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The electric field is radial, so the flux through the Gaussian surface is
Φ=
4
2
π
rE
g
, where
E
is the magnitude of the field. Gauss’ law yields
42
0
22
2
ππ
ε
Er
q
A r
a
gg
=+
−
di
.
We solve for
E
:
E
q
r
A
Aa
r
−
L
N
M
M
O
Q
P
P
1
4
2
2
0
2
2
2
π
π
π
For the field to be uniform, the first and last terms in the brackets must cancel. They do if
q
– 2
π
Aa
2
= 0 or
A = q
/2
π
a
2
. With
a
= 2.00
×
10
−
2
m and
q
= 45.0
×
10
−
15
C, we have
11
2
1.79 10
C/m .
A
−
=×
49. To find an expression for the electric field inside the shell in terms of
A
and the
distance from the center of the shell, select
A
so the field does not depend on the distance.
We use a Gaussian surface in the form of a sphere with radius
r
g
, concentric with the
spherical shell and within it (
a < r
g
< b
). Gauss’ law will be used to find the magnitude of
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 Spring '08
 Any
 Physics

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