charged ring attracts the negatively charged electron and accelerates it. The
electron has its maximum speed at this point. When the electron moves past the center of the ring the force
on it is opposite to its motion and it slows down.
23.30.
IDENTIFY: Example 23.10 shows that for a line of charge, Va − Vb =
λ
2π ⑀ 0
ln(rb /ra ). Apply conservation
of energy to the motion of the proton.
SET UP: Let point a be 18.0 cm from the line and let point b be at the distance of closest approach, where
Kb = 0.
EXECUTE: (a) K a = 12 mv 2 = 12 (1.67 × 10−27 kg)(1.50 × 103 m/s) 2 = 1.88 × 10−21 J.
(b) K a + qVa = Kb + qVb . Va − Vb =
Kb − K a −1.88 × 10−21 J
=
= −0.01175 V.
q
1.60 × 10−19 C
⎛ 2π ⑀ 0 ⎞
ln( rb /ra ) = ⎜
⎟ (−0.01175 V).
⎝ λ ⎠
⎛ 2π ⑀ 0 (0.01175 V) ⎞
⎛ 2π ⑀ 0 (−0.01175 V) ⎞
rb = ra exp ⎜
⎟⎟ = 0.158 m.
⎟ = (0.180 m)exp ⎜⎜ −
−12
λ
⎝
⎠
⎝ 5.00 × 10 C/m ⎠
23.31.
EVALUATE: The potential increases with decreasing distance from the line of charge. As the positively
charged proton approaches the line of charge it gains electrical potential energy and loses kinetic energy.
IDENTIFY: The voltmeter measures the potential difference between the two points. We must relate this
quantity to the linear charge density on the wire.
SET UP: For a very long (infinite) wire, the potential difference between two points is
ΔV =
λ
ln( rb /ra ).
2π ⑀ 0
EXECUTE: (a) Solving for λ gives
23.32.
( ΔV )2π ⑀0
=
ln(rb /ra )
575 V
= 9.49 × 10−8 C/m
⎛ 3.50 cm ⎞
(18 × 10 N ⋅ m /C )ln ⎜
⎟
⎝ 2.50 cm ⎠
(b) The meter will read less than 575 V because the electric field is weaker over this 1.00-cm distance than
it was over the 1.00-cm distance in part (a).
(c) The potential difference is zero because both probes are at the same distance from the wire, and hence
at the same potential.
EVALUATE: Since a voltmeter measures potential difference, we are actually given ΔV , even though that
is not stated explicitly in the problem.
IDENTIFY: The voltmeter reads the potential difference between the two points where the probes are
placed. Therefore we must relate the potential difference to the distances of these points from the center of
the cylinder. For points outside the cylinder, its electric field behaves like that of a line of charge.
λ=
9
2
2
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Electric Potential
SET UP: Using ΔV =
λ
2π ⑀ 0
23-15
ln (rb /ra ) and solving for rb , we have rb = ra e 2π ⑀0 ΔV/λ .
⎛
⎞
1
(175 V)
⎜⎜
9
2 2⎟
2 × 9.00 × 10 N ⋅ m /C ⎟⎠
⎝
EXECUTE: The exponent is
= 0.648, which gives
15.0 × 10−9 C/m
rb = (2.50 cm) e0.648 = 4.78 cm.
23.33.
The distance above the surface is 4.78 cm − 2.50 cm = 2.28 cm.
EVALUATE: Since a voltmeter measures potential difference, we are actually given ΔV , even though that
is not stated explicitly in the problem. We must also be careful when using the formula for the potential
difference because each r is the distance from the center of the cylinder, not from the surface.
IDENTIFY: For points outside the cylinder, its electric field behaves like that of a line of charge. Since a
voltmeter reads potential difference, that is what we need to calculate.
SET UP: The potential difference is ΔV =
EXECUTE:
ln (rb /ra ).
(a) Substituting numbers gives
ΔV =
23.34.
λ
2π ⑀0
λ
2π ⑀ 0
⎛ 10.0 cm ⎞
ln (rb /ra ) = (8.50 × 10−6 C/m)(2 × 9.00 × 109 N ⋅ m 2 /C2 ) ln ⎜
⎟
⎝ 6.00 cm ⎠
ΔV = 7.82 × 104 V = 78,200 V = 78.2 kV
(b) E = 0 inside the cylinder, so the potential is constant there, meaning that the voltmeter reads zero.
EVALUATE: Caution! The fact that the voltmeter reads zero in part (b) does not mean that V = 0 inside
the cylinder. The electric field is zero, but the potential is constant and equal to the potential at the surface.
IDENTIFY: The work required is equal to the change in the electrical potential energy of the charge-ring
system. We need only look at the beginning and ending points, since the potential difference is
independent of path for a conservative field.
⎛ 1 Q
⎞
− 0⎟
SET UP: (a) W = ΔU = qΔV = q (Vcenter − V∞ ) = q ⎜
4
a
π
⑀
0
⎝
⎠
EXECUTE: Substituting numbers gives
ΔU = (3.00 × 10−6 C)(9.00 × 109 N ⋅ m 2 /C2 )(5.00 × 10−6 C)/(0.0400 m) = 3.38 J
(b) We can take any path since the potential is independent of path.
(c) SET UP: The net force is away from the ring, so the ball will accelerate away. Energy conservation
gives U 0 = K max = 12 mv 2 .
EXECUTE: Solving for v gives
v=
23.35.
2U 0
2(3.38 J)
=
= 67.1 m/s.
m
0.00150 kg
EVALUATE: Direct calculation of the work from the electric field would be extremely difficult, and we
would need to know the path followed by the charge. But, since the electric field is conservative, we can
bypass all this calculation just by looking at the end points (infinity and the center of the ring) using the
potential.
IDENTIFY: The electric field of the line of charge does work on the sphere, increasing its kinetic energy.
λ
⎛r ⎞
SET UP: K1 + U1 = K 2 + U 2 and K1 = 0. U = qV so qV1 = K 2 + qV2 . V =
ln ⎜ 0 ⎟ .
2π ⑀ 0 ⎝ r ⎠
EXECUTE: V1 =
⎛r ⎞
⎛r ⎞
λ
λ
ln ⎜ 0 ⎟ . V2 =
ln ⎜ 0 ⎟ .
2π ⑀ 0 ⎝ r1 ⎠
2π ⑀ 0 ⎝ r2 ⎠
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23-16
Chapter 23
K 2 = q(V1 − V2 ) =
⎛ r ⎞ ⎞ λq
⎛r ⎞
λ ⎛ ⎛ r0 ⎞
λq
(ln r2 − ln r1 ) =
ln ⎜ 2 ⎟ .
⎜⎜ ln ⎜ ⎟ − ln ⎜ 0 ⎟ ⎟⎟ =
2π ⑀ 0 ⎝ ⎝ r1 ⎠
2π ⑀ 0 ⎝ r1 ⎠
⎝ r2 ⎠ ⎠ 2π ⑀0
(3.00 × 10−6 C/m)(8.00 × 10−6 C)
⎛ 4.50 ⎞
ln ⎜
⎟ = 0.474 J.
2π (8.854 × 10 C /(N ⋅ m )
⎝ 1.50 ⎠
EVALUATE: The potential due to the line of charge does not go to zero at infinity but is defined to be zero
at an arbitrary distance r0 from the line.
K2 =
23.36.
−12
2
2
IDENTIFY: If the small sphere is to have its minimum speed, it must just stop at 8.00 cm from the surface
of the large sphere. In that case, the initial kinetic energy of the small sphere is all converted to electrical
potential energy at its point of closest approach.
SET UP: K1 + U1 = K 2 + U 2 . K 2 = 0. U1 = 0. Therefore, K1 = U 2 . Outside a spherical charge
distribution the potential is the same as for a point charge at the location of the center of the sphere, so
U = kqQ/r. K = 12 mv 2 .
EXECUTE: U 2 =
v1 =
23.37.
kqQ
1
kqQ
.
, with r2 = 12.0 cm + 8.0 cm = 0.200 m. mv12 =
2
r2
r2
2kqQ
2(8.99 × 109 N ⋅ m 2 /C2 )(3.00 × 10−6 C)(5.00 × 10−6 C)
=
= 150 m/s.
mr2
(6.00 × 10−5 kg)(0.200 m)
EVALUATE: If the small sphere had enough initial speed to actually penetrate the surface of the large
sphere, we could no longer treat the large sphere as a point charge once the small sphere was inside.
IDENTIFY: We can model the axon membrane as a large sheet having equal but opposite charges on its
opposite faces.
SET UP: For two oppositely charged sheets of charge, Vab = Ed . The positively charged sheet is the one
at higher potential.
Vab 70 × 10−3 V
=
= 9.3 × 106 V/m. The electric field is directed inward, toward the
d
7.5 × 10−9 m
G
interior of the axon, since the outer surface of the membrane has positive charge and E points away from
positive charge and toward negative charge.
(b) The outer surface has positive charge so it is at higher potential than the inner surface.
EVALUATE: The electric field is quite strong compared to ordinary laboratory fields in devices such as
student oscilloscopes. The potential difference is only 70 mV, but it occurs over a distance of only 7.5 nm.
IDENTIFY and SET UP: For oppositely charged parallel plates, E = σ /⑀ 0 between the plates and the
potential difference between the plates is V = Ed .
EXECUTE: (a) E =
23.38.
EXECUTE: (a) E =
σ
⑀0
=
47.0 × 10−9 C/m 2
⑀0
= 5310 N/C.
(b) V = Ed = (5310 N/C)(0.0220 m) = 117 V.
23.39.
(c) The electric field stays the same if the separation of the plates doubles. The potential difference
between the plates doubles.
EVALUATE: The electric field of an infinite sheet of charge is uniform, independent of distance from the
sheet. The force on a test charge between the two plates is constant because the electric field is constant.
The potential difference is the work per unit charge on a test charge when it moves from one plate to the
other. When the distance doubles, the work, which is force times distance, doubles and the potential
difference doubles.
IDENTIFY and SET UP: Use the result of Example 23.9 to relate the electric field between the plates to the
potential difference between them and their separation. The force this field exerts on the particle is given
by Eq. (21.3). Use the equation that precedes Eq. (23.17) to calculate the work.
V
360 V
EXECUTE: (a) From Example 23.9, E = ab =
= 8000 V/m.
d
0.0450 m
(b) F = q E = (2.40 × 10− 9 C)(8000 V/m) = +1.92 × 10−5 N
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Electric Potential
23-17
(c) The electric field between the plates is shown in Figure 23.39.
Figure 22.39
The plate with positive charge (plate a) is at higher potential. The electric field is directed from high
G
G
potential toward low potential (or, E is from + charge toward − charge), so E points from a to b. Hence
G
the force that E exerts on the positive charge is from a to b, so it does positive work.
G
b G
W = ∫ F ⋅ dl = Fd , where d is the separation between the plates.
a
W = Fd = (1.92 × 10−5 N)(0.0450 m) = +8.64 × 10−7 J
(d) Va − Vb = +360 V (plate a is at higher potential)
ΔU = U b − U a = q (Vb − Va ) = (2.40 × 10−9 C)( −360 V) = −8.64 × 10−7 J.
