Re: capacitance of a disk

["John S. Denker" <jsd@MONMOUTH.COM>]

At 04:41 PM 2/8/01 -0500, Carl E. Mungan wrote:

> So what is the capacitance of an isolated disk? Feel free to make
> whatever simplifications or approximations you want and otherwise
> change the problem to make it more tractable or meaningful.

Wow, what a cool question.

Here's how I'd solve it.

First, I'm going to forget the equation that says
        Q=CV
and I'm even going to forget the formula for energy at constant voltage
        U = .5 C V^2
Instead I'm going to concentrate on the energy at constant charge:
        U = .5 Q^2 / C                          (eq 3)

Second, I'm going to invoke the principle of least action;  that is, I'm
going to argue that the charge will rearrange itself into the lowest-energy
arrangement permitted by the constraints.

Here's a ranking of some interesting ways of constraining a constant charge Q:
  a) A point charge.
  b) A uniformly charged ring of radius R made out of
     thin wire with a tiny radius r.
  c) A uniformly charged thin disk of radius R.
  d) A conducting thin disk of radius R
  e) A uniformly charged sphere of radius R.

I argue that the charge moves to progressively lower energy as we move down
the list from (a) to (c), because we are progressively relaxing the
constraints.  Therefore the capacitance gets progressively higher as we
move down the list.  Note the reciprocal energy-versus-capacitance
relationship in equation (3).

The assignment asks about case (d).  To give a good answer, I want to give
a controlled approximation;  that is, rigorous upper and lower bounds.  I'm
too lazy to do case (d) exactly, but I can bracket it with case (e) which
is an upper bound on the capacitance, and cases (abc) each of which is a
lower bound.

So let's take inventory.
  a) This case is doable exactly.  Unfortunately the energy is infinite
     so the capacitance is zero and this lower bound is not a very
     interesting lower bound.  Rats.
  b) Unfortunately also infinite energy.
  c) This looks good.  We should be able to get a non-infinite upper bound
     on the energy of this critter, and hence a useful lower bound
     on the case of interest.
  d) This is the case of interest.  Finding the charge distribution
     on the conducting disk is hard, so we won't touch this case
     directly.  We rely on the other cases to bracket it.
  e) This is doable exactly.  This provides a lower bound on the
     energy and an upper bound on the capacitance.


============

To summarize up to this point:  The capacitance of the conducting disk is
bounded above by the capacitance of a sphere of the same radius, and
bounded below by the capacitance of a uniformly-charged disk of the same
radius.

The sphere is easy and well known.  So all we need is an estimate of the
energy of a uniformly charged disk.  Any reasonable over-estimate will do.

Now the energy of a capacitor is just the energy in its electric
field.  The energy density goes like the square of the field strength, and
then you integrate the energy density over all space to get the energy.

At surfaces, the electric field strength is proportional to the charge per
unit area.  For the uniformly-charged disk (case c) this is
        sigma  = Q / (2 pi R^2)                 (eq 4)
where the factor of 2 is because the disk has two sides.  It is instructive
to compare this with the sphere (case e) for which we have
        sigma' = Q / (4 pi R^2)                 (eq 5)
so the field near the disk is twice the field near the sphere.  For points
not near the surfaces, the field is everywhere smaller than these values.

We can estimate the field-energy of the disk in two pieces:

1) Draw an imaginary sphere around the disk, with the same radius, and
consider the field outside thereof.  The _largest_ field that could
possibly exist at this imaginary sphere is proportional to sigma which is
given by equation (4).  That's twice the field of a real sphere with charge
Q.  When we square this and integrate over all volume outside the imaginary
sphere, this contributes to the energy at most two^2 times Z, where Z
denotes the energy of a real sphere with charge Q.

2) Now consider the field _inside_ the imaginary sphere.  This field has
energy also.  Again we approximate this field by the biggest value it could
possibly have.  We square and integrate over the volume inside the
sphere.  This gives an energy of at most (4/3) Z.

****

Combining the pieces, we find that the energy is at most (4+4/3)Z, so the
capacitance of a disk is at least 3/16 of the capacitance of a sphere.

I believe this is a rigorous bound, but not a particularly tight bound.  If
I had to guess, I would guess 1/2 is a better estimate than 3/16.

And as mentioned above, 1 is a rigorous upper bound.