(b) We differentiate the potential with respect to
x
to find the
x
component of the electric
field:
2
00
0
92
2
1
5
42
1
ln
44
4
(
)
(8.99 10 N m C )(43.6 10
C)
(3.92 10
N m C)
,
(
0.135 m)
(
0.135 m)
x
V
Q
xL
Q
x
Q
E
x
Lx
x
Lx Lx x
xx L
xx
εε
ε
−−
∂∂
−
−
⎛⎞
⎛
⎞
=−
−
⎜⎟
⎜
⎟
−
−
⎝⎠
⎝
⎠
×⋅
×
×
⋅
++
ππ
π
or
(3.92 10
N m C)

.
(
0.135 m)
x
E
−
=
+
(c) Since
0
x
E
<
, its direction relative to the positive
x
axis is 180 .
°
(d) At
x = d
= 6.20 cm, we obtain
(3.92 10
N m C)


0.0321 N/C.
(0.0620 m)(0.0620 m 0.135 m)
x
E
−
==
+
(e) Consider two points an equal infinitesimal distance on either side of
P
1
, along a line
that is perpendicular to the
x
axis. The difference in the electric potential divided by their
separation gives the transverse component of the electric field. Since the two points are
This is the end of the preview. Sign up
to
access the rest of the document.