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Re: A funny capacitor.

William Beaty wrote:

... Yes, my thinking is very visual and handwavy. It's informed
guessing while thinking aloud. I start talking, then monitor my
own "bullshit detector" to see if half-subconscious analysis agrees
with what I'm saying. I end up with an understanding of what's
happening, but it doesn't give me total certainty. ....

We are in the same boat and your arguments make sense. But
let me approach the question differently. First I notice that there
is left-right symmetry in A and B (see the original question at
the end). Thus if a battery is connected to the leads the charges
Q, on the vertical plates, will be identical in each setups.

Suppose I want to measure C1 and C2 by connecting them to
a battery of known V and by measuring Q which it supplies
to charge each capacitor. This can be done with a scope (by
numerically integrating under the I(t) trace) or with a ballistic
galvanometer calibrated in units of charge. The ratio of Q/V
will give the value of C for each setup.

Instead of asking "which of the two setup has a larger C?"
I can now ask "which of the setups should take more Q from
the battery?" Why should the presence of the third plate
require a larger Q?

I tried to start from the most general description but got
stuck. In that approach we have three metallic objects.
Their potentials (V1, V2 and V3) and their charges (Q1,
Q2 and Q3) are related in the following way:

Q1=b12*V1 + b12*V2 + b13*V3
Q2=b21*V1 + b22*V2 + b23*V3
Q3=b31*V1 + b22*V2 + b33*V3

I am not certain how to express a capacitance, for example
C12 (between the leads connected to objects 1 2) in terms
of b coefficients. The net Q3 (on the object not connected
to the battery) must be zero. Furthermore, Q1=Q2 in each
setup (because of the left-right symmetry). But V3 is not zero.

In principle the answer to my question is in the above
equations but I can not get it. Perhaps somebody will help.
What follows is the first message of this thread. I am now
less certain of the correctness of my answer.


Let me present a Hewitt-like question.

Consider a parallel plates capacitor as at A below.


* *
* *
******* ******* C1
* *
* *

The floor, the ceiling and the walls are very far
away. The capacitance is C1. Then bring a third
plate and position it as in B below.


* *
* *
******* ******* C2
* *
* *


Will the new capacitance, C2, be larger, smaller
or the same as C1? In my opinion it will be smaller.
Do you agree? Please justify the answer. [Plates
are in fixed positions with respect to each other.]
Ludwik Kowalski