I recently mentioned a couple of new things I’d learned in teaching an electricity and magnetism class this semester. One is the answer to this question:
Charge Q is placed on an infinitely thin conducting disk of radius R. How does it distribute itself over the disk?
The answer turns out to be that the charge distribution is the same as if you started with a uniform distribution of charge over a spherical shell, and compressed the shell down to a disk by smashing each piece of the surface straight up or down in the direction perpendicular to the disk.
Although I know of various proofs of this, particularly one provided by my friend and colleague Ovidiu Lipan, none of them seemed to me like a satisfying explanation of why the result was true. Of course, there might not be such a thing, but when the final answer has such a simple description (compared to most problems in E&M, which have incredibly messy solutions), it seems like there ought to be a nice reason for it.
Later I came across some notes by Kirk McDonald of Princeton provide a somewhat intuitive answer. I’ll summarize the idea here.
Start with something every E&M student should know. Take a spherical shell of radius R and thickness dR, and fill it uniformly with charge. Then the electric field inside the shell is zero. The slick way to prove this is with Gauss’s Law, but you can prove it with more basic geometric reasoning as follows. (In fact, I believe that the argument goes all the way back to Newton, although of course he wasn’t talking about electric fields, since they hadn’t been discovered / invented yet).
Say that you’re located at an arbitrary point inside the spherical shell. Draw two cones with infinitesimal solid angle going out in opposite directions. These intersect the shell giving two little “charge elements” shown in red below.
You can convince yourself that the electric field contributions from these charge elements exactly cancel each other. The volume, and hence the amount of charge, of each is proportional to the square of its distance from you, so by the inverse square law they give equal but opposite fields.
Since you can chop up the whole shell into such pairs, the total electric field vanishes.
The clever insight in McDonald’s notes is that the same argument holds even if you take the spherical shell and compress it into an ellipsoid (i.e., rescale all of the z coordinates by some constant factor, while leaving x and y the same):
With a bit of effort, you can convince yourself that all the various distances and volumes scale in such a way that the two field contributions remain equal and opposite.
Now that we know that the field inside this ellipsoid is zero, what can we conclude? First, take the limit as dR goes to zero, so we have a two-dimensional surface charge distribution. The result must be the same as the surface charge distribution on a solid, conducting ellipsoid. After all, the conducting ellipsoid has to have charge only on its surface and have zero field inside. The usual electrostatic uniqueness theorems apply here, which say that there’s only one charge distribution that leads to this result. Since we’ve found a charge distribution that does so, it must be the charge distribution.
Keep smashing the ellipsoid down until it lies in a plane, and you’ve got the solution for the conducting disk.