Patel, Kinal – Homework 5 – Due: Feb 5 2008, 7:00 pm – Inst: Weathers
1
This printout should have 12 questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
be±ore answering.
The due time is Central
time.
001
(part 1 o± 1) 10 points
On a clear, sunny day, there is a vertical elec
trical feld o± about 146 N
/
C pointing down
over ²at ground or water.
What is the magnitude o± the sur±ace charge
density on the ground ±or these conditions?
Correct answer: 1
.
29271
×
10

9
C
/
m
2
.
Explanation:
Let :
E
=

146 N
/
C
.
By Gauss’ law,
I
~
E
·
d
~
A
=
q
²
0
.
I± we consider the electric feld as being im
mediately above the ground or water, then we
can think o± the ground or water as an inf
nite sheet o± charge. The question is, should
the electric feld beneath the sur±ace be the
same as the feld above? Should it be zero?
The real answer is quite complicated. In or
der to work this problem, we assume the feld
is zero beneath the sur±ace. This makes sense:
since water is a conductor, the feld inside the
water should be zero. Using Gauss’ Law,
E A
=
σ A
²
0
,
we have
σ
=
E ²
0
= (

146 N
/
C)
×
(
8
.
85419
×
10

12
C
2
/
N
·
m
2
)
=

1
.
29271
×
10

9
C
/
m
2
,
with a magnitude o± 1
.
29271
×
10

9
C
/
m
2
.
keywords:
002
(part 1 o± 2) 10 points
A conducting spherical shell having an inner
radius o± 4
.
1 cm and outer radius o± 5
.
4 cm
carries a net charge o± 7
.
2
μ
C. A conducting
sphere is placed at the center o± this shell
having a radius o± 0
.
9 cm carring a net charge
o± 1
.
7
μ
C.
0
.
9 cm
4
.
1 cm
,
Q
0
2
inside
5
.
4 cm
Q
00
2
outside
1
.
7
μ
C
7
.
2
μ
C
P
Determine the sur±ace charge density on the
inner sur±ace o± the shell.
Correct answer:

8
.
04769
×
10

5
C
/
m
2
.
Explanation:
Let :
r
= 0
.
9 cm
,
not required
a
= 4
.
1 cm
= 0
.
041 m
,
b
= 5
.
4 cm
= 0
.
054 m
,
and
q
= 1
.
7
μ
C
= 1
.
7
×
10

6
C
.
Basic Concept:
E
= 0 inside a Conductor.
Gauss’ Law,
I
~
E
·
d
~
A
=
Q
²
0
.
Solution:
Since the electric feld is zero
inside any conductor in electrostatic equi
librium, the net charge is zero inside any
spherical Gaussian sur±ace o± radius
r
, where
a < r < b
. Thus the charge on the inner sur
±ace o± the sphere must be

q
. I± we call the
charge density on the inner sur±ace
σ
in
, then

q
= 4
π a
2
σ
in
σ
in
=

q
4
π a
2
=

1
.
7
×
10

6
C
4
π
(0
.
041 m)
2
=

8
.
04769
×
10

5
C
/
m
2
.
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View Full DocumentPatel, Kinal – Homework 5 – Due: Feb 5 2008, 7:00 pm – Inst: Weathers
2
003
(part 2 of 2) 10 points
Determine the surface charge density on the
outer surface of the shell.
Correct answer: 0
.
000242881 C
/
m
2
.
Explanation:
Let :
Q
= 7
.
2
μ
C
= 7
.
2
×
10

6
C
.
The total charge contained in any spherical
Gaussian surface of radius
r > b
must be
Q
+
q
. Since the charge at the origin and the
charge on the inner surface of the sphere are
equal and opposite, the charge on the outer
surface must be
Q
+
q
.
We ±nd
σ
out
in a
manner similar to that used in Part 1
Q
+
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 Spring '08
 Weathers
 Electrostatics, Potential Energy, Work, Correct Answer, Electric charge, patel

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