EVALUATE: We see that Wa →b = −(U b − U a ) = U a − U b .
23.40.
IDENTIFY and SET UP: Vab = Ed for parallel plates.
EXECUTE: d =
23.41.
Vab
1.5 V
=
= 1.5 × 106 m = 1.5 × 103 km.
E 1.0 × 10−6 V/m
EVALUATE: The plates would have to be nearly a thousand miles apart with only a AA battery across
them! This is a small field!
IDENTIFY and SET UP: Consider the electric field outside and inside the shell and use that to deduce the
potential.
EXECUTE: (a) The electric field outside the shell is the same as for a point charge at the center of the
shell, so the potential outside the shell is the same as for a point charge:
V=
q
for r > R.
4π ⑀0r
The electric field is zero inside the shell, so no work is done on a test charge as it moves inside the shell
q
for r ≤ R.
and all points inside the shell are at the same potential as the surface of the shell: V =
4π ⑀ 0 R
kq
RV (0.15 m)(−1200 V)
so q =
=
= −20 nC
R
k
k
(c) EVALUATE: No, the amount of charge on the sphere is very small. Since U = qV the total amount of
(b) V =
electric energy stored on the balloon is only (20 nC)(1200 nC) = 2.4 × 10−5 J.
23.42.
IDENTIFY: The electric field is zero inside the sphere, so the potential is constant there. Thus the potential
at the center must be the same as at the surface, where it is equivalent to that of a point-charge.
SET UP: At the surface, and hence also at the center of the sphere, the potential is that of a point-charge,
V = Q/(4π ⑀ 0 R).
EXECUTE: (a) Solving for Q and substituting the numbers gives
Q = 4π ⑀ 0 RV = (0.125 m)(1500 V)/(9.00 × 109 N ⋅ m 2 /C2 ) = 2.08 × 10−8 C = 20.8 nC
23.43.
(b) Since the potential is constant inside the sphere, its value at the surface must be the same as at the
center, 1.50 kV.
EVALUATE: The electric field inside the sphere is zero, so the potential is constant but is not zero.
IDENTIFY: Example 23.8 shows that the potential of a solid conducting sphere is the same at every point
inside the sphere and is equal to its value V = q/4π ⑀ 0 R at the surface. Use the given value of E to find q.
SET UP: For negative charge the electric field is directed toward the charge.
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23-18
Chapter 23
For points outside this spherical charge distribution the field is the same as if all the charge were
concentrated at the center.
q
(3800 N/C)(0.200 m) 2
2
EXECUTE: E =
and
q
π
⑀
Er
=
4
=
= 1.69 × 10−8 C.
0
4π ⑀0r 2
8.99 × 109 N ⋅ m 2 /C2
Since the field is directed inward, the charge must be negative. The potential of a point charge, taking ∞
q
(8.99 × 109 N ⋅ m 2 /C2 )(−1.69 × 10−8 C)
as zero, is V =
=
= −760 V at the surface of the sphere.
4π ⑀0r
0.200 m
23.44.
Since the charge all resides on the surface of a conductor, the field inside the sphere due to this
symmetrical distribution is zero. No work is therefore done in moving a test charge from just inside the
surface to the center, and the potential at the center must also be –760 V.
EVALUATE: Inside the sphere the electric field is zero and the potential is constant.
IDENTIFY: By the definition of electric potential, if a positive charge gains potential along a path, then the
potential along that path must have increased. The electric field produced by a very large sheet of charge is
uniform and is independent of the distance from the sheet.
(a) SET UP: No matter what the reference point, we must do work on a positive charge to move it away
from the negative sheet.
EXECUTE: Since we must do work on the positive charge, it gains potential energy, so the potential
increases.
(b) SET UP: Since the electric field is uniform and is equal to σ /2ε 0 , we have ΔV = Ed =
σ
2⑀0
d.
EXECUTE: Solving for d gives
d=
23.45.
2⑀ 0ΔV
2(8.85 × 10−12 C 2 /N ⋅ m 2 )(1.00V)
= 0.00295 m = 2.95 mm
6.00 × 10−9 C/m 2
EVALUATE: Since the spacing of the equipotential surfaces (d = 2.95 mm) is independent of the distance
from the sheet, the equipotential surfaces are planes parallel to the sheet and spaced 2.95 mm apart.
G
IDENTIFY and SET UP: Use Eq. (23.19) to calculate the components of E .
σ
=
EXECUTE: V = Axy− Bx 2 + Cy
(a) E x = −
Ey = −
∂V
= − Ay + 2 Bx
∂x
∂V
= − Ax − C
∂y
∂V
=0
∂z
(b) E = 0 requires that E x = E y = E z = 0.
Ez =
E z = 0 everywhere.
E y = 0 at x = −C/A.
And E x is also equal to zero for this x, any value of z and y = 2 Bx /A = (2 B /A)( −C /A) = −2 BC /A2 .
EVALUATE: V doesn’t depend on z so Ez = 0 everywhere.
23.46.
IDENTIFY: Apply Eq. (23.19).
G
1 q
SET UP: Eq. (21.7) says E =
rˆ is the electric field due to a point charge q.
4π ⑀ 0 r 2
⎞
kQ
kQx
kQx
∂V
∂⎛
⎟=
=− ⎜
= 3 .
∂x
∂x ⎜ x 2 + y 2 + z 2 ⎟ ( x 2 + y 2 + z 2 )3/ 2
r
⎝
⎠
kQy
kQz
Similarly, E y = 3 and E z = 3 .
r
r
EXECUTE: (a) E x = −
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Electric Potential
23-19
kQ ⎛ xiˆ yˆj zkˆ ⎞ kQ
⎜ + + ⎟⎟ = 2 rˆ, which agrees with Eq. (21.7).
r
r ⎠ r
r 2 ⎜⎝ r
G
EVALUATE: V is a scalar. E is a vector and has components.
kq
IDENTIFY and SET UP: For a solid metal sphere or for a spherical shell, V =
outside the sphere and
r
kq
at all points inside the sphere, where R is the radius of the sphere. When the electric field is radial,
V=
R
∂V
E=−
.
∂r
⎛1 1⎞
kq kq
−
= kq ⎜ − ⎟ .
EXECUTE: (a) (i) r < ra : This region is inside both spheres. V =
ra rb
⎝ ra rb ⎠
⎛1 1 ⎞
kq kq
(ii) ra < r < rb : This region is outside the inner shell and inside the outer shell. V =
−
= kq ⎜ − ⎟ .
r rb
⎝ r rb ⎠
(iii) r > rb : This region is outside both spheres and V = 0 since outside a sphere the potential is the same
(b) From part (a), E =
23.47.
as for a point charge. Therefore the potential is the same as for two oppositely charged point charges at the
same location. These potentials cancel.
⎛1 1⎞
1 ⎛q q⎞
1
(b) Va =
q ⎜ − ⎟.
⎜ − ⎟ and Vb = 0, so Vab =
4π ⑀ 0 ⎝ ra rb ⎠
4π ⑀0 ⎝ ra rb ⎠
⎛1 1 ⎞
(c) Between the spheres ra < r < rb and V = kq ⎜ − ⎟ .
⎝ r rb ⎠
∂V
q ∂ ⎛1 1 ⎞
1 q
Vab
1
=−
=
E=−
.
⎜ − ⎟ =+
∂r
4π ⑀ 0 ∂r ⎝ r rb ⎠
4π ⑀ 0 r 2 ⎛ 1 1 ⎞ r 2
⎜ − ⎟
⎝ ra rb ⎠
(d) From Eq. (23.23): E = 0, since V is constant (zero) outside the spheres.
(e) If the outer charge is different, then outside the outer sphere the potential is no longer zero but is
1 q
1 Q
1 (q − Q)
V=
−
=
. All potentials inside the outer shell are just shifted by an amount
4π ⑀ 0 r 4π ⑀ 0 r
4π ⑀ 0
r
V=
1 Q
. Therefore relative potentials within the shells are not affected. Thus (b) and (c) do not
4π ⑀ 0 rb
change. However, now that the potential does vary outside the spheres, there is an electric field there:
∂V
∂ ⎛ kq − kQ ⎞ kq ⎛ Q ⎞ k
=− ⎜ +
E=−
⎟ = ⎜1 − ⎟ = 2 (q − Q).
r ⎠ r2 ⎝
q⎠ r
∂r
∂r ⎝ r
EVALUATE: In part (a) the potential is greater than zero for all r < rb .
23.48.
⎛1 1⎞
⎛1 1 ⎞
IDENTIFY: Exercise 23.47 shows that V = kq ⎜ − ⎟ for r < ra , V = kq ⎜ − ⎟ for ra < r < rb and
r
r
⎝ a b⎠
⎝ r rb ⎠
⎛1 1⎞
Vab = kq ⎜ − ⎟ .
⎝ ra rb ⎠
kq
SET UP: E = 2 , radially outward, for ra ≤ r ≤ rb .
r
⎛1 1⎞
500 V
EXECUTE: (a) Vab = kq ⎜ − ⎟ = 500 V gives q =
= 7.62 × 10−10 C.
⎛
⎞
1
1
⎝ ra rb ⎠
−
k⎜
⎟
⎝ 0.012 m 0.096 m ⎠
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23-20
Chapter 23
(b) Vb = 0 so Va = 500 V. The inner metal sphere is an equipotential with V = 500 V.
1 1 V
= + .
r ra kq
V = 400 V at r = 1.45 cm, V = 300 V at r = 1.85 cm, V = 200 V at r = 2.53 cm, V = 100 V at
r = 4.00 cm, V = 0 at r = 9.60 cm. The equipotential surfaces are sketched in Figure 23.48.
EVALUATE: (c) The equipotential surfaces are concentric spheres and the electric field lines are radial, so
the field lines and equipotential surfaces are mutually perpendicular. The equipotentials are closest at
smaller r, where the electric field is largest.
Figure 23.48
23.49.
IDENTIFY: Outside the cylinder it is equivalent to a line of charge at its center.
SET UP: The difference in potential between the surface of the cylinder (a distance R from the central
axis) and a general point a distance r from the central axis is given by ΔV =
λ
ln(r/R).
2π ⑀ 0
EXECUTE: (a) The potential difference depends only on r, and not direction. Therefore all points at the
same value of r will be at the same potential. Thus the equipotential surfaces are cylinders coaxial with the
given cylinder.
(b) Solving ΔV =
λ
2π ⑀ 0
ln( r/R ) for r, gives r = Re 2π ⑀0 ΔV/λ .
For 10 V, the exponent is (10 V)/[(2 × 9.00 × 109 N ⋅ m 2 /C2 )(1.50 × 10−9 C/m)] = 0.370, which gives
r = (2.00 cm)e0.370 = 2.90 cm. Likewise, the other radii are 4.20 cm (for 20 V) and 6.08 cm (for 30 V).
(c) Δr1 = 2.90 cm − 2.00 cm = 0.90 cm; Δr2 = 4.20 cm − 2.90 cm = 1.30 cm; Δr3 = 6.08 cm − 4.20 cm = 1.88 cm
23.50.
