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34. (a) According to the result of problem 28, the electric potential at a point with
coordinate
x
is given by
0
ln
.
4
Qx
L
V
Lx
ε
−
§·
=
¨¸
©¹
π
At
x = –d
we obtain (in SI units)
91
5
0
3
(8.99 10 )(43.6 10
)
0.135
ln
ln 1
4
0.135
0.135
(2.90 10 V)ln 1
.
Qd
L
V
Ld
d
d
−
−
+×
×
§
·
==
+
¨
¸
©
¹
=×
+
π
(b) We differentiate the potential with respect to
x
to find the
x
component of the electric
field (in SI units):
2
00
0
5
4
1
ln
44
4
(
)
(8.99 10 )(43.6 10
)
(3.92 10 )
,
(
0.135)
(
0.135)
x
V
Q
xL
Q
x
Q
E
x
x
L
x
L
x
x
x
x
L
xx
εε
−−
∂∂
−
−
§
·
=−
−
¨
¸
−
−
©
¹
××
×
++
ππ
π
or
4
(3.92 10 )

.
(
0.135)
x
E
−
×
=
+
(c) Since
0
x
E
<
, its direction relative to the positive
x
axis is 180 .
°
(d) At
x = –d
we obtain (in SI units)
4
(3.92 10 )


0.0321 N/C.
(0.0620)(0.0620 0.135)
x
E
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This note was uploaded on 01/25/2011 for the course PHYSICS 17029 taught by Professor Rebello,nobels during the Spring '10 term at Kansas State University.
 Spring '10
 Rebello,NobelS
 Electric Potential

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