118 Week 3: Potential Energy and Potential
We integrate both sides, the right hand side over the entire solid angle:
V=
∫
dV=
∫
kedq
s =
∫ 1
− 1
∫ 2 π
0
keσr^2 dcos(θ)dφ
(R^2 +r^2 − 2 Rrcos(θ))^1 /^2
(173)
We can do theφintegral immediately and factor out all the constants:
V= 2πr^2 σke
∫ 1
− 1
dcos(θ)
(R^2 +r^2 − 2 Rrcos(θ))^1 /^2
(174)
This is much easier to integrate than the vector relation of the field chapter example:
V = 2πr^2 σke
∫ 1
− 1
dcos(θ)
(R^2 +r^2 − 2 Rrcos(θ))^1 /^2
=
2 πr^2 σke
− 2 Rr
∫ 1
− 1
− 2 Rrdcos(θ)
(R^2 +r^2 − 2 Rrcos(θ))^1 /^2
=^2 πr
(^2) σke
− 2 Rr
2
(
R^2 +r^2 − 2 Rrcos(θ)
) 1 / 2 ∣∣
∣
1
− 1
=
2 πr^2 σke
− 2 Rr
2 ((R−r)−(R+r))
=
2 πr^2 σke
− 2 Rr (−^2 r)
=
ke(4πr^2 σ)
R
=
keQ
R
(175)
Much, much easier!
Example 3.4.6: Potential of a Uniform Ball of Charge
Q
r r
R
S(outer)
S(inner)
Figure 33: A solid sphere of uniform charge densityρand radiusR.
Find the fieldandthe potential at all points in space of a solid insulating sphere with
uniform charge densityρand radiusR.
If you will recall, finding the field of a solid sphere of charge isbothan example in the text above
and was a homework assignment a couple of weeks ago – so by now youshould have gone over it
repeatedly and made it your own. The result was:
Er=
ke
( 4 πR (^3) ρ
3
)
r^2
=
keQ
r^2
r > R