Re: [Phys-L] spherical capacitor problem

[John Denker <jsd@av8n.com>]

On 06/29/2015 05:42 AM, Carl Mungan wrote:
> My objection is the formula for an isolated sphere assumes the other
> plate is a concentric sphere at infinity. (That’s how one derives the
> formula for spherical capacitance after all.) But that means the
> other sphere is located “inside” the first capacitor. How can one say
> the two spheres are in series in that case?

There are lots of ways of getting the right answer.

For me, the easiest way is to use the energy formula
      E = 0.5 C V^2

Two spheres makes twice as much energy per unit voltage,
so that corresponds to half as much capacitance.  I can
figure this out in my head in less time than it takes
to ask the question.

It takes slightly longer to fully /explain/ the answer.
We know there is twice as much energy because:

1) For each sphere, most of the energy is located very
 near the sphere.  

2) For d ≫ a, the distribution of field lines near one
 electrode is "almost" insensitive to the size, shape,
 and location of the counterelectrode.

So I guess knowing the answer to the original question 
comes down to knowing facts (1) and (2).  So, at the
next level of detail:

  1a) In 3D the field strength falls off like 1/r^2 so
   the energy density falls off like 1/r^4 ... so the 
   integral converges.  In spherical coordinates the
   measure is r^2 dr, so the convergence isn't super-
   quick, but it's quick enough.  (In 2D we would be 
   having a different conversation.)

  2a) The field layout is intuitive and familiar to me
   based on experience playing with field-line simulators.
     https://www.av8n.com/physics/laplace.htm

  2b) You can prove theoretically that (2) must be so,
   as follows.  You can find the field-line distribution
   by means of a variational principle.  As a starting
   point for the variational search for the two-sphere
   field, the simple superposition of two one-sphere
   fields is not a bad guess.  The search will make
   changes relative to this initial guess, but in the
   limit d ≫ a the changes will be small.

   Furthermore, as you can see from the Hellmann-Feynman
   theorem or otherwise, if the error in the field
   distribution is small, the error in the energy is
   small squared ... so taking the two-sphere energy
   to be twice the one-sphere energy is an even better
   approximation than you might have expected.