W9_parallel_resonance.eps

120 Week 3: Potential Energy and Potential

r

k Qe

R

R

V(r)

Figure 34: The potential produced by a uniform sphere of charge both inside and outside, as a
function ofr.

Suppose we have built a ball of radiusrand total charge:

Q(r) =

4 π

3

ρr^3 (178)

(so far). We know (or can figure out easily given the results just above) that the potential on its
surface is just:

V(r) =

keQ(r)

r =

4 πke

3 ρr

(^2). (179)
Now imagine bringing in a differential chunk of chargedQand spreading it around on the surface,
increasing the radius of the ball just a bit. The workwehave to do bringing the charge from∞to
the surface of the ball (which is also the increase in the potential energy of the ball) is:
dW(us)=dU(ball)=V(r)dQ=V(r)ρ 4 πr^2 dr (180)
where we use the fact that the charge of a thin shell of radiusrand thicknessdris just the volume
of the shell times the charge per unit volume. We can now add up this increment of energy by
integrating to “build a ball”:

U =

∫R

0

V(r)ρ 4 πr^2 dr

=

∫R

0

4 πke

3 ρr

(^2) ρ 4 πr (^2) dr
= ke^16 π
(^2) ρ 2
3

∫R

0

r^4 dr

= ke^16 π

(^2) ρ 2
3

R^5

5

=

3

5

ke

(

4 πR^3

3 ρ

) 2

R

=

3

5

keQ^2

R

=

3

5

V(R)Q (181)

This is an extremely interesting result. Note first that if we knewnothingabout how the charge
was distributed and were asked to estimate its energy, the only sensible answer we can give (that
makes dimensional sense) isU=V×Q. Charge times potential equals potential energy. Of course
we don’t expect the energy to beexactlythis – we expect it to be less, because we can bring in the
first bits of charge “for free” and do ever more work as we build up the ball – we expect it to be
somethinglessthan this estimate.

Later we’ll do more examples of this sort of integral when we discusscapacitance, and will find
that theformof this result is quite general, but (as one might expect) the leading fraction will vary