101. (a) For
r > r
2
the field is like that of a point charge and
V
Q
r
=
1
4
0
π
ε
,
where the zero of potential was taken to be at infinity.
(b) To find the potential in the region
r
1
<
r < r
2
, first use Gauss’s law to find an
expression for the electric field, then integrate along a radial path from
r
2
to
r
. The
Gaussian surface is a sphere of radius
r
, concentric with the shell. The field is radial and
therefore normal to the surface. Its magnitude is uniform over the surface, so the flux
through the surface is
Φ
= 4
π
r
2
E
. The volume of the shell is
43
2
3
1
3
π
b
gc
h
rr
−
, so the
charge density is
ρ
=
−
3
4
2
3
1
3
Q
π
c
h
,
and the charge enclosed by the Gaussian surface is
qr
r
Q
=
F
H
G
I
K
J
−=
−
−
F
H
G
I
K
J
4
3
3
1
3
3
1
3
2
3
1
3
π
c
h
.
Gauss’ law yields
4
4
0
2
3
1
3
2
3
1
3
0
3
1
3
2
2
3
1
3
π
π
rE Q
E
Qr
r
rr r
=
−
−
F
H
G
I
K
J
¡
=
−
−
c
h
.
If
V
s
is the electric potential at the outer surface of the shell (
r = r
2
) then the potential a
distance
r
from the center is given by
VV
E
d
rV
Q
r
r
r
dr
V
Q
rrr
r
r
r
ss
r
r
r
r
s
=−
−
−
F
H
G
I
K
J
−
−+−
F
H
G
I
K
J
z
z
4
1
4
1
22
02
3
1
3
1
3
2
3
1
3
2
2
2
1
3
1
3
2
π
π
.
The potential at the outer surface is found by placing
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 Spring '10
 Rebello,NobelS
 Charge, Electrostatics, Magnetic Field, Electric charge, Fundamental physics concepts, 3k

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