tovar (jdt436) – homework 11 – Turner – (59070)
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
001
(part 1 oF 2) 10.0 points
A long coaxial cable consists oF an inner cylin
drical conductor with radius
R
1
and an outer
cylindrical conductor shell with inner radius
R
2
and outer radius
R
3
as shown. The ca
ble extends out perpendicular to the plane
shown. The charge on the inner conductor
per unit length along the cable is
λ
and the
corresponding charge on the outer conductor
per unit length is

λ
(same in magnitudes
but with opposite signs) and
λ >
0.
Q
R
1
R
2
R
3
b

Q
±ind the magnitude oF the electric feld at
the point a distance
r
1
From the axis oF the
inner conductor, where
R
1
< r
1
< R
2
.
1.
None oF these.
2.
E
=
λ R
1
4
π ǫ
0
r
1
2
3.
E
=
λ
2
R
1
4
π ǫ
0
r
1
2
4.
E
=
λ
√
2
π ǫ
0
r
1
5.
E
=
λ
√
3
π ǫ
0
r
1
6.
E
= 0
7.
E
=
λ
2
π ǫ
0
r
1
correct
8.
E
=
2
λ
√
3
π ǫ
0
r
1
9.
E
=
λ R
1
3
π ǫ
0
r
1
2
10.
E
=
λ
2
π ǫ
0
R
1
Explanation:
Pick a cylindrical Gaussian surFace with the
radius
r
1
and apply the Gauss’s law; we obtain
E
·
ℓ
·
2
π r
1
=
Q
ǫ
0
E
=
λ
2
π ǫ
0
r
1
002
(part 2 oF 2) 10.0 points
±or a 100 m length oF coaxial cable with
inner radius 0
.
834467 mm and outer radius
2
.
02108 mm.
±ind the capacitance
C
oF the cable.
Correct answer: 6
.
28905 n±.
Explanation:
Let :
ℓ
= 100 m
,
R
1
= 0
.
834467 mm
,
and
R
2
= 2
.
02108 mm
.
We calculate the potential across the capaci
tor by integrating
E ·
d
s. We may choose
a path oF integration along a radius;
i.e.,
E ·
d
s =
E
dr
.
V
=

1
2
π ǫ
0
q
l
i
R
1
R
2
dr
r
=

1
2
π ǫ
0
q
l
ln
r
v
v
v
v
R
1
R
2
=
q
2
π ǫ
0
l
ln
R
2
R
1
.
Since
C
=
q
V
, we obtain the capacitance
C
=
2
π ǫ
0
l
ln
p
R
2
R
1
P
=
2
π
(8
.
85419
×
10
−
12
c
2
/
N
·
m
2
)
ln
p
2
.
02108 mm
0
.
834467 mm
P
×
(100 m)
=
6
.
28905 n±
.
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2
003
10.0 points
Given a spherical capacitor with radius of the
inner conducting sphere
a
and the outer shell
b
. The outer shell is grounded. The charges
are +
Q
and

Q
. A point
C
is located at
r
=
R
2
, where
R
=
a
+
b
.
a
A
B
C
+
Q

Q
b
What is the capacitance of this spherical
capacitor?
1.
C
=
b
k
e
2.
C
=
1
k
e
(
a
+
b
)
3.
C
=
b
2
4
k
e
(
b

a
)
4.
C
=
k
e
b
5.
C
=
a
k
e
6.
C
=
b

a
2
k
e
ln
p
b
a
P
7.
C
=
1
k
e
p
1
a

1
b
P
correct
8.
C
=
k
e
a
9.
C
=
1
k
e
(
a

b
)
10.
C
=
a
+
b
k
e
Explanation:
Δ
V
=
V
a

V
b
=
k
e
Q
p
1
a

1
b
P

0
since
V
b
is grounded.
The charge on the
inside of the shell doesn’t aFect the grounded
potential.
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 Spring '08
 Turner
 Capacitance, Work, Electric charge, μF, equivalent capacitance, Tovar

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