EVALUATE: As we can see, Δr increases, so the surfaces get farther apart. This is very different from a
sheet of charge, where the surfaces are equally spaced planes.
IDENTIFY: As the sphere approaches the point charge, the speed of the sphere decreases because it loses
kinetic energy, but its acceleration increases because the electric force on it increases. Its mechanical
energy is conserved during the motion, and Newton’s second law and Coulomb’s law both apply.
SET UP: K a + U a = Kb + U b , K = 12 mv 2 , U = kq1q2 /r , F = kq1q2 /r 2 , and F = ma.
EXECUTE: Find the distance between the two charges when v2 = 25.0 m/s.
K a + U a = Kb + U b .
1
1
K a = mva2 = (4.00 × 10−3 kg)(40.0 m/s) 2 = 3.20 J.
2
2
1 2 1
Kb = mvb = (4.00 × 10−3 kg)(25.0 m/s)2 = 1.25 J.
2
2
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Electric Potential
23-21
qq
(8.99 × 109 N ⋅ m 2 /C2 )(5.00 × 10−6 C)(5.00 × 10−6 C)
Ua = k 1 2 =
= 1.498 J.
ra
0.0600 m
qq
U b = K a + U a − Kb = 3.20 J + 1.498 J − 1.25 J = 3.448 J. U b = k 1 2 and
rb
kq1q2 (8.99 × 109 N ⋅ m 2 /C2 )(5.00 × 10−6 C)(2.00 × 10−6 C)
=
= 0.02607 m.
Ub
3.448 J
rb =
Fb =
a=
23.51.
kq1q2
rb2
=
(8.99 × 109 N ⋅ m 2 / C2 )(5.00 × 10−6 C)(2.00 × 10−6 C)
(0.02607 m) 2
= 132.3 N.
F
132.3 N
=
= 3.31 × 104 m/s 2 .
m 4.00 × 10− 3 kg
EVALUATE: As the sphere approaches the point charge, its speed decreases but its acceleration keeps
increasing because the electric force on it keeps increasing.
⎛qq
qq q q ⎞
IDENTIFY: U = k ⎜ 1 2 + 1 3 + 2 3 ⎟
r13
r23 ⎠
⎝ r12
SET UP: In part (a), r12 = 0.200 m, r23 = 0.100 m and r13 = 0.100 m. In part (b) let particle 3 have
coordinate x, so r12 = 0.200 m, r13 = x and r23 = 0.200 − x.
⎛ (4.00 nC)(−3.00 nC) (4.00 nC)(2.00 nC) (−3.00 nC)(2.00 nC) ⎞
−7
+
+
EXECUTE: (a) U = k ⎜
⎟⎠ = −3.60 × 10 J
(0.200 m)
(0.100 m)
(0.100 m)
⎝
⎛qq
qq
qq ⎞
(b) If U = 0, then 0 = k ⎜ 1 2 + 1 3 + 2 3 ⎟ . Solving for x we find:
x
r12 − x ⎠
⎝ r12
8
6
⇒ 60 x 2 − 26 x + 1.6 = 0 ⇒ x = 0.074 m, 0.360 m. Therefore, x = 0.074 m since it is
0 = −60 + −
x 0.2 − x
the only value between the two charges.
EVALUATE: U13 is positive and both U 23 and U12 are negative. If U = 0, then U13 = U 23 + U12 . For
23.52.
x = 0.074 m, U13 = +9.7 × 10−7 J, U 23 = −4.3 × 10−7 J and U12 = −5.4 × 10−7 J. It is true that U = 0 at
this x.
IDENTIFY: Two forces do work on the sphere as it falls: gravity and the electrical force due to the sheet.
The energy of the sphere is conserved.
SET UP: The gravity force is mg, downward. The electric field of the sheet is E =
σ
2ε 0
upward, and the
force it exerts on the sphere is F = qE. The sphere gains kinetic energy K = 12 mv 2 as it falls.
EXECUTE: mg = 4.90 × 10−6 N. E =
σ
8.00 × 10−12 C/m 2
=
= 0.4518 N/C. The electric force
2ε 0 2(8.854 × 10−12 C2 /(N ⋅ m 2 )
is qE = (3.00 × 10−6 C)(0.4518 N/C) = 1.355 × 10−6 N, upward. The net force is downward, so the sphere
moves downward when released. Let y = 0 at the sheet. U grav = mgy. For the electric force,
Wa →b
= Va − Vb . Let point a be at the sheet and let point b be a distance y above the sheet. Take Va = 0.
q
The force on q is qE , upward, so
Wa →b
= Ey and Vb = − Ey. U b = − Eyq. K1 + U1 = K 2 + U 2 . K1 = 0.
q
y1 = 0.400 m, y2 = 0.100 m. K 2 = U1 − U 2 = mg ( y1 − y2 ) − E ( y1 − y2 )q.
K 2 = (5.00 × 10−7 kg)(9.8 m/s 2 )(0.300 m) − (0.4518 N/C)(0.300 m)(3.00 × 10−6 C).
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23-22
Chapter 23
1
K 2 = 1.470 × 10−6 J − 0.407 × 10−6 J = 1.063 × 10−6 J. K 2 = mv22 so
2
2K2
2(1.063 × 10−6 J)
=
= 2.06 m/s.
m
5.00 × 10−7 kg
v2 =
23.53.
EVALUATE: Because the weight is greater than the electric force, the sphere will accelerate downward,
but if it were light enough the electric force would exceed the weight. In that case it would never get closer
to the sheet after being released.
IDENTIFY: The remaining nucleus (radium minus the ejected alpha particle) repels the alpha particle,
giving it 4.79 MeV of kinetic energy when it is far from the nucleus. The mechanical energy of the system
is conserved.
qq′
SET UP: U = k
. U a + K a = U b + Kb . The charge of the alpha particle is +2e and the charge of the
r
radon nucleus is +86e.
EXECUTE: (a) The final energy of the alpha particle, 4.79 MeV, equals the electrical potential energy of
the alpha-radon combination just before the decay. U = 4.79 MeV = 7.66 × 10−13 J.
(b) r =
23.54.
kqq′ (8.99 × 109 N ⋅ m 2 /C2 )(2)(86)(1.60 × 10−19 C) 2
=
= 5.17 ×10−14 m.
−13
U
7.66×10
J
EVALUATE: Although we have made some simplifying assumptions (such as treating the atomic nucleus
as a spherically symmetric charge, even when very close to it), this result gives a fairly reasonable estimate
for the size of a nucleus.
IDENTIFY: The charged particles repel each other and therefore accelerate away from one another,
causing their speeds and kinetic energies to continue to increase. They do not have equal speeds because
they have different masses. The mechanical energy and momentum of the system are conserved.
SET UP: The proton has charge qp = + e and mass mp = 1.67 × 10−27 kg. The alpha particle has charge
qa = + 4e and mass ma = 4mp = 6.68 × 10−27 kg. We can apply both conservation of energy and
q1q2
F
, where F = k 2 .
m
r
EXECUTE: Acceleration: The maximum force and hence the maximum acceleration occurs just after they
(2)(1.60 ×10−19 C) 2
= 9.09 × 10−9 N.
are released, when r = 0.225 nm. F = (8.99 × 109 N ⋅ m 2 /C2 )
(0.225×10−9 m) 2
conservation of linear momentum to the system. a =
ap =
F
9.09× 10−9 N
F
9.09 × 10−9 N
18
2
=
=
.
×
=
=
= 1.36 × 1018 m/s 2 . The
5
44
10
m/s
;
a
a
mp 1.67 × 10−27 kg
ma 6.68 × 10−27 kg
acceleration of the proton is larger by a factor of ma /mp .
Speed: Conservation of energy says U1 + K1 = U 2 + K 2 . K1 = 0 and U 2 = 0, so K 2 = U1 .
U1 = k
qq′
(2)(1.60 × 10−19 C) 2
= (8.99 × 109 N ⋅ m 2 /C2 )
= 2.05 × 10−18 J, so the total kinetic energy of the
r
0.225 × 10−9 m
two particles when they are far apart is K 2 = 2.05 × 10−18 J. Conservation of linear momentum says how
⎛ mp ⎞
this energy is divided between the proton and alpha particle. p1 = p2 . 0 = mpvp − ma va so va = ⎜
⎟ vp .
⎝ ma ⎠
2
mp ⎞
⎛ mp ⎞ 2 1
2⎛
K 2 = 12 mpvp2 + 12 ma va2 = 12 mpvp2 + 12 ma ⎜
⎟ vp = 2 mpvp ⎜1 +
⎟.
⎝ ma ⎠
⎝ ma ⎠
vp =
2K2
2(2.05 × 10−18 J)
=
= 4.43 × 104 m/s.
mp (1 + (mp /ma ))
(1.67 × 10−27 kg) 1 + 14
(
)
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Electric Potential
23.55.
23-23
⎛ mp ⎞
4
4
1
va = ⎜
⎟ vp = 4 (4.43 × 10 m/s) = 1.11 × 10 m/s. The maximum acceleration occurs just after they are
⎝ ma ⎠
released. The maximum speed occurs after a long time.
EVALUATE: The proton and alpha particle have equal momenum, but proton has a greater acceleration
and more kinetic energy.
(a) IDENTIFY: Apply the work-energy theorem, Eq. (6.6).
SET UP: Points a and b are shown in Figure 23.55a.
Figure 23.55a
EXECUTE: Wtot = ΔK = K b − K a = K b = 4.35 × 10−5 J
The electric force FE and the additional force F both do work, so that Wtot = WFE + WF .
WFE = Wtot − WF = 4.35 × 10−5 J − 6.50 × 10−5 J = −2.15 × 10−5 J
EVALUATE: The forces on the charged particle are shown in Figure 23.55b.
Figure 23.55b
The electric force is to the left (in the direction of the electric field since the particle has positive charge).
The displacement is to the right, so the electric force does negative work. The additional force F is in the
direction of the displacement, so it does positive work.
(b) IDENTIFY and SET UP: For the work done by the electric force, Wa →b = q (Va − Vb ).
EXECUTE: Va − Vb =
Wa →b −2.15 × 10−5 J
=
= −2.83 × 103 V.
q
7.60 × 10−9 C
EVALUATE The starting point (point a) is at 2.83 × 103 V lower potential than the ending point (point b).
We know that Vb > Va because the electric field always points from high potential toward low potential.
(c) IDENTIFY: Calculate E from Va − Vb and the separation d between the two points.
SET UP: Since the electric field is uniform and directed opposite to the displacement
Wa →b = − FE d = − qEd , where d = 8.00 cm is the displacement of the particle.
EXECUTE: E = −
Wa →b
V −V
−2.83 × 103 V
=− a b =−
= 3.54 × 104 V/m.
0.0800 m
qd
d
EVALUATE: In part (a), Wtot is the total work done by both forces. In parts (b) and (c) Wa →b is the work
done just by the electric force.
23.56.
IDENTIFY: The electric force between the electron and proton is attractive and has magnitude F =
For circular motion the acceleration is arad = v 2 /r. U = − k
ke 2
r2
.
e2
.
r
SET UP: e = 1.60 × 10−19 C. 1 eV = 1.60 × 10−19 J.
