973
Chapter 24
1. (a) An ampere is a coulomb per second, so
84 A
h
84
C
h
s
3600
s
h
3.0
10
5
C.
(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. If the electric potential is zero at infinity then at the surface of a uniformly charged
sphere it is
V = q
/4
0
R
, where
q
is the charge on the sphere and
R
is the sphere radius.
Thus
q
= 4
0
RV
and the number of electrons is
6
5
9
2
2
19
1.0
10
m
400V
4
2.8
10 .
8.99
10 N m
C
1.60
10
C
q
R V
n
e
e
4. (a)
15
19
4
4
3.3 10
N
1.60
10
C
2.1 10 N C
2.1 10 V/m.
E
F e
(b)
V
E
s
2.4
10
4
N C
0.12m
2.9
10
3
V.
5. 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
s
E dx
0
x
V
s
Ex
,
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
2
0
V
2 8.85
10
12
C
2
N
m
2
50V
0.10
10
6
C m
2
8.8
10
3
m.

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CHAPTER 24
974
6. (a)
V
B
– V
A
=
U
/
q
= –
W
/(–
e
) = – (4.78
10
–19
J)/(–1.60
10
–19
C) = 2.99 V.
(b)
V
C
– V
A
= V
B
– V
A
= 2.99 V.
(c)
V
C
– V
B
= 0 (since
C
and
B
are on the same equipotential line).
7. We connect
A
to the origin with a line along the
y
axis, along which there is no change
of potential (Eq. 24-18:
E
d
r
s
0
). Then, we connect the origin to
B
with a line along
the
x
axis, along which the change in potential is
V
r
E
d
r
s
4.00
xdx
4.00
4
2
2
0
4
0
x
4
which yields
V
B
– V
A
= –32.0 V.
8. (a) By Eq. 24-18, the change in potential is the negative of the “area” under the curve.
Thus, using the area-of-a-triangle formula, we have
2
0
1
10
2
10
2
x
V
E ds
which yields
V
= 20 V.
(b) For any region within
0
3m,
x
E ds
r
r
is positive, but for any region for which
x
> 3 m it is negative. Therefore,
V = V
max
occurs at
x
= 3 m.
3
0
1
10
3
10
2
x
V
E ds
which yields
V
max
= 25 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 25 V. Using the formula for a
triangle (3 <
x
< 4) and a rectangle (4 <
x < X
), we require
1
1
10
4
10
25.

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