Physics Worksheet With Answers - Terry Honan - Blinn College Page 8
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88
Chapter B - Problems
Solution to B.13
Q
`
inside
(a) For spherical symmetry Gauss's law gives: E = k
r
.
e
2
r
For r < a the uniform charge distribution gives
3
Q
4
r
3
˛ r
=
= Q
Q
.
inside
4
3
3
3
a
˛ a
3
`
Q
Which gives: E = k
Hr < aL.
r r
e
3
a
`
Q
= Q so E = k
Ha < r < bL.
Between a and b: Q
r
e
inside
2
r
Inside the conductor, between b and c, the electric field is zero.
E = 0 Hb < r < cL.
`
Q +q
Outside the conductor Q
= Q +q ï E = k
r
Hr > cL.
e
inside
2
r
Q
(b) Inside the insulating sphere the charge density is r =
Hfor r < aL.
4
3
˛ a
3
Charge on a conductor is on its surface. In this case there are two surfaces.
Since the electric field is zero inside a conductor, Gauss's law applied to a gaussian surface enclosing the hole in the conductor
(between b and c) gives Q
= 0. It must be true that the charge on the inside surface cancels the charge Q at the center. The
inside
charge on the inside surface (at r = b) is -Q . Since the total charge on the conductor is q the charge on the outside surface is Q + q .
Q
The surface charge density is the charge per area. For a spherical surface: s =
.
2
4 p r
-Q
On the inside surface at b: s =
.
2
4 p b
Q+q
On the outside surface at a: s =
.
2
4 p a
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