EXECUTE: (a)
mv 2 ke 2
ke 2
= 2 and v =
.
r
mr
r
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23-24
Chapter 23
1
1 ke 2
1
(b) K = mv 2 =
=− U
2
2 r
2
1
1 ke 2
1 k (1.60 × 10−19 C) 2
=−
= −2.17 × 10−18 J = −13.6 eV.
(c) E = K + U = U = −
2
2 r
2 5.29 × 10−11 m
23.57.
EVALUATE: The total energy is negative, so the electron is bound to the proton. Work must be done on
the electron to take it far from the proton.
G
G
G
IDENTIFY and SET UP: Calculate the components of E from Eq. (23.19). Eq. (21.3) gives F from E .
EXECUTE: (a) V = Cx 4/3
C = V/x 4/3 = 240 V/(13.0 × 10−3 m)4/3 = 7.85 × 104 V/m 4/3
∂V
4
= − Cx1/3 = −(1.05 × 105 V/m 4/3 ) x1/3
∂x
3
G
The minus sign means that E x is in the − x -direction, which says that E points from the positive anode
(b) E x = −
toward the negative cathode.
G
G
(c) F = qE so Fx = − eE x = 43 eCx1/3
Halfway between the electrodes means x = 6.50 × 10−3 m.
Fx = 43 (1.602 × 10−19 C)(7.85 × 104 V/m 4/3 )(6.50 × 10−3 m)1/3 = 3.13 × 10−15 N
Fx is positive, so the force is directed toward the positive anode.
G
EVALUATE: V depends only on x, so E y = Ez = 0. E is directed from high potential (anode) to low
23.58.
potential (cathode). The electron has negative charge, so the force on it is directed opposite to the electric
field.
IDENTIFY: At each point (a and b), the potential is the sum of the potentials due to both spheres. The
voltmeter reads the difference between these two potentials. The spheres behave like point charges since
the meter is connected to the surface of each one.
SET UP: (a) Call a the point on the surface of one sphere and b the point on the surface of the other
sphere, call r the radius of each sphere and call d the center-to-center distance between the spheres. The
potential difference Vba between points a and b is then
Vb – Va = Vba =
1 ⎡ −q
q
−q ⎞ ⎤
2q ⎛ 1
1⎞
⎛q
+
−⎜ +
− ⎟.
⎟ =
⎜
4π ⑀0 ⎢⎣ r d − r ⎝ r d − r ⎠ ⎥⎦ 4π ⑀ 0 ⎝ d − r r ⎠
EXECUTE: Substituting the numbers gives
⎛
⎞
1
1
6
Vb – Va = 2(250 μ C) (9.00 × 109 N ⋅ m 2 /C2 ) ⎜
−
⎟ = –12.0 × 10 V. The meter reads 12.0 MV.
0.750
m
0.250
m
⎝
⎠
(b) Since Vb – Va is negative, Va > Vb , so point a is at the higher potential.
23.59.
EVALUATE: An easy way to see that the potential at a is higher than the potential at b is that it would
require positive work to move a positive test charge from b to a since this charge would be attracted by the
negative sphere and repelled by the positive sphere.
kq q
IDENTIFY: U = 1 2
r
SET UP: Eight charges means there are 8(8 − 1)/2 = 28 pairs. There are 12 pairs of q and − q separated by
d, 12 pairs of equal charges separated by
2d and 4 pairs of q and −q separated by
3d .
2
4 ⎞
12kq ⎛
1
1 ⎞
⎛ 12 12
2
EXECUTE: (a) U = kq 2 ⎜ − +
−
+
⎟ = − d ⎜1 −
⎟ = −1.46q /π ⑀ 0d
d
2d
3d ⎠
2 3 3⎠
⎝
⎝
EVALUATE: (b) The fact that the electric potential energy is less than zero means that it is energetically
favorable for the crystal ions to be together.
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Electric Potential
23.60.
23-25
kq1q2
. For part (b) apply conservation of energy.
r
SET UP: Let q1 = 2.00 μ C and q2 = −3.50 μ C. Let ra = 0.250 m and rb → ∞.
IDENTIFY: For two small spheres, U =
EXECUTE: (a) U =
(8.99 × 109 N ⋅ m 2 /C2 )(2.00 × 10−6 C)(−3.50 × 10−6 C)
= −0.252 J
0.250 m
(b) Kb = 0. U b = 0. U a = −0.252 J. K a + U a = Kb + U b gives K a = 0.252 J. K a = 12 mva2 , so
va =
23.61.
2Ka
2(0.252 J)
=
= 18.3 m/s
m
1.50 × 10−3 kg
EVALUATE: As the sphere moves away, the attractive electrical force exerted by the other sphere does
negative work and removes all the kinetic energy it initially had. Note that it doesn’t matter which sphere is
held fixed and which is shot away; the answer to part (b) is unaffected.
(a) IDENTIFY: Use Eq. (23.10) for the electron and each proton.
SET UP: The positions of the particles are shown in Figure 23.61a.
r = (1.07 × 10−10 m)/2 = 0.535 × 10−10 m
Figure 23.61a
EXECUTE: The potential energy of interaction of the electron with each proton is U =
1 ( −e 2 )
, so the
4π ⑀ 0 r
total potential energy is
U =−
2e 2
2(8.988 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C)2
=−
= −8.60 × 10−18 J
4π ⑀0r
0.535 × 10−10 m
U = −8.60 × 10−18 J(1 eV/1.602 × 10−19 J) = −53.7 eV
EVALUATE: The electron and proton have charges of opposite signs, so the potential energy of the system
is negative.
(b) IDENTIFY and SET UP: The positions of the protons and points a and b are shown in Figure 23.61b.
rb = ra2 + d 2
ra = r = 0.535 × 10−10 m
Figure 23.61b
Apply K a + U a + Wother = Kb + U b with point a midway between the protons and point b where the
electron instantaneously has v = 0 (at its maximum displacement d from point a).
EXECUTE: Only the Coulomb force does work, so Wother = 0.
U a = −8.60 × 10−18 J (from part (a))
K a = 12 mv 2 = 12 (9.109 × 10−31 kg)(1.50 × 106 m/s) 2 = 1.025 × 10−18 J
Kb = 0
U b = −2ke 2 /rb
Then U b = K a + U a − Kb = 1.025 × 10−18 J − 8.60 × 10−18 J = −7.575 × 10−18 J.
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23-26
Chapter 23
rb = −
2ke 2
2(8.988 × 109 N ⋅ m 2 /C2 )(1.60 × 10−19 C)2
=−
= 6.075 × 10−11 m
Ub
−7.575 × 10−18 J
Then d = rb2 − ra2 = (6.075 × 10−11 m) 2 − (5.35 × 10−11 m) 2 = 2.88 × 10−11 m.
23.62.
EVALUATE: The force on the electron pulls it back toward the midpoint. The transverse distance the
electron moves is about 0.27 times the separation of the protons.
IDENTIFY: Apply ∑ Fx = 0 and ∑ Fy = 0 to the sphere. The electric force on the sphere is Fe = qE. The
potential difference between the plates is V = Ed .
SET UP: The free-body diagram for the sphere is given in Figure 23.62.
EXECUTE: T cosθ = mg and T sin θ = Fe gives
Fe = mg tanθ = (1.50 × 10−3 kg)(9.80 m/s2 )tan(30°) = 0.0085 N.
Fe = Eq =
Vq
Fd (0.0085 N)(0.0500 m)
and V =
=
= 47.8 V.
d
q
8.90 × 10−6 C
EVALUATE: E = V/d = 956 V/m. E = σ /⑀ 0 and σ = E⑀ 0 = 8.46 × 10−9 C/m 2 .
Figure 23.62
23.63.
(a) IDENTIFY: The potential at any point is the sum of the potentials due to each of the two charged
conductors.
SET UP: From Example 23.10, for a conducting cylinder with charge per unit length λ the potential
outside the cylinder is given by V = (λ /2π ⑀0 )ln( r0 /r ) where r is the distance from the cylinder axis and r0
is the distance from the axis for which we take V = 0. Inside the cylinder the potential has the same value
as on the cylinder surface. The electric field is the same for a solid conducting cylinder or for a hollow
conducting tube so this expression for V applies to both. This problem says to take r0 = b.
EXECUTE: For the hollow tube of radius b and charge per unit length −λ: outside V = −(λ /2π ⑀ 0 )ln(b/r );
inside V = 0 since V = 0 at r = b.
For the metal cylinder of radius a and charge per unit length λ:
outside V = (λ /2π ⑀ 0 )ln(b/r ), inside V = (λ /2π ⑀ 0 )ln(b/a ), the value at r = a.
(i) r < a; inside both V = (λ /2π ⑀ 0 )ln(b/a )
(ii) a < r < b; outside cylinder, inside tube V = (λ /2π ⑀0 )ln(b/r )
(iii) r > b; outside both the potentials are equal in magnitude and opposite in sign so V = 0.
(b) For r = a, Va = (λ /2π ⑀ 0 )ln(b/a ).
For r = b, Vb = 0.
Thus Vab = Va − Vb = (λ /2π ⑀0 )ln(b/a ).
(c) IDENTIFY and SET UP: Use Eq. (23.23) to calculate E.
∂V
λ ∂ ⎛b⎞
λ ⎛ r ⎞⎛ b ⎞
Vab 1
EXECUTE: E = −
=−
ln ⎜ ⎟ = −
.
⎜ ⎟⎜ − ⎟ =
2π ⑀ 0 ∂r ⎝ r ⎠
2π ⑀ 0 ⎝ b ⎠⎝ r 2 ⎠ ln(b/a ) r
∂r
(d) The electric field between the cylinders is due only to the inner cylinder, so Vab is not changed,
Vab = (λ /2π ⑀ 0 )ln(b/a ).
EVALUATE: The electric field is not uniform between the cylinders, so Vab ≠ E (b − a ).
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Electric Potential
23.64.
IDENTIFY: The wire and hollow cylinder form coaxial cylinders. Problem 23.63 gives E (r ) =
23-27
Vab 1
.
ln(b /a ) r
SET UP: a = 145 × 10−6 m, b = 0.0180 m.
EXECUTE: E =
Vab 1
and
ln(b /a ) r
Vab = E ln(b /a )r = (2.00 × 104 N/C)(ln (0.018 m/145 × 10−6 m))0.012 m = 1157 V.
23.65.
EVALUATE: The electric field at any r is directly proportional to the potential difference between the wire
and the cylinder.
G
G
G
G
IDENTIFY and SET UP: Use Eq. (21.3) to calculate F and then F = ma gives a. E = V/d .
G
G
G
G
G
EXECUTE: (a) FE = qE . Since q = −e is negative FE and E are in opposite directions; E is upward so
G
V
22.0 V
FE is downward. The magnitude of E is E = =
= 1.10 × 103 V/m = 1.10 × 103 N/C. The
d 0.0200 m
magnitude of FE is FE = q E = eE = (1.602 × 10−19 C)(1.10 × 103 N/C) = 1.76 × 10−16 N.
