Assignment 3 Solutions
Chpt 24
24.
. The potential is
92
2
1
2
2
rod
rod
00
0
1
1
(
8
.
9
91
0Nm C)
(
2
5
.
61
0 C
)
44
4
3
.
7
1
1
0
m
6.20 V.
P
dq
Q
Vd
q
RR
R
εε
ε
−
−
−×
⋅×
==
=
=
−
ππ
π
×
=−
∫∫
We note that the result is exactly what one would expect for a pointcharge –
Q
at a
distance
R
. This
“coincidence”
is due, in part, to the fact that
V
is a scalar quantity.
30.
Consider an infinitesimal segment of the rod, located between
x
and
x + dx
. It has
length
dx
and contains charge
dq =
λ
dx
, where
λ
=
Q
/
L
is the linear charge density of the
rod. Its distance from
P
1
is
d
+
x
and the potential it creates at
P
1
is
x
d
dx
x
d
dq
dV
+
=
+
=
λ
πε
0
0
4
1
4
1
To find the total potential at
P
1
, we integrate over the length of the rod and obtain:
0
0
0
2
1
5
3
ln(
)
ln 1
4
(8.99 10 N m C )(56.1 10
C)
0.12 m
ln 1
7.39 10 V.
0.12 m
0.025 m
L
L
dx
Q
L
x
dx
L
d
λλ
−
−
⎛⎞
+
=
+
⎜⎟
+
⎝⎠
×⋅
×
=+
∫
π
=
×
38.
(a) From the result of Problem 2430, the electric potential at a point with coordinate
x
is given by
0
ln
.
4
Qx
L
V
Lx
−
=
p
At
x = d
we obtain
2
1
5
0
3
(8.99 10 N m C )(43.6 10
C)
0.135 m
ln
ln 1
4
0.135 m
0.135 m
(2.90 10 V)ln 1
.
Qd
L
V
Ld
d
d
−
−
+×
⋅
×
⎛
+
⎜
⎝
=×
+
π
⎞
⎟
⎠
1
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View Full Document(b) We differentiate the potential with respect to
x
to find the
x
component of the electric
field:
2
00
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
0
(
)
x
Lx
x
LxLx x
xxL
xx
εε
−−
∂∂
−
−
⎛⎞
⎛
⎞
=−
−
⎜⎟
⎜
⎟
−
⎝⎠
⎝
⎠
×⋅
×
×
⋅
++
ππ
π
ε
−
or
(3.92 10
N m C)

(
0.135 m)
x
E
−
=
+
.
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 Spring '08
 Parsons
 Charge, Electric Potential, Potential Energy, Electric charge, rod R

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