Re: A funny capacitor.

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Referring to the "funny capacitor" Ludwik asked a question;
see part 1 below. JohnD provided a hint; see part 2 below.
This hint was not sufficient; see part 3 below. I am asking
for additional help.

PART 1
*******

I would know how to calculate C of the funny capacitor, at
least in principle, if the potential of the bottom plate were
specified, for example zero. In that case I would impose
potentials on vertical plates, for example, -50 and +50 V,
and used the Laplace equation to find surface charges
on each vertical plate. Then C=Q/V. The boundary
conditions (potentials on three objects) is given and the
rest is just a matter of number crunching.

               *           *
           *           *
 -50 *******           ******* +50
           *           *
           *           *

             *********
                *
                * "floating"

But how can the problem be solved when the lower plate is
floating. That plate will certainly find its own potential, that
is it will adjust to potentials on other plates. But I do not
know to find that potential. Therefore the Laplace equation
has an undefined boundary condition. Is this problem
solvable? Note that plates are in fixed positions with
respect to each other.

PART 2
*******  JohnD responded, in part:

> In that case I would impose potentials on vertical plates,
> and used the Laplace equation to find surface charges
> on each vertical plate. Then C=Q/V.

1) Q = C V where C is the 3x3 capacitance matrix. You can
walk through the columns by changing the imposed voltage
on each of the elements.

> The boundary conditions (potentials on three objects) is
> given and the rest is just a matter of number crunching.

2) Yes.  There exist many ways to do the crunching, including
the spreadsheets discussed in this connection last month.

> But how can the problem be solved when the lower plate
> is floating.

3) V = C^-1 Q, for imposed Q.

Inverting a 3x3 matrix isn't very hard.  A fellow named Gauss
had something to say about it.

PART 3
*******

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? Assume, for example, that all plates are
squares of size L=8 cm, that the distance d between the
vertical plates d1=10 cm and that the horizontal plate is
at the distance d2=3 cm below the bottoms of vertical
plates.

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?

2) Yes, I learned a lot from messages on how to solve
electrostatic problems using the Laplace method.

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

a) That would be useful if C12 and C13 were given.
But in this problem the goal is to calculate these
capacitances for a specified geometric arrangement.

b) Inverting a 3x3 matrix isn't very hard.  A fellow
named Gauss had something to say about it.

It is true that Gaussian elimination is more efficient
than Cramer's rule (as JohnD wrote in another
message) but that is a different matter.

I still do not know how to correctly predict the
potential V3 on the electrically-floating horizontal
plate when potentials V1=-50 volts and V2=+50
volts, for the specified geometry. The left-right
symmetry was implied What is the value of V3
for the geometry specified above?

If the bottom plate was not a conductor then the
potentials from left to right would be something
like from -45 volts to +45 volts. Intuitively I feel
that V3 on the metallic plate would depend on its
vertical location; it certainly zero if the location
is very far away from vertical plates.

On the other hand I was tempted to assume that
V3=0 (at any location due the left-right symmetry).
This is probably not correct. Why not? What is
the correct answer about V3?
Ludwik Kowalski