(b) Calculate the acceleration of the electron produced by the electric force:
F
1.76 × 10−16 N
= 1.93 × 1014 m/s 2 .
a= =
m 9.109 × 10−31 kg
EVALUATE: This acceleration is much larger than g = 9.80 m/s 2 , so the gravity force on the electron can
G
G
be neglected. FE is downward, so a is downward.
(c) IDENTIFY and SET UP: The acceleration is constant and downward, so the motion is like that of a
projectile. Use the horizontal motion to find the time and then use the time to find the vertical
displacement.
EXECUTE: x-component: v0 x = 6.50 × 106 m/s; a x = 0; x − x0 = 0.060 m; t = ?
x − x0 = v0 xt + 12 a xt 2 and the a x term is zero, so t =
x − x0
0.060 m
=
= 9.231 × 10−9 s.
v0 x
6.50 × 106 m/s
y-component: v0 y = 0; a y = 1.93 × 1014 m/s 2 ; t = 9.231 × 10−9 m/s; y − y0 = ?
y − y0 = v0 yt + 12 a yt 2 . y − y0 = 12 (1.93 × 1014 m/s 2 )(9.231× 10−9 s)2 = 0.00822 m = 0.822 cm.
(d) The velocity and its components as the electron leaves the plates are sketched in Figure 23.65.
vx = v0 x = 6.50 × 106 m/s (since a x = 0 )
v y = v0 y + a yt
v y = 0 + (1.93 × 1014 m/s 2 )(9.231 × 10−9 s)
v y = 1.782 × 106 m/s
Figure 23.65
tan α =
vy
1.782 × 106 m/s
= 0.2742 so α = 15.3°.
6.50 × 106 m/s
EVALUATE: The greater the electric field or the smaller the initial speed the greater the downward
deflection.
(e) IDENTIFY and SET UP: Consider the motion of the electron after it leaves the region between the
plates. Outside the plates there is no electric field, so a = 0. (Gravity can still be neglected since the
electron is traveling at such high speed and the times are small.) Use the horizontal motion to find the time
it takes the electron to travel 0.120 m horizontally to the screen. From this time find the distance downward
that the electron travels.
vx
=
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23-28
Chapter 23
EXECUTE: x-component: v0 x = 6.50 × 106 m/s; a x = 0; x − x0 = 0.120 m; t = ?
x − x0 = v0 xt + 12 axt 2 and the a x term is term is zero, so t =
x − x0
0.120 m
=
= 1.846 × 10−8 s.
v0 x
6.50 × 106 m/s
y-component: v0 y = 1.782 × 106 m/s (from part (b)); a y = 0; t = 1.846 × 10−8 m/s; y − y0 = ?
y − y0 = v0 yt + 12 a yt 2 = (1.782 × 106 m/s)(1.846 × 10−8 s) = 0.0329 m = 3.29 cm.
23.66.
EVALUATE: The electron travels downward a distance 0.822 cm while it is between the plates and a
distance 3.29 cm while traveling from the edge of the plates to the screen. The total downward deflection is
0.822 cm + 3.29 cm = 4.11 cm. The horizontal distance between the plates is half the horizontal distance
the electron travels after it leaves the plates. And the vertical velocity of the electron increases as it travels
between the plates, so it makes sense for it to have greater downward displacement during the motion after
it leaves the plates.
IDENTIFY: The charge on the plates and the electric field between them depend on the potential difference
across the plates.
σ
Qd
(a) SET UP: For two parallel plates, the potential difference between them is V = Ed = d =
.
⑀0
⑀0 A
EXECUTE: Solving for Q gives Q = ⑀ 0 AV /d =
(8.85 × 10−12 C2 /N ⋅ m 2 )(0.030 m)2 (25.0 V)
.
0.0050 m
Q = 3.98 × 10 –11 C = 39.8 pC.
(b) E = V/d = (25.0 V)/(0.0050 m) = 5.00 × 103 V/m.
(c) SET UP: Energy conservation gives 12 mv 2 = eV .
EXECUTE: Solving for v gives v =
23.67.
2eV
2(1.60 × 10−19 C)(25.0 V)
=
= 2.96 × 106 m/s.
m
9.11 × 10−31 kg
EVALUATE: Typical voltages in student laboratory work run up to around 25 V, so typical reasonable
values for the charge on the plates is about 40 pC and a reasonable value for the electric field is about
5000 V/m, as we found here. The electron speed would be about 3 million m/s.
V
1
(a) IDENTIFY and SET UP: Problem 23.63 derived that E = ab
, where a is the radius of the inner
ln(b/a ) r
cylinder (wire) and b is the radius of the outer hollow cylinder. The potential difference between the two
cylinders is Vab . Use this expression to calculate E at the specified r.
EXECUTE: Midway between the wire and the cylinder wall is at a radius of
r = ( a + b)/2 = (90.0 × 10−6 m + 0.140 m)/2 = 0.07004 m.
E=
Vab 1
50.0 × 103 V
=
= 9.71 × 104 V/m
ln(b/a ) r ln(0.140 m/90.0 × 10−6 m)(0.07004 m)
(b) IDENTIFY and SET UP: The electric force is given by Eq. (21.3). Set this equal to ten times the weight
of the particle and solve for q , the magnitude of the charge on the particle.
EXECUTE: FE = 10mg
10mg 10(30.0 × 10−9 kg)(9.80 m/s 2 )
=
= 3.03 × 10−11 C
4
E
9.71 × 10 V/m
EVALUATE: It requires only this modest net charge for the electric force to be much larger than the
weight.
(a) IDENTIFY: Calculate the potential due to each thin ring and integrate over the disk to find the
potential. V is a scalar so no components are involved.
SET UP: Consider a thin ring of radius y and width dy. The ring has area 2π y dy so the charge on the ring
is dq = σ (2π y dy ).
q E = 10mg and q =
23.68.
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Electric Potential
23-29
EXECUTE: The result of Example 23.11 then says that the potential due to this thin ring at the point on the
axis at a distance x from the ring is
V = ∫ dV =
σ
dV =
1
4π ⑀ 0
R
y dy
2⑀ 0 ∫0
dq
2
x +y
2
x +y
2
=
=
2
2πσ
4π ⑀ 0
y dy
x2 + y 2
σ ⎡
σ
R
x2 + y2 ⎤ =
⎥⎦ 0 2⑀ 0
2⑀ 0 ⎢⎣
( x + R − x)
2
2
EVALUATE: For x R this result should reduce to the potential of a point charge with Q = σπ R 2 .
x 2 + R 2 = x(1 + R 2 /x 2 )1/2 ≈ x(1 + R 2 /2 x 2 ) so
Then V ≈
x 2 + R 2 − x ≈ R 2 /2 x
σ R 2 σπ R 2
Q
=
=
, as expected.
2⑀0 2 x 4π ⑀ 0 x 4π ⑀0 x
(b) IDENTIFY and SET UP: Use Eq. (23.19) to calculate E x .
⎞ σx ⎛1
⎞
∂V
σ ⎛
x
1
=−
− 1⎟ =
⎜
⎜ −
⎟.
⎜
⎟
⎜
⎟
2
2
2
2
∂x
2⑀ 0 ⎝ x + R
x +R ⎠
⎠ 2⑀ 0 ⎝ x
EVALUATE: Our result agrees with Eq. (21.11) in Example 21.11.
G
b G
(a) IDENTIFY: Use Va − Vb = ∫ E ⋅ dl .
EXECUTE: E x = −
23.69.
a
SET UP: From Problem 22.42, E (r ) =
E (r ) =
λ
2π ⑀0 r
λr
2π ⑀ 0 R 2
for r ≤ R (inside the cylindrical charge distribution) and
for r ≥ R. Let V = 0 at r = R (at the surface of the cylinder).
EXECUTE: r > R
G
G G
G
Take point a to be at R and point b to be at r, where r > R. Let dl = dr . E and dr are both radially
G G
r
r
outward, so E ⋅ d r = E dr. Thus VR − Vr = ∫ E dr. Then VR = 0 gives Vr = − ∫ E dr. In this interval
R
R
(r > R), E (r ) = λ /2π ⑀ 0 r , so
Vr = − ∫
λ
λ r dr
λ
⎛r⎞
dr = −
=−
ln ⎜ ⎟ .
∫
R 2π ⑀ r
R r
2
2
π
π
⑀
⑀
⎝R⎠
0
0
0
r
EVALUATE: This expression gives Vr = 0 when r = R and the potential decreases (becomes a negative
number of larger magnitude) with increasing distance from the cylinder.
EXECUTE: r < R
G G
R
Take point a at r, where r < R, and point b at R. E ⋅ dr = E dr as before. Thus Vr − VR = ∫ E dr. Then
r
R
VR = 0 gives Vr = ∫ E dr. In this interval (r < R ), E ( r ) = λ r/2π ⑀0 R 2 , so
r
Vr = ∫
R
λr
r 2π ⑀ R
0
dr =
2
λ
∫
R
2π ⑀ 0 R 2 r
r dr =
⎛ R2 r 2 ⎞
− ⎟.
⎜
2 ⎟⎠
2π ⑀ 0 R 2 ⎜⎝ 2
λ
λ ⎛ ⎛r⎞ ⎞
⎜1 − ⎜ ⎟ ⎟ .
4π ⑀ 0 ⎜⎝ ⎝ R ⎠ ⎟⎠
2
Vr =
EVALUATE: This expression also gives Vr = 0 when r = R. The potential is λ /4π ⑀0 at r = 0 and
decreases with increasing r.
(b) EXECUTE: Graphs of V and E as functions of r are sketched in Figure 23.69.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
23-30
Chapter 23
Figure 23.69
EVALUATE: E at any r is the negative of the slope of V (r ) at that r (Eq. 23.23).
23.70.
IDENTIFY: Divide the rod into infinitesimal segments with charge dq. The potential dV due to the segment
1 dq
is dV =
. Integrate over the rod to find the total potential.
4π ⑀ 0 r
SET UP: dq = λ dl , with λ = Q/π a and dl = a dθ .
EXECUTE: dV =
23.71.
dq
1 λ dl
1 Q dl
1 Q dθ
1 π Q dθ
1 Q
=
=
=
=
. V=
.
∫
0
4π ⑀ 0 r
4π ⑀ 0 a
4π ⑀0 π a a 4π ⑀ 0 π a
4π ⑀0
πa
4π ⑀ 0 a
1
EVALUATE: All the charge of the ring is the same distance a from the center of curvature.
IDENTIFY: We must integrate to find the total energy because the energy to bring in more charge depends
on the charge already present.
SET UP: If ρ is the uniform volume charge density, the charge of a spherical shell or radius r and
thickness dr is dq = ρ 4π r 2 dr , and ρ = Q/(4/3 π R3 ). The charge already present in a sphere of radius r is
q = ρ (4/3 π r 3 ). The energy to bring the charge dq to the surface of the charge q is Vdq, where V is the
potential due to q, which is q/4π ⑀ 0r.
EXECUTE: The total energy to assemble the entire sphere of radius R and charge Q is sum (integral) of the
tiny increments of energy.
