W9_parallel_resonance.eps

Week 2: Continuous Charge and Gauss’s Law 83

Q

r r

R

S(outer)

S(inner)

Figure 18: A solid sphere of uniform charge densityρand radiusR.

We begin by writing Gauss’s Law for the outer surface in the figure??:

∮

Souter

E~·nˆdA = 4πke

∫

V/S

ρdV

Er 4 πr^2 = 4πke

{∫R

0

∫ 2 π

0

∫π

0

ρr^2 sin(θ)dθ dφ dr

+

∫r

R

0 dV

}

= 4πke(2πρ)

∫R

0

r^2 dr

∫ 1

− 1

d(cos(θ))

= 4πke(

4 πR^3

3

ρ)

= 4πkeQtotal (108)

We divide both sides by 4πr^2 and get:

Er=

keQ

r^2 r > R (109)

or (as by now you should come to expect) the spherical distributionof charge creates a fieldoutside
of the sphere that is identical to that of a point charge of the sametotal value at the origin.

Note that we did a bunch of stuff that we didn’t really “have” to do – in an actual solution you’d
be tempted to skip those steps or do them by inspection, which is fine, but that risks confusing
at least some of you whodon’tjust see what we are skipping and why it is OK to do so. So note
well – to find the total charge insideSouter, we integrated over the charge distribution from 0 tor
including the region where it was zero– getting, of course, a zero value for that value. Zero regions
drop out, and we’d usually just integrate over thesupportofρ(the volume where it is nonzero)
without thinking about it. Note also that this integral explicitly illustrates doing multiple integrals
of a symmetric function – we just do the integrals over each coordinate independently (which is then
really easy).

Finally, note theclever trickfor integratingθin spherical coordinates. sin(θ)dθ=−d(cos(θ)), so
we change variables fromθ→cos(θ) (and change and swap order of the limits to get rid of the minus
sign). It isvery oftenmuch easier to integrate with cos(θ) as the variable instead ofθin spherical
coordinates – in this case one can just look at it and see that one gets “2” from the integral in your
head, for example.