homework 07 – KANUNGO, ANIL – Due: Sep 18 2007, 4:00 am
1
Question 1, chap 24, sect 3.
part 1 of 4
10 points
A long coaxial cable consists of an inner
cylindrical conductor with radius
R
1
and an
outer cylindrical conductor shell with inner
radius
R
2
and outer radius
R
3
as shown. The
cable extends out perpendicular to the plane
shown. The charge per unit length along the
cable on the inner conductor is
λ
and the
corresponding charge per unit length on the
outer conductor is
−
λ
(same magnitudes but
with opposite signs) and
λ >
0.
Q
R
1
R
2
R
3
−
Q
Find the magnitude of the electric field at
the point a distance
r
1
from the axis of the
inner conductor, where
R
1
< r
1
< R
2
.
1.
E
=
λ
2
π ǫ
0
r
1
correct
2.
E
=
λ R
1
3
π ǫ
0
r
1
2
3.
E
= 0
4.
E
=
λ
2
π ǫ
0
R
1
5.
E
=
λ
√
3
π ǫ
0
r
1
6.
E
=
λ R
1
4
π ǫ
0
r
1
2
7.
None of these.
8.
E
=
λ
√
2
π ǫ
0
r
1
9.
E
=
2
λ
√
3
π ǫ
0
r
1
10.
E
=
λ
2
R
1
4
π ǫ
0
r
1
2
Explanation:
Pick a cylindrical Gaussian surface with
radius
r
1
and apply Gauss’s law:
E
·
d A
=
E
·
ℓ
(2
π r
1
) =
Q
ǫ
0
E
=
λ
2
π ǫ
0
r
1
.
Question 2, chap 24, sect 3.
part 2 of 4
10 points
The electric field vector points
1.
in the negative ˆ
r
direction.
2.
in the positive ˆ
r
direction.
correct
Explanation:
The field points from positive charge to
negative change.
Since the center conductor
is negatively charged, the electric field vector
points in the negative ˆ
r
direction.
Question 3, chap 24, sect 3.
part 3 of 4
10 points
Find the magnitude of the electric field at
the point a distance
r
2
from the axis of the
inner conductor, where
R
3
< r
2
.
1.
E
=
λ
2
π ǫ
0
r
2
2.
E
=
λ
2
R
1
4
π ǫ
0
r
2
2
3.
E
=
λ
√
3
π ǫ
0
r
2
4.
None of these.
5.
E
=
2
λ
√
3
π ǫ
0
r
2
6.
E
=
λ R
1
4
π ǫ
0
r
2
2
7.
E
=
λ
2
π ǫ
0
R
1
8.
E
=
λ R
1
3
π ǫ
0
r
2
2
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homework 07 – KANUNGO, ANIL – Due: Sep 18 2007, 4:00 am
2
9.
E
= 0
correct
10.
E
=
λ
√
2
π ǫ
0
r
2
Explanation:
Pick a cylindrical Gaussian surface with
radius
r
2
and apply Gauss’ law.
Because
there is no net charge inside the Gaussian
surface, the electric field
E
= 0
.
Question 4, chap 24, sect 3.
part 4 of 4
10 points
For a 100 m length of coaxial cable with in
ner radius 1 mm and outer radius 1
.
9 mm,
find the capacitance
C
of the cable.
The
value of the permittivity of free space
is
8
.
85419
×
10
−
12
c
2
/
N
·
m
2
.
Correct answer: 8
.
66747
nF (tolerance
±
1
%).
Explanation:
Let :
ℓ
= 100 m
,
R
1
= 1 mm
,
R
2
= 1
.
9 mm
,
and
ǫ
0
= 8
.
85419
×
10
−
12
c
2
/
N
·
m
2
.
We calculate the potential across the capac
itor 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
integraldisplay
R
1
R
2
dr
r
=
−
1
2
π ǫ
0
q
ℓ
ln
r
vextendsingle
vextendsingle
vextendsingle
vextendsingle
R
1
R
2
=
q
2
π ǫ
0
ℓ
ln
parenleftbigg
R
2
R
1
parenrightbigg
.
Since
C
=
q
V
, the capacitance is
C
=
q
V
=
2
π ǫ
0
ℓ
ln
parenleftbigg
R
2
R
1
parenrightbigg
=
2
π
(8
.
85419
×
10
−
12
c
2
/
N
·
m
2
)
ln
parenleftbigg
1
.
9 mm
1 mm
parenrightbigg
×
(100 m)
=
8
.
66747 nF
.
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 Fall '08
 Turner
 Electrostatics, Work, Electric charge, Anil

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