U = ∫ Vdq = ∫
23.72.
q
4π ⑀ 0r
dq = ∫
4
3
ρ π r3
3 ⎛ 1 Q2 ⎞
( ρ 4π r 2dr ) = ⎜
⎟
0 4π ⑀ r
5 ⎜⎝ 4π ⑀ 0 R ⎟⎠
0
R
where we have substituted ρ = Q/(4/3 π R3 ) and simplified the result.
EVALUATE: For a point charge, R → 0 so U → ∞, which means that a point charge should have infinite
self-energy. This suggests that either point charges are impossible, or that our present treatment of physics
is not adequate at the extremely small scale, or both.
G
G G
b G
IDENTIFY: Va − Vb = ∫ E ⋅ dl . The electric field is radially outward, so E ⋅ dl = E dr.
a
SET UP: Let a = ∞, so Va = 0.
EXECUTE: From Example 22.9, we have the following. For r > R: E =
For r < R: E =
kQr
R3
kQ
r
2
and V = − kQ ∫
r dr ′
∞ r ′2
=
kQ
.
r
and
R G
r G
G
G kQ kQ r
kQ kQ 1 2 r kQ kQ kQr 2 kQ ⎡
r2 ⎤
− 3 ∫ r ′dr ′ =
− 3 r′
=
+
−
=
−
V = − ∫ E ⋅ dr ′ − ∫ E ⋅ dr ′ =
3
⎢
⎥.
∞
R
R
R R R
R R 2
R 2 R 2 R3 2 R ⎢⎣
R 2 ⎥⎦
(b) The graphs of V and E versus r are sketched in Figure 23.72.
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No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Potential
23-31
EVALUATE: For r < R the potential depends on the electric field in the region r to ∞.
Figure 23.72
23.73.
IDENTIFY: The sphere no longer behaves as a point charge because we are inside of it. We know how the
electric field varies with distance from the center of the sphere and want to use this to find the potential
difference between the center and surface, which requires integration.
kQ ⎛
r2 ⎞
SET UP: Use the result of Problem 23.72. For r < R, V =
⎜⎜ 3 − 2 ⎟⎟ .
2R ⎝
R ⎠
3kQ
EXECUTE: At the center of the sphere, r = 0 and V1 =
. At the surface of the sphere, r = R and
2R
kQ (8.99 × 109 N ⋅ m 2 /C2 )(4.00 × 10−6 C)
kQ
=
= 3.60 × 105 V.
. The potential difference is V1− V2 =
R
2R
2(0.0500 m)
EVALUATE: To check our answer, we could actually do the integration. We can use the fact that
R
kQr
kQ R
kQ ⎛ R 2 ⎞ kQ
E = 3 so V1 − V2 = ∫ Edr = 3 ∫ rdr = 3 ⎜
.
⎟=
0
R
R 0
R ⎜⎝ 2 ⎟⎠ 2 R
IDENTIFY: For r < c, E = 0 and the potential is constant. For r > c, E is the same as for a point charge
kq
and V = .
r
SET UP: V∞ = 0
V2 =
23.74.
EXECUTE: (a) Points a, b and c are all at the same potential, so Va − Vb = Vb − Vc = Va − Vc = 0.
kq (8.99 × 109 N ⋅ m 2 /C 2 )(150 × 10−6 C)
=
= 2.25 × 106 V
R
0.60 m
(b) They are all at the same potential.
(c) Only Vc − V∞ would change; it would be −2.25 × 106 V.
Vc − V∞ =
23.75.
EVALUATE: The voltmeter reads the potential difference between the two points to which it is connected.
IDENTIFY and SET UP: Apply Fr = − dU/dr and Newton’s third law.
EXECUTE: (a) The electrical potential energy for a spherical shell with uniform surface charge density
and a point charge q outside the shell is the same as if the shell is replaced by a point charge at its center.
Since Fr = − dU/dr , this means the force the shell exerts on the point charge is the same as if the shell
were replaced by a point charge at its center. But by Newton’s third law, the force q exerts on the shell is
the same as if the shell were a point charge. But q can be replaced by a spherical shell with uniform
surface charge and the force is the same, so the force between the shells is the same as if they were both
replaced by point charges at their centers. And since the force is the same as for point charges, the
electrical potential energy for the pair of spheres is the same as for a pair of point charges.
(b) The potential for solid insulating spheres with uniform charge density is the same outside of the sphere
as for a spherical shell, so the same result holds.
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23-32
23.76.
Chapter 23
(c) The result doesn’t hold for conducting spheres or shells because when two charged conductors are
brought close together, the forces between them cause the charges to redistribute and the charges are no
longer distributed uniformly over the surfaces.
qq
kq q
EVALUATE: For the insulating shells or spheres, F = k 1 22 and U = 1 2 , where q1 and q2 are the
r
r
charges of the objects and r is the distance between their centers.
IDENTIFY: Apply Newton's second law to calculate the acceleration. Apply conservation of energy and
conservation of momentum to the motions of the spheres.
qq
kq q
SET UP: Problem 23.75 shows that F = k 1 22 and U = 1 2 , where q1 and q2 are the charges of the
r
r
objects and r is the distance between their centers.
EXECUTE: Maximum speed occurs when the spheres are very far apart. Energy conservation gives
kq1q2 1
1
2
2
= m50v50
+ m150v150
. Momentum conservation gives m50v50 = m150v150 and v50 = 3v150 .
r
2
2
r = 0.50 m. Solve for v50 and v150: v50 = 12.7 m/s, v150 = 4.24 m/s. Maximum acceleration occurs just
after spheres are released. ∑ F = ma gives
kq1q2
r2
= m150a150 .
(9 × 109 N ⋅ m 2 /C 2 )(10−5 C)(3 × 10−5 C)
23.77.
= (0.15 kg)a150 . a150 = 72.0 m/s 2 and a50 = 3a150 = 216 m/s 2 .
(0.50 m) 2
EVALUATE: The more massive sphere has a smaller acceleration and a smaller final speed.
IDENTIFY: Use Eq. (23.17) to calculate Vab .
SET UP: From Problem 22.45, for R ≤ r ≤ 2 R (between the sphere and the shell) E = Q/4π ⑀ 0r 2 .
Take a at R and b at 2R.
EXECUTE: Vab = Va − Vb = ∫
2R
R
E dr =
Q
2 R dr
Vab =
23.78.
Q ⎡ 1⎤
2R
Q ⎛1
1 ⎞
=
−
=
⎜ −
⎟
4π ⑀0 ∫R r 2 4π ⑀0 ⎢⎣ r ⎥⎦ R
4π ⑀ 0 ⎝ R 2 R ⎠
Q
8π ⑀0 R
EVALUATE: The electric field is radially outward and points in the direction of decreasing potential, so
the sphere is at higher potential than the shell.
G
b G
IDENTIFY: Va − Vb = ∫ E ⋅ dl
a
G
G G
SET UP: E is radially outward, so E ⋅ dl = E dr. Problem 22.44 shows that E ( r ) = 0 for r ≤ a,
E ( r ) = kq/r 2 for a < r < b, E ( r ) = 0 for b < r < c and E ( r ) = kq/r 2 for r > c.
kq
.
c
c G
G b G G kq
kq
(b) At r = b: Vb = − ∫ E ⋅ dr − ∫ E ⋅ dr =
−0= .
∞
c
c
c
G
G
G
c
a dr
G b
G a
G kq
⎡1 1 1 ⎤
(c) At r = a: Va = − ∫ E ⋅ dr − ∫ E ⋅ dr − ∫ E ⋅ dr =
− kq ∫ 2 = kq ⎢ − + ⎥
∞
c
b
b r
c
⎣c b a ⎦
EXECUTE: (a) At r = c: Vc = − ∫
c kq
∞ r2
dr =
⎡1 1 1 ⎤
(d) At r = 0: V0 = kq ⎢ − + ⎥ since it is inside a metal sphere, and thus at the same potential as its
⎣c b a ⎦
surface.
⎡1 1⎤
EVALUATE: The potential difference between the two conductors is Va − Vb = kq ⎢ − ⎥ .
⎣a b⎦
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Electric Potential
23.79.
23-33
IDENTIFY: Slice the rod into thin slices and use Eq. (23.14) to calculate the potential due to each slice.
Integrate over the length of the rod to find the total potential at each point.
(a) SET UP: An infinitesimal slice of the rod and its distance from point P are shown in Figure 23.79a.
Figure 23.79a
Use coordinates with the origin at the left-hand end of the rod and one axis along the rod. Call the axes
x′ and y′ so as not to confuse them with the distance x given in the problem.
EXECUTE: Slice the charged rod up into thin slices of width dx′. Each slice has charge dQ = Q (dx′/a )
and a distance r = x + a − x′ from point P. The potential at P due to the small slice dQ is
dV =
1 ⎛ dQ ⎞
1 Q ⎛ dx′ ⎞
⎜
⎟=
⎜
⎟.
4π ⑀ 0 ⎝ r ⎠ 4π ⑀ 0 a ⎝ x + a − x′ ⎠
Compute the total V at P due to the entire rod by integrating dV over the length of the rod ( x′ = 0 to x′ = a ):
V = ∫ dV =
a
Q
dx′
Q
Q
⎛x+a⎞
=
[ − ln( x+ a − x′)]0a =
ln ⎜
⎟.
4π ⑀ 0a ∫0 ( x + a − x′) 4π ⑀ 0a
4π ⑀ 0a ⎝ x ⎠
⎛x⎞
ln ⎜ ⎟ = 0.
4π ⑀0a ⎝ x ⎠
(b) SET UP: An infinitesimal slice of the rod and its distance from point R are shown in Figure 23.79b.
EVALUATE: As x → ∞, V →
Q
Figure 23.79b
dQ = (Q/a )dx′ as in part (a)
Each slice dQ is a distance r = y 2 + ( a − x′) 2 from point R.
EXECUTE: The potential dV at R due to the small slice dQ is
dV =
1 ⎛ dQ ⎞
1 Q
⎜
⎟=
4π ⑀ 0 ⎝ r ⎠ 4π ⑀ 0 a
V = ∫ dV =
Q
a
4π ⑀ 0a ∫ 0
dx′
2
y + ( a − x′) 2
dx′
y 2 + ( a − x′) 2
.
.
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23-34
Chapter 23
In the integral make the change of variable u = a − x′; du = −dx′
V =−
Q
du
0
4π ⑀ 0a ∫ a
y2 + u2
=-
Q
(
)
0
⎡ln u + y 2 + u 2 ⎤ .
⎥⎦ a
4π ⑀ 0a ⎢⎣
⎡ ⎛
2
2 ⎞⎤
⎡ln y − ln a + y 2 + a 2 ⎤ = Q ⎢ln ⎜ a + a + y ⎟ ⎥ .
⎥⎦ 4π ⑀ a ⎢ ⎜
⎟⎥
4π ⑀ 0a ⎣⎢
y
0
⎠⎦
⎣ ⎝
(The expression for the integral was found in Appendix B.)
⎛ y⎞
Q
EVALUATE: As y → ∞, V →
ln ⎜ ⎟ = 0.
4π ⑀0a ⎝ y ⎠
V =−
(
Q
(c) SET UP: part (a): V =
)
Q
⎛ x+a⎞
⎛ a⎞
ln ⎜
ln ⎜1 + ⎟ .
