Week 3: Potential Energy and Potential 119
and
Er=ke(
4 πρ
3)
r= ρr
3 ǫ 0r < Rfor the exterior and interior of the sphere (where we used 4πke= 1/ǫ 0 in the last equation just so
you don’t completely forget this relation as we prefer to work withkebut one day you’ll need to be
able to work withǫ 0 ). So just to humor me, get out paper and prove (to yourself, if nobody else)
that you can still get this result, starting with Gauss’s Law andwithout looking.
With the field(s) in hand, we now recapitulate the reasoning of the previous example. The
distribution of charge has compact support, so we can integrate infrom infinity to find the potential
(relative to infinity):
V(r) = −∫r∞E~·d~l=−∫r∞Er > Rdr= −
∫r∞keQ r′−^2 dr′= keQ
rr > R (176)and we find, as hopefully you had already anticipated, that the potential of the solid sphereoutside
was that of a point charge with the same total charge at the origin,in perfect correspondance with
the field.
The place things get more interesting is when we try to evaluate the potentialinsidethe sphere.
The potential is defined as an integral in from∞, but thefield changes functional formatr=R.
We therefore have to do the integralpiecewise, doing first the integral from∞toR, then fromRto
r. This is why we wrote out both terms in the spherical shell example above, even though the field
inside was zero (and so was that part of the integral) – we want to get in the habit ofalwaysdoing
the integral piecewise and simply being happy when one or another piece is zero, rather than either
expecting it or forgetting that this is what we are really doing. Thus:
V(r) = −∫r∞E~·d~l=−∫R
∞Er > Rdr−∫rREr < Rdr= −
∫R
∞ke(
4 πR^3 ρ
3)
r′−^2 dr′−∫rRke(
4 πρ
3)
r′dr′= ke(
4 πR^2 ρ
3)
+ke(
2 πρ
3){
R^2 −r^2}
= 2πkeρR^2 −ke(
2 πρ
3)
r^2 r < R (177)Let’s think a teensy bit about this result, and then plot it (as we did for the field) to help us
remember it, as (recall) the uniform ball of charge is the basis of thesimplest model for an atom
and hence the key to easily understanding lots of things such as polarization, ionization, and more.
First of all, note that the potential is (by the meaning of integrals in the first place) theareaunder
theEr(r) curve fromrto∞.E~is continuous but not smooth (look back at figure??and note the
cusp atr=R), butV(r) is continuous andsmoothatr=R– the function and its first derivative
match at the point, although the second derivatives differ. Outsidethe potential drops off like 1/r,
a monopolar potential that corresponds to the monopolar field. Inside, the potentialincreases like
an upside down quadraticall the way to the origin, where it has its maximum value!
There is one more thing that we need to do before abandoning the ball of charge. Suppose we
are handed such a ball. A perfectly reasonable question for any physics groupie is “How much work
did it take to assemble all of this charge?” After all, the charge is mutually repulsive – every bit of
charge we put into the ball had to be brought in “from infinity” against the field of the charge that
is already there. This latter insight is the key to writing down a simple integral to tell us how much
work was done, and hencewhat the potential energy of a uniform ball of charge is.