Re: A funny capacitor.

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

At 09:07 AM 3/1/01 -0500, Ludwik Kowalski wrote:
> 1) Here is the capacitor matrix:
>
>  Q1=b12*V1 + b12*V2 + b13*V3
>  Q2=b21*V1 + b22*V2 + b23*V3
>  Q3=b31*V1 + b22*V2 + b33*V3
>
> How would you express the bij coefficients in terms of
> what is given? [geometry, etc.]

a) A word about notation:  Given the matrix equation
        Q = C V                         (equation 1)
we can expand it in components as

  Q1=C12*V1 + C12*V2 + C13*V3
  Q2=C21*V1 + C22*V2 + C23*V3           (equation 2)
  Q3=C31*V1 + C22*V2 + C33*V3

Where obviously Cij is the (i,j) element of the C matrix.  So what Ludwik
writes as bij is precisely this matrix element, Cij.

b) To find these values, start by setting V1=0 and V2=0 and V3=nonzero.
Calculate the charge on each object using Laplace's equation.  This tells
us, among other things, Q1.  Equation (1) implies equation (2) which in
this case reduces to Q1 = C13 V3.  It is not very hard to solve this
equation for C13.  Similarly for other matrix elements.

You will discover the following facts, which will soon come in handy:

Gauge invariance requires Cij to be symmetric:
         Cij = Cji                              (equation 3)

Charge conservation requires each row (and/or column) to sum to zero
         SUM_j Cij = 0    for all i     (equation 4)
assuming (!) the sum runs over all relevant objects.


> Suppose the values of these coefficients are given.
> How to express C12 in terms of these coefficients?
> By C12 I mean the capacitance a battery would see
> if it were connected to leads attached to vertical plates?

c) That is an unhelpful notation. IMHO C12 should denote the (1,2) element
of the C matrix, equivalent to the b12 that Ludwik wrote.  It represents
the charge on element 1 induced by a voltage on element 2.  The symbol Cij
should NOT be redefined to represent the two-terminal capacitance.

d)  The two-terminal capacitance is not even well-defined unless we specify
what is happening to the third element.  If it is floating, that's one
thing.  If it is grounded, that's another.  Since Ludwik mentioned a
battery, he probably intended that elements 1 and 2 (and the battery)
should be floating relative to element 3.  We will handle that case in a
moment, but first let's do the other case, with element 3 grounded.

This may seem like pure contrariness, but it illustrates some good physics.

Also, let's arrange that element 3 is NOT symmetric w.r.t. the others;
otherwise it's too easy.  As an example, consider the following attempt at
a perspective drawing:

.                      ///////////////
.             V1------///////////////
.                    ///////////////
.
.
.                     ///////////////
.             V2-----///////////////
.                   ///////////////
.
.
.                       ////////
.             V3-------////////
.                     ////////
.


The capacitance matrix is

.          3   -3   0
.    C =  -3    5  -2                   (equation 5)
.          0   -2   2

Note that plate 1 has no direct interactions with plate 3.

Anyway, if we apply V1 = +1 volt and V2 = -1 volt, the charge distribution is

  Q1    6
  Q2 = -8
  Q2    2

So you can see that the notion of two-terminal capacitance doesn't even
make sense in this case.  Is it 6, or is it 8?  You get 6 coulombs per volt
on one plate and 8 coulombs per volt on the other!

OK, enough of that.  Let's stop being contrary and return to the case that
was probably intended, where we have
        V1 = +1 = known, Q1 unknown
        V2 = -1 = known, Q2 unknown
        V3 unknown, Q3 = 0 = known <==== since it is "floating"


Let's start by solving for V3.  From equation (2) we immediately obtain
        Q3 = C31 - C32  C33 V3          (equation 8)
and then we set Q3 = 0.  It's not very hard to solve 1 linear equation in
one unknown.  We obtain
        V3 = (C32 - C31) / C33          (equation 9)

Now that we know V3 (and still know V1 and V2) we can immediately apply
equation 2 to obtain expressions for Q1 and Q2.

        Q1 = C11 - C12 + C13 (C32 - C31) / C33

and since this is the charge resulting from a unit voltage it is also the
sought-after expression for the two-terminal capacitance.  The formula for
Q2 can also be found from equation 2.  We get

        Q2 = C21 - C22 + C23 (C32 - C31) / C33

As a useful check, you can verify that Q1 = -Q2 even though that's not
super-obvious from the foregoing expressions.  You will need to use gauge
invariance and charge conservation to show this.

> 3) JohnD provided a hint "V = C^-1 Q, for imposed Q"

That wasn't a very useful hint.  In fact, it's totally wrong.  The
capacitance matrix doesn't have an inverse.  It's singular.  It has to be.
Gauge invariance guarantees you can't possibly calculate the voltage given
the charge in this way.  Sorry.