⎟=
4π ⑀0 a ⎝ x ⎠ 4π ⑀ 0a ⎝
x⎠
Q
From Appendix B, ln(1 + u ) = u − u 2 /2 . . . , so ln(1 + a/x ) = a/x − a 2 /2 x 2 and this becomes a/x when x is
large.
EXECUTE: Thus V →
part (b): V =
Q ⎛a⎞
Q
. For large x, V becomes the potential of a point charge.
⎜ ⎟=
4π ⑀ 0a ⎝ x ⎠ 4π ⑀ 0 x
⎡ ⎛ a + a2 + y 2 ⎞⎤
2 ⎞
⎛
⎢ ln ⎜
⎟ ⎥ = Q ln ⎜ a + 1 + a ⎟ .
⎟ ⎥ 4π ⑀0a ⎜ y
4π ⑀0a ⎢ ⎜
y
y 2 ⎟⎠
⎝
⎠⎦
⎣ ⎝
Q
From Appendix B,
1 + a 2 /y 2 = (1 + a 2 /y 2 )1/2 = 1 + a 2 /2 y 2 + …
Thus a/y + 1 + a 2 /y 2 → 1 + a/y + a 2 /2 y 2 + … → 1 + a/y. And then using ln(1 + u ) ≈ u gives
V→
Q
Q ⎛a⎞
Q
ln(1 + a/y ) →
.
⎜ ⎟=
4π ⑀ 0a
4π ⑀0a ⎝ y ⎠ 4π ⑀ 0 y
EVALUATE: For large y, V becomes the potential of a point charge.
23.80.
IDENTIFY: The potential at the surface of a uniformly charged sphere is V =
kQ
.
R
4
SET UP: For a sphere, V = π R3. When the raindrops merge, the total charge and volume are conserved.
3
EXECUTE: (a) V =
kQ k (−3.60 × 10−12 C)
=
= −49.8 V.
R
6.50 × 10−4 m
(b) The volume doubles, so the radius increases by the cube root of two: Rnew = 3 2 R = 8.19 × 10−4 m and
the new charge is Qnew = 2Q = −7.20 × 10−12 C. The new potential is
Vnew =
23.81.
kQnew k (−7.20 × 10−12 C)
=
= −79.0 V.
Rnew
8.19 × 10−4 m
EVALUATE: The charge doubles but the radius also increases and the potential at the surface increases by
2
only a factor of 1/3 = 22/3 ≈ 1.6.
2
(a) IDENTIFY and SET UP: The potential at the surface of a charged conducting sphere is given by
1 q
Example 23.8: V =
. For spheres A and B this gives
4π ⑀0 R
VA =
QA
QB
and VB =
.
4π ⑀ 0 RA
4π ⑀ 0 RB
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Electric Potential
23-35
EXECUTE: VA = VB gives QA/4π ⑀ 0 RA = QB /4π ⑀ 0 RB and QB /QA = RB /RA . And then RA = 3RB implies
QB /QA = 1/3.
(b) IDENTIFY and SET UP: The electric field at the surface of a charged conducting sphere is given in
Example 22.5:
E=
1 q
.
4π ⑀ 0 R 2
EXECUTE: For spheres A and B this gives
EA =
QA
4π ⑀0 RA2
and EB =
QB
4π ⑀0 RB2
.
EB ⎛ QB ⎞⎛ 4π ⑀0 RA2 ⎞
2
2
=⎜
⎟⎜
⎟⎟ = QB /QA ( RA /RB ) = (1/3)(3) = 3.
E A ⎜⎝ 4π ⑀ 0 RB2 ⎟⎜
Q
A
⎠⎝
⎠
23.82.
EVALUATE: The sphere with the larger radius needs more net charge to produce the same potential. We
can write E = V/R for a sphere, so with equal potentials the sphere with the smaller R has the larger E.
IDENTIFY: Apply conservation of energy, K a + U a = Kb + U b .
SET UP: Assume the particles initially are far apart, so U a = 0. The alpha particle has zero speed at the
distance of closest approach, so Kb = 0. 1 eV = 1.60 × 10−19 J. The alpha particle has charge +2e and the
lead nucleus has charge +82e.
EXECUTE: Set the alpha particle’s kinetic energy equal to its potential energy: K a = U b gives
11.0 MeV =
23.83.
k (2e)(82e)
k (164)(1.60 × 10−19 C) 2
and r =
= 2.15 × 10−14 m.
r
(11.0 × 106 eV)(1.60 × 10−19 J/eV)
EVALUATE: The calculation assumes that at the distance of closest approach the alpha particle is outside
the radius of the lead nucleus.
IDENTIFY and SET UP: The potential at the surface is given by Example 23.8 and the electric field at the
surface is given by Example 22.5. The charge initially on sphere 1 spreads between the two spheres such as
to bring them to the same potential.
1 Q1
1 Q1
EXECUTE: (a) E1 =
= R1E1
, V1 =
4π ⑀ 0 R12
4π ⑀0 R1
(b) Two conditions must be met:
1) Let q1 and q2 be the final charges of each sphere. Then q1 + q2 = Q1 (charge conservation)
2) Let V1 and V2 be the final potentials of each sphere. All points of a conductor are at the same potential,
so V1 = V2 .
V1 = V2 requires that
1 q1
1 q2
and then q1/R1 = q2 /R2
=
4π ⑀0 R1 4π ⑀ 0 R2
q1R2 = q2 R1 = (Q1 − q1 ) R1
This gives q1 = ( R1/[R1 + R2 ])Q1 and q2 = Q1 − q1 = Q1(1 − R1/[R1 + R2 ]) = Q1( R2 /[R1 + R2 ]).
(c) V1 =
1 q1
Q1
1 q2
Q1
=
and V2 =
=
, which equals V1 as it should.
4π ⑀ 0 R1 4π ⑀ 0 ( R1 + R2 )
4π ⑀ 0 R2 4π ⑀ 0 ( R1 + R2 )
(d) E1 =
V1
Q1
V
Q1
=
. E2 = 2 =
.
R1 4π ⑀ 0 R1 ( R1 + R2 )
R2 4π ⑀ 0 R2 ( R1 + R2 )
EVALUATE: Part (a) says q2 = q1( R2 /R1). The sphere with the larger radius needs more charge to produce
the same potential at its surface. When R1 = R2 , q1 = q2 = Q1/2. The sphere with the larger radius has the
smaller electric field at its surface.
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23-36
23.84.
Chapter 23
G
b G
IDENTIFY: Apply Va − Vb = ∫ E ⋅ dl .
a
SET UP: From Problem 22.65, for r ≥ R, E =
EXECUTE: (a) r ≥ R: E =
kQ
2
⇒ V = −∫
r kQ
∞ r ′2
kQ
r
2
dr′ =
. For r ≤ R, E =
kQ ⎡ r 3
r4 ⎤
4 3 − 3 4 ⎥.
2 ⎢
r ⎢⎣ R
R ⎥⎦
kQ
, which is the potential of a point charge.
r
r
3
⎡
kQ r
r4 ⎤
(b) r ≤ R: E = 2 ⎢ 4 3 − 3 4 ⎥ and
r ⎢⎣ R
R ⎥⎦
⎤
R
r
kQ ⎡
r2
R 2 r 3 R3 ⎤ kQ ⎡ r 3
r2
− 2 2 + 2⎥ .
V = − ∫ Edr ′ − ∫ Edr′ =
1− 2 2 + 2 2 + 3 − 3 ⎥ =
⎢
⎢
3
∞
R
R ⎣⎢
R
R
R
R ⎦⎥ R ⎣⎢ R
R
⎦⎥
EVALUATE: At r = R, V =
23.85.
kQ
2kQ
. The electric field is radially outward and
. At r = 0, V =
R
R
V increases as r decreases.
IDENTIFY: Apply conservation of energy: E1 = E2 .
SET UP: In the collision the initial kinetic energy of the two particles is converted into potential energy at
the distance of closest approach.
EXECUTE: (a) The two protons must approach to a distance of 2rp , where rp is the radius of a proton.
2
k (1.60 × 10−19 C) 2
⎡1
⎤ ke
= 7.58 × 106 m/s.
E1 = E2 gives 2 ⎢ mpv 2 ⎥ =
and v =
2(1.2 × 10−15 m)(1.67 × 10−27 kg)
⎣2
⎦ 2rp
(b) For a helium-helium collision, the charges and masses change from (a) and
v=
k (2(1.60 × 10−19 C))2
(3.5 × 10−15 m)(2.99)(1.67 × 10−27 kg)
= 7.26 × 106 m/s.
2
(c) K =
mpv
3kT mv 2
(1.67 × 10−27 kg)(7.58 × 106 m/s) 2
=
=
= 2.3 × 109 K.
. TP =
2
2
3k
3(1.38 × 10−23 J/K)
THe =
23.86.
mHev 2 (2.99)(1.67 × 10−27 kg)(7.26 × 106 m/s) 2
=
= 6.4 × 109 K.
3k
3(1.38 × 10−23 J/K)
(d) These calculations were based on the particles’ average speed. The distribution of speeds ensures that
there is always a certain percentage with a speed greater than the average speed, and these particles can
undergo the necessary reactions in the sun’s core.
EVALUATE: The kinetic energies required for fusion correspond to very high temperatures.
G
b G
W
IDENTIFY and SET UP: Apply Eq. (23.20). a →b = Va − Vb and Va − Vb = ∫ E ⋅ dl .
a
q0
G
∂V ˆ ∂V ˆ ∂V ˆ
EXECUTE: (a) E = −
i−
j−
k = −2 Axiˆ + 6 Ayˆj − 2 Azkˆ
∂x
∂y
∂z
(b) A charge is moved in along the z-axis. The work done is given by
0 G
0
W
6.00 × 10−5 J
= 640 V/m 2 .
W = q ∫ E ⋅ kˆdz = q ∫ (−2 Az )dz = + ( Aq ) z02 . Therefore, A = a →2 b =
−6
2
z0
z0
qz0
(1.5 × 10 C)(0.250 m)
G
2
ˆ
ˆ
(c) E (0,0,0.250) = −2(640 V/m )(0.250 m) k = −(320 V/m)k .
(d) In every plane parallel to the xz -plane, y is constant, so V ( x,y,z ) = Ax 2 + Az 2 − C , where C = 3 Ay 2 .
V +C
= R 2 , which is the equation for a circle since R is constant as long as we have constant
A
potential on those planes.
x2 + z 2 =
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Electric Potential
(e) V = 1280 V and y = 2.00 m, so x 2 + z 2 =
1280 V + 3(640 V/m 2 )(2.00 m) 2
640 V/m 2
23-37
= 14.0 m 2 and the radius of
the circle is 3.74 m.
23.87.
G
EVALUATE: In any plane parallel to the xz-plane, E projected onto the plane is radial and hence
perpendicular to the equipotential circles.
IDENTIFY: Apply conservation of energy to the motion of the daughter nuclei.
SET UP: Problem 23.72 shows that the electrical potential energy of the two nuclei is the same as if all
their charge was concentrated at their centers.
EXECUTE: (a) The two daughter nuclei have half the volume of the original uranium nucleus, so their
7.4 × 10−15 m
radii are smaller by a factor of the cube root of 2: r =
= 5.9 × 10−15 m.
