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Re: A funny capacitor.

I accept your argument based on energy considerations.
Ludwik Kowalski

William Beaty wrote:

On Tue, 27 Feb 2001, Ludwik Kowalski wrote:

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

We don't need to measure V to determine the sign of the changing
electrostatic potential energy of the system. We only need to note that
the 3rd plate is attracted, therefore the total electrostatic PE of the
system must decrease as that plate is allowed to approach the two others.

Suppose we charge a 2-plate capacitor, disconnect it from the charging
source, then bring the 3rd plate nearby. We *know* that the stored energy
has decreased. Suppose we ignore the voltages between the third plate and
each of the two parallel plates, and instead focus attention only on the
voltage between the two parallel plates. Could it conceivably rise as the
energy decreases? Since PE = 1/2 * C * V^2, the voltage can only rise if
the total capacitance (as measured across the plates) decreases MORE than
in inverse proportion to the voltage rise (since the V^2 term is present.)
However, the parallel plate capacitor must also obey Q = C * V, which
insists that voltage and capacitance can only vary in inverse proportion
to V. Therefore V cannot rise as the 3rd plate approaches. It can only
change in inverse proportion to C, therefore if the stored PE decreases,
we are forcibly led to conclude that C increases and V decreases. (Note
that the presence of the third plate does not affect Q=C*V because the
charge Q on each of the parallel plates never varies. )

My reasoning is totally ignoring the voltages measured between the 3rd
plate and each of the parallel plates. They aren't needed.

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.

That was correct, but the presence of the two new capacitors then
increases the value of C2 by some amount. My reasoning above shows
that this increase is greater than the decrease caused by the reduced
number of field lines linking the two plates, so the net value of the C2
capacitor has increased

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