W9_parallel_resonance.eps

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