Re: A funny capacitor.

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

William Beaty wrote:

> Start with the 2-plate capacitor and the extra plate far away.
> As the extra plate approaches from a distance, there is attraction
> by induction and work is done (the plate accelerates.)  The
> mechanical energy had to come from the electrostatic energy
> stored in the system, PE = 1/2 V^2 / C, so the approaching
> extra plate caused V to decrease, or C to increase, or both.
> Since Q = C * V, and since Q is constant (nothing connected
> to the plates), any change in C causes a simultaneous opposite
> change in V, so if the stored energy decreased, then V *had*
> to decrease and C *had* to increase.  (C is the total capacitance
> as measured between the two terminals.)

To determine C experimentally I would connect the leads
to a battery and watch for the Q. If Q goes up when the third
plate is approaching then C2 is larger than C1. If the Q goes
down then C2 is smaller that C1. It is only a matter of looking
at the direction of a transient current with a galvanometer.

You suggest to charge C1 from a battery of known Vo (to
get Q=C1*Vo) and then to disconnect the leads. After that
you watch what happens to V as the horizontal plate is moved
a little closer. Presumably a gold-leaf electrometer shows if V
goes up or down. If V goes up then C becomes smaller, if V
goes down then C becomes larger.

The two approaches are equally valid. But we are not
discussing an experiment. We want to make a theoretical
prediction. So it is a gedanken experiment. I agree that
the horizontal plate is going to be polarized and will be
attracted to vertical plates. The quazi-static work done
by you (to prevent acceleration) will be negative and
the potential energy will be decreasing. From this you
come to a conclusion that V must go down.

I would agree with you if one of the two parallel plates
was moved toward another (and the third plate was
not present). Why?  Because in this case V is the measure
of the total potential energy. But I am not sure that your
conclusion is justified for the "funny capacitor" where two
values of V must be known to determine the potential
energy.

We recognized that when the horizontal plate is allowed
to come closer (in a virtual displacement), the number
of field lines linking the horizontal plates directly becomes
smaller. That is why my original prediction was C2<C1.
But I am no longer convinced it was correct. Perhaps I
should think it over when I am less tired.
Ludwik Kowalski