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1
Chapter 24 Class Examples
7.
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
.
.
which yields
V
B
– V
A
= –32.0 V.
18.
When the charge
q
2
is infinitely far away, the potential at the origin is due only to the
charge
q
1
:
V
1
=
1
0
4
q
d
=
5.76
×
10
7
V.
Thus,
q
1
/
d
= 6.41
×
10
17
C/m.
Next, we note that when
q
2
is located at
x
= 0.080 m, the
net potential vanishes (
V
1
+
V
2
= 0).
Therefore,
21
0
0.08 m
kq
kq
d
Thus, we find
q
2
=
1
(
/ )(0.08 m)
qd
= –5.13
×
10
18
C =
–32
e
.
24.
The potential is
9
2
2
12
2
rod
rod
0
0
0
1
1
(8.99 10 N m C )(25.6 10
C)
4
4
4
3.71 10 m
6.20 V.
P
dq
Q
V
dq
R
R
R
We note that the result is exactly what one would expect for a pointcharge –
Q
at a
distance
R
. This
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 Fall '08
 IASHVILI

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