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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
the electric field a distance
r
g
from the shell center. The charge that is both in the shell
and within the Gaussian sphere is given by the integral
qd
V
s
=
z
ρ
over the portion of the
shell within the Gaussian surface. Since the charge distribution has spherical symmetry,
we may take
dV
to be the volume of a spherical shell with radius
r
and infinitesimal
thickness
dr
:
dV
r dr
=
4
2
π
. Thus,
()
22
2
2
4
4
4
2
.
gg
g
rr
r
s
g
aa
a
A
qr
d
r
r
d
r
A
r
d
r
A
r
a
r
πρ
π
==
=
=
−
³³
³
The total charge inside the Gaussian surface is
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 Spring '10
 Rebello,NobelS

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