Re: A funny capacitor.

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

ANOTHER QUESTION

Suppose I have a lot of time and motivation to calculate the
Cij coefficients for an arbitrary set of three floating objects.

  Q1=C12*V1 + C12*V2 + C13*V3
  Q2=C21*V1 + C22*V2 + C23*V3                            Equations (2)
  Q3=C31*V1 + C32*V2 + C33*V3

Here I am facing 9 unkowns. The equation 3 (Cij=Cji)  reduces
the number of unknown to 6. Suppose it is correct to interpret
Equation 4 as "sum of Cij is zero in each column". Now I have
6 equations with six unknown. What remains is to calculate Q1,
Q2 and Q3 for an arbitrary set of V1, V2 and V3; more or less
as in illustrations shown here last week.

Suppose this step is over and all Cij were calculated (in Farads).
What is next? We want to get the capacitance seen by the battery
connected to vertical plates of our funny capacitor.

"John S. Denker" wrote:

> ... Let's return to the case that was 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. ....

Using the notation of the original message I would write

         C2 = C11 - C12 + C13 [(C32 - C31) / C33]

Then I would move the horizontal plate further away and do the
same thing for the new setup. The new capacitance C2' should
be smaller than C2. I finally know what to do if I wanted to
compare C2 and C2' numerically. I would like to thank JohnD
again for being a patient tutor. Mission accomplished; I wish I
had time and motivation to do all the number crunching.

Aha, one more question. At the end of his message John wrote:

> The capacitance matrix doesn't have an inverse.  It's singular.

What is wrong in saying that

  V1=B11*Q1 + B12*Q2 + B13*Q3
  V2=B21*Q1 + B22*Q2 + B23*Q3
  V3=B31*Q1 + B32*Q2 + B33*Q3

is the inverse of equation 2 (and vice versa)?
Ludwik Kowalski