Re: [Phys-l] defn of capacitance

[John Denker <jsd@av8n.com>]

In the context of   Qi = Cij Vj        [1]


On 02/07/2008 02:46 PM, Carl Mungan wrote:

> In any case, that aside, I would be glad if you could actually 
> demonstrate how to use the formula to calculate a standard 
> two-conductor capacitor. 

That's easy.  The full capacitance matrix is 2x2 and the
matrix elements of C are:

      [  a  -a ]
      [        ]
      [ -a   a ] 

That's really the only possible form it could take, given the
required symmetries (rows sum to zero, columns sum to zero,
symmetric Cij = Cji).

We define the differential-mode voltage
  ΔV := V2 - V1
and the common-mode voltage
  Vc := V2 + V1

Then V1 = (Vc-ΔV)/2
     V2 = (Vc_ΔV)/2


Then Q1 = a V1 - a V2
        = a (Vc-ΔV)/2 - a (Vc+ΔV)/2
        = - a ΔV

and  Q2 = + a ΔV


A diminished capacitance matrix is formed by dropping one row and
one column from the full capacitance matrix.  If we do that in the
usual way we are left with the 1x1 matrix
     [ a ]
as discussed at
  http://www.av8n.com/physics/laplace.htm#sec-calc

A 1x1 matrix is tantamount to a scalar, so we can see that the matrix
element "a" is just the prosaic two-terminal capacitance.  In 20/20
hindsight I could have called it "c" all along, but then you might
have suspected I was cheating.

This example is so simple that you might well ask why we bothered.
The point here is to show how the machinery works.  It works just 
fine for N-terminal capacitors for any N from 2 on up.

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

If you want an exercise that is designed to be about as easy as a
three-terminal capacitor can be, consider the setup shown in
  http://www.av8n.com/physics/img48/spherical-capacitor.png

Start with two hemispheres of radius R, separated by a gap of
width g.  All of that is surrounded by a Faraday cage (shown as 
an octagon, using dashed lines).  The size of the cage is huge
compared to R, and R is huge compared to g.

Calculate the full 3x3 capacitance matrix.