Solution 7 - homework 07 KANUNGO ANIL Due 4:00 am Question...

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

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 .