Chapter 24
1. (a) An Ampere is a Coulomb per second, so
84
84
3600
30
10
5
A h
C h
s
s
h
C
F
H
G
I
K
J
F
H
G
I
K
J
.
.
(b) The change in potential energy is
U = q
V
= (3.0
10
5
C)(12 V) = 3.6
10
6
J.
2. The magnitude is
U = e
V
= 1.2
10
9
eV = 1.2 GeV.
3. The electric field produced by an infinite sheet of charge has magnitude
E
=
/2
0
,
where
is the surface charge density. The field is normal to the sheet and is uniform.
Place the origin of a coordinate system at the sheet and take the
x
axis to be parallel to the
field and positive in the direction of the field. Then the electric potential is
V
V
E dx
V
Ex
s
x
s
z
0
,
where
V
s
is the potential at the sheet. The equipotential surfaces are surfaces of constant
x
; that is, they are planes that are parallel to the plane of charge. If two surfaces are
separated by
x
then their potentials differ in magnitude by
V = E
x
= (
/2
0
)
x
.
Thus,
x
V
2
2 885
10
50
010
10
88
10
0
12
2
6
3
.
.
.
.
C
N m
V
C m
m
2
2
c
hb g
4. (a)
V
B
– V
A
=
U
/
q
= –
W
/(–
e
) = – (3.94
10
–19
J)/(–1.60
10
–19
C) = 2.46 V.
(b)
V
C
– V
A
= V
B
– V
A
= 2.46 V.
(c)
V
C
– V
B
= 0 (Since
C
and
B
are on the same equipotential line).
5. (a)
E
F e
39
10
160
10
2 4
10
15
19
4
.
.
.
.
N
C
N C
c
hc
h
(b)
V
E s
2 4
10
012
2 9
10
4
3
.
.
.
.
N C
m
V
c
hb
g
6. (a) By Eq. 2418, the change in potential is the negative of the
“area”
under the curve.
Thus, using the areaofatriangle formula, we have
995
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CHAPTER 24
V
E ds
x
z
10
1
2
2
20
0
2
bgbg
which yields
V
= 30 V.
(b) For any region within
0
3
z
x
E ds
m,
is positive, but for any region for which
x
> 3 m it is negative. Therefore,
V = V
max
occurs at
x
= 3 m.
V
E ds
x
z
10
1
2
3
20
0
3
bgbg
which yields
V
max
= 40 V.
(c) In view of our result in part (b), we see that now (to find
V
= 0) we are looking for
some
X
> 3 m such that the
“area”
from
x
= 3 m to
x = X
is 40 V. Using the formula for a
triangle (3 <
x
< 4) and a rectangle (4 <
x < X
), we require
1
2
1
20
4
20
40
bgbgb
gbg
X
.
Therefore,
X
= 5.5 m.
7. (a) The work done by the electric field is (in SI units)
19
12
21
0
0
0
12
0
0
0
(1.60 10
)(5.80 10
)(0.0356)
1.87 10
J.
2
2
2(8.85 10
)
f
d
i
q
q
d
W
q E ds
dz
ﾴ
ﾴ
�
�
ﾴ
�
�
(b) Since
V – V
0
= –
W
/
q
0
= –
z
/2
0
, with
V
0
set to be zero on the sheet, the electric
potential at
P
is (in SI units)
12
2
12
0
(5.80 10
)(0.0356)
1.17
10
V.
2
2(8.85 10
)
z
V
ﾴ
ﾴ
ﾴ
8. We connect
A
to the origin with a line along the
y
axis, along which there is no change
of potential (Eq. 2418:
E ds
z
0
). Then, we connect the origin to
B
with a line along
the
x
axis, along which the change in potential is
V
E ds
xdx
x
F
H
G
I
K
J
z
z
4 00
4 00
4
2
2
0
4
0
4
.
.
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 Fall '08
 Many
 Physics, Electrostatics, Conservation Of Energy, Energy, Potential Energy, Electric charge

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