3
2
(b) U =
k (46e) 2 k (46) 2 (1.60 × 10−19 C) 2
=
= 4.14 × 10−11 J. U = 2 K , where K is the final kinetic energy of
2r
1.18 × 10−14 m
each nucleus. K = U/2 = (4.14 × 10−11 J)/2 = 2.07 × 10−11 J.
(c) If we have 10.0 kg of uranium, then the number of nuclei is
10.0 kg
= 2.55 × 1025 nuclei. And each releases energy U, so
n=
(236 u)(1.66 × 10−27 kg/u)
E = nU = (2.55 × 1025 )(4.14 × 10−11 J) = 1.06 × 1015 J = 253 kilotons of TNT.
23.88.
(d) We could call an atomic bomb an “electric” bomb since the electric potential energy provides the
kinetic energy of the particles.
EVALUATE: This simple model considers only the electrical force between the daughter nuclei and
neglects the nuclear force.
∂V
IDENTIFY and SET UP: In part (a) apply E = −
. In part (b) apply Gauss’s law.
∂r
∂V
ρ a2 ⎡ r
r2 ⎤ ρ a ⎡ r r2 ⎤
∂V
= − 0 ⎢ −6 2 + 6 3 ⎥ = 0 ⎢ − 2 ⎥ . For r ≥ a, E = −
EXECUTE: (a) For r ≤ a, E = −
= 0.
18⑀ 0 ⎢⎣ a
∂r
∂r
a ⎥⎦ 3⑀ 0 ⎢⎣ a a ⎥⎦
G
E has only a radial component because V depends only on r.
Q
ρ a ⎡ r r2 ⎤
(b) For r ≤ a, Gauss’s law gives Er 4π r 2 = r = 0 ⎢ − 2 ⎥ 4π r 2 and
3⑀ 0 ⎣⎢ a a ⎦⎥
⑀0
Er + dr 4π ( r 2 + 2rdr ) =
Qr + dr
=
⑀0
2
⎢
3⑀ 0 ⎣⎢
a
−
( r 2 + 2rdr ) ⎤
2
⎥ 4π ( r + 2rdr ). Therefore,
a2
⎦⎥
2
2 2r 1 ⎤
ρ ⎡ 4r ⎤
⎡ 4r ⎤
− 2 + − 2 + ⎥ and ρ (r ) = 0 ⎢3 − ⎥ = ρ0 ⎢1 − ⎥ .
⎢
⑀0
⑀0
a a
a⎦
a⎦
3⑀0
3 ⎣
⎣ 3a ⎦
⎣ a
(c) For r ≥ a, ρ (r ) = 0, so the total charge enclosed will be given by
Qr + dr − Qr
≈
ρ0a 4π r dr ⎡ 2r
2
a⎡ 2
a
⎡1
4r 3 ⎤
r4 ⎤
Q = 4π ∫ ρ (r )r dr = 4πρ0 ∫ ⎢ r −
dr = 4πρ0 ⎢ r 3 − ⎥ = 0.
⎥
0
0
3a ⎥⎦
3a ⎥⎦
⎢⎣
⎢⎣ 3
0
EVALUATE: Apply Gauss’s law to a sphere of radius r > R. The result of part (c) says that Qencl = 0, so
E = 0. This agrees with the result we calculated in part (a)
IDENTIFY: Angular momentum and energy must be conserved.
SET UP: At the distance of closest approach the speed is not zero. E = K + U . q1 = 2e, q2 = 82e.
a
23.89.
=
ρ ( r )4π r dr
ρ0a ⎡ r + dr
EXECUTE: mv1b = mv2 r2 . E1 = E2 gives E1 =
1
kq q
mv2 2 + 1 2 . E1 = 11 MeV = 1.76 × 10−12 J. r2 is the
2
r2
⎛b⎞
b 2 kq q
distance of closest approach. Substituting in for v2 = v1 ⎜ ⎟ we find E1 = E1 2 + 1 2 .
r2
r2
⎝ r2 ⎠
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23-38
Chapter 23
( E1 )r22 − (kq1q2 ) r2 − E1b 2 = 0. For b = 10−12 m, r2 = 1.01 × 10−12 m. For b = 10−13 m, r2 = 1.11 × 10−13 m.
And for b = 10−14 m, r2 = 2.54 × 10−14 m.
23.90.
EVALUATE: As b decreases the collision is closer to being head-on and the distance of closest approach
decreases. Problem 23.82 shows that the distance of closest approach is 2.15 × 10−14 m when b = 0.
IDENTIFY: Consider the potential due to an infinitesimal slice of the cylinder and integrate over the length
of the cylinder to find the total potential. The electric field is along the axis of the tube and is given by
∂V
E=−
.
∂x
SET UP: Use the expression from Example 23.11 for the potential due to each infinitesimal slice. Let the
slice be at coordinate z along the x-axis, relative to the center of the tube.
EXECUTE: (a) For an infinitesimal slice of the finite cylinder, we have the potential
k dQ
kQ
dz
=
. Integrating gives
dV =
2
2
L
( x − z) + R
( x − z )2 + R 2
V=
kQ L / 2
dz
kQ L /2− x
du
=
where u = x − z. Therefore,
∫
∫
−
− L /2 − x
L
/2
2
2
2
L
L
( x − z) + R
u + R2
V=
kQ ⎡ ( L/2 − x)2 + R 2 + ( L/2 − x) ⎤
⎥ on the cylinder axis.
ln ⎢
L ⎢ ( L/2 + x) 2 + R 2 − L/2 − x ⎥
⎣
⎦
(b) For L << R, V ≈
V≈
kQ ⎡ ( L/2 − x) 2 + R 2 + L/2 − x ⎤ kQ ⎡ x 2 − xL + R 2 + L/2 − x ⎤
⎥≈
⎥.
ln ⎢
ln ⎢
L ⎢ ( L/2 + x)2 + R 2 − L/2 − x ⎥ L ⎢ x 2 + xL + R 2 − L/2 − x ⎥
⎣
⎦
⎣
⎦
kQ ⎡ 1 − xL/(R 2 + x 2 ) + ( L/2 − x)/ R 2 + x 2 ⎤ kQ ⎡ 1 − xL/2( R 2 + x 2 ) + ( L/2 − x)/ R 2 + x 2 ⎤
⎥=
⎥.
ln ⎢
ln ⎢
L ⎢ 1 + xL/(R 2 +x 2 ) + ( − L/2 − x)/ R 2 + x 2 ⎥ L ⎢1 + xL/2( R 2 + x 2 ) + (− L/2 − x)/ R 2 + x 2 ⎥
⎣
⎦
⎣
⎦
⎤
⎡
⎤⎞
kQ ⎡1 + L/2 R 2 + x 2 ⎤ kQ ⎛ ⎡
L
L
⎥=
⎜ ln ⎢1 +
ln ⎢
⎥ − ln ⎢1 −
⎥ ⎟.
2
2 ⎟
L ⎢1 − L/2 R 2 + x 2 ⎥ L ⎜⎝ ⎣⎢ 2 R 2 + x 2 ⎦⎥
⎣⎢ 2 R + x ⎦⎥ ⎠
⎣
⎦
kQ
2L
kQ
V≈
, which is the same as for a ring.
=
2
L 2 x2 + R2
x + R2
V≈
∂V
=
(c) E x = −
∂x
2kQ
( ( L − 2 x) + 4 R − ( L + 2 x) + 4 R )
2
2
2
2
( L − 2 x) 2 + 4 R 2 ( L + 2 x) 2 + 4 R 2
EVALUATE: For L << R the expression for E x reduces to that for a ring of charge, as given in Example
23.91.
23.14.
IDENTIFY: When the oil drop is at rest, the upward force q E from the electric field equals the
downward weight of the drop. When the drop is falling at its terminal speed, the upward viscous force
equals the downward weight of the drop.
4
SET UP: The volume of the drop is related to its radius r by V = π r 3.
3
EXECUTE: (a) Fg = mg =
(b)
4π r 3
4π ρ r 3 gd
.
ρ g. Fe = q E = q VAB /d . Fe = Fg gives q =
3
3 VAB
4π r 3
9η vt
. Using this result to replace r in the expression in part (a) gives
ρ g = 6πη rvt gives r =
3
2ρ g
3
q =
4π ρ gd ⎡ 9η vt ⎤
d η 3vt3
.
⎢
⎥ = 18π
3 VAB ⎢⎣ 2 ρ g ⎥⎦
VAB 2 ρ g
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Electric Potential
(c) q = 18π
23-39
10−3 m (1.81 × 10−5 N ⋅ s/m 2 )3 (1.00 × 10−3 m/39.3 s)3
= 4.80 × 10−19 C = 3e. The drop has
9.16 V
2(824 kg/m3 )(9.80 m/s 2 )
acquired three excess electrons.
r=
9(1.81 × 10−5 N ⋅ s/m 2 )(1.00 × 10−3 m/39.3 s)
2(824 kg/m3 )(9.80 m/s 2 )
= 5.07 × 10−7 m = 0.507 μ m.
⎛ 4π r 3 ⎞
EVALUATE: The weight of the drop is ⎜
ρ g = 4.4 × 10−15 N. The density of air at room temperature
⎜ 3 ⎟⎟
⎝
⎠
is 1.2 kg/m3 , so the buoyancy force is ρairVg = 6.4 × 10−18 N and can be neglected.
23.92.
IDENTIFY: vcm =
m1v1 + m2v2
m1 + m2
SET UP: E = K1 + K 2 + U , where U =
EXECUTE: (a) vcm =
kq1q2
.
r
(6 × 10−5 kg)(400m/s) + (3 × 10−5 kg)(1300 m/s)
6.0 × 10−5 kg + 3.0 × 10−5 kg
= 700 m/s
1
1
kq q
1
2
. After expanding the center of mass velocity and
(b) Erel = m1v12 + m2v2 2 + 1 2 − (m1 + m2 )vcm
2
2
2
r
1 m1m2
kq q
1
kq q
[v12 + v22 − 2v1v2 ] + 1 2 = μ (v1 − v2 ) 2 + 1 2 .
collecting like terms Erel =
2 m1 + m2
r
2
r
1
k (2.0 × 10−6 C)(−5.0 × 10−6 C)
= −1.9 J
(c) Erel = (2.0 × 10−5 kg)(900 m/s) 2 +
2
0.0090 m
(d) Since the energy is less than zero, the system is “bound.”
kq q
(e) The maximum separation is when the velocity is zero: −1.9 J = 1 2 gives
r
r=
k (2.0 × 10−6 C)(−5.0 × 10−6 C)
= 0.047 m.
−1.9 J
(f) Now using v1 = 400 m/s, and v2 = 1800 m/s, we find Erel = +9.6 J. The particles do escape, and the
final relative velocity is v1 − v2 =
2 Erel
μ
=
2(9.6 J)
2.0 × 10−5 kg
= 980 m/s.
EVALUATE: For an isolated system the velocity of the center of mass is constant and the system must
retain the kinetic energy associated with the motion of the center of mass.
© Copyright 2012 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.
No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.