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Re: funny capacitor



TOO BAD THAT THE SPRING BREAK ENDS BEFORE I
BECOME COMFORTABLE WITH THE V(Q) PROBLEMS.
I WILL CONTINUE PROBING, AS MUCH AS I CAN.

We agreed on the Cij coefficients for the funny capacitor:

+2.86, -0.46, -1.15, -1.25
-0.46, +2.86, -1.15, -1.25
-1.15, -1.15, +3.26, -0.96
-1.25, -1.25, -0.96, +3.46

I am not comfortable with C44=3.46. Here is the
reasoning. Where am I wrong?

John wrote "we have four independent voltages. We
can set voltages on anything we want, and turn the
crank on Laplace's equation." We did this to find the
Cij coefficients. At the same time we found that an
arbitrarily chosen set of potentials (V1=-100, V2=+100
and V3=+20) produced the well defined charges
(Q1=-198, Q2=143 and Q3=+65.2). In other words
we found the "one and only one" solution of the Q(V)
problem. We think that no other combination of Vs
can produce the same set of Qs.

As often emphasized by John, V1, V2 and V3 are really
potential differences with respect to object #4; we call them
potentials but this is only a verbal shortcut. The gauge lead
of a voltmeter would always go to the object #4. In other
words, the object #4 is associated with the gauge. John
wrote: "the V(Q) problem does have four independent
variables, namely three charges and one gauge."

The entire universe, in the context of this problem, consists
of four objects which either receive or give away charges.
The first three objects are plates of our funny capacitor. Let
the fourth object be an enclosure. Why not? By knowing Q1,
Q2 and Q3 we can always calculate Q4 as -(Q1+Q2+Q3).
The value of V4 remains zero no matter how large is Q4. A
difference of potentials "with respect to itself" is always zero.

A non-gauge object i whose potential does not change when
its net Q changes would have Cii=infinity. So what does it
mean that C44=3.46? Doesn't this imply that V4 must change
when Q4 changes? Something is not right. What is it?
Ludwik Kowalski