Week 3: Potential Energy and Potential 117
A spherical shell of charge thus produces a potentialoutsidethat looks like the potential of a
point charge at the origin to match its field that looks like that of a point charge at the origin.
Inside, its potential is constant, the value it had on the shell itself coming infrom the outside.
Now, a bit of warning based on my many years of teaching this class. For some of you, the first
time you see a problem like this on a quiz with a region where the field is zero, the Devil is going
to whisper into your ear “C’mon, dude. The field in these is zero, so the potential in there must
be zero too. Put down zero and let’s move on.” Unfortunately, if youlisten to the Devil, you’ll
be condemned to Physics Quiz Hell, because this would bewrong! Remember that the electrical
field is basically the derivative of the potential. The derivative ofany constantis zero, not just the
particularconstant whosevalueis zero.
Think of it in terms of the tops of mesas, flat mountains. Anyplace that is “flat” in potential has
no field. A charge placed there doesn’t gain energy moving around. But that doesn’t mean that the
heightof the mesa is sea-level, or that one doesn’t have to climb a steep slope from sea-level to reach
the flat part. Similarly, we may have to do quite a bit of work to push a test charge from infinity to
the edge of a spherical shell of charge, but once we go inside the field vanishes and we can move it
anywhere without doing work. The potential inside is constant, butthat constant has to reflect the
totalwork done coming in from infinity (per unit charge) and is not particularly likely to bezero.
Example 3.4.5: Advanced: Spherical Shell of Charge
syxθdq
rφR−rRzPFigure 32: Geometry for finding the potential of a uniform sphericalshellof constant charge density
σby direct integration.
Consider figure 32. You should recognize it has being almost exactly the same geometry as was
used to integrate to find the (much more difficult)electric fieldof the spherical shell last week in a
similarly advanced example. In a way, it would be a lot easier to just do these two examples in the
opposite order, as it is a lot easier to integrate to find the potentialthan the field in the first place,
and once we have done so we can always find the field by differentiating.
As before, we lose nothing by putting a pointP at a distanceRfrom the origin. We consider
the chargedqof a tiny patchdAon the surface of the sphere, and write down the potential of this
patch atP:
dV=kedq
s= keσr(^2) dcos(θ)dφ
(R^2 +r^2 − 2 Rrcos(θ))^1 /^2