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Re: [Phys-l] Capacitor problem reprise



At 12:05 PM 3/28/2008, M R Meyer, you wrote:

For those who followed the original capacitor problem posted earlier this
week: I shared your solutions and responses with the author (Bryan Suits)
and after some thought, he has asked me to post the following as a
follow-up for your continued comment:

In physics classes we often use schematic symbols to represent
real objects rather than terms of an equation. Schematics are
just easier to draw on a blackboard. It is an interesting
difference of perspective but is not the essential physics.
I also wish to avoid issues of small effects versus big as
this distracts from the understanding. Therefore I present
a closely related problem without symbols where numerical
values are not an issue.

We are talking electrostatics here and you are welcome to
add resistors to keep currents from becoming infinite while
equilibrium is being established, etc.

You have an (effectively) infinite conducting ground and a
capacitor characterized with capacitance C which can be
as large as you like. One side of the capacitor is connected
to the ground. A battery with EMF V > 0 is attached between
ground and the other side of the capacitor. You then have
a charge Q = CV on one plate and a charge of equal
magnitude and opposite sign on the other. In this 1st case,
the battery is oriented so that the positively charged plate
of the capacitor has a potential relative to ground of V while
the other is at ground potential.

Now disconnect the battery and then sever the ground connection.

One question is, now what are the potentials of the two plates
relative to ground?

Before you answer, consider a 2nd situation which is
identical except with the battery turned around so that
you start with a capacitor with the same charges (+Q and -Q),
though on opposite plates, and now the negatively charged
plate is at a potential of -V relative to ground and the
positive plate is at ground potential. When you are done
can you tell whether the capacitor was prepared using
the 1st or 2nd method? If not, then the potentials
relative to ground must have changed. If so, then there
must be something (physically) different that you can
detect which "stores" information from the preparation
process. Knowledge of the capacitance value C does not
help you explain the answer in either case, there is more
to the problem.

- BHS


To model the scenario, I presume the device is mounted on high quality
standoffs in a metal gauze box. I mark the capacitor with a
red blob at the relatively positively connected end.

To realistically model the effect of disconnecting both
sides of the cap from its charging voltage I account for the
stray capacitance from either end of the subject capacitor.
This is likely to be of the order of 3 pF per end to the gauze 'ground.'

I add a standoff leakage resistor of 1E12 ohms from each end of the cap
to the gauze ground, and I finally add an internal leakage resistor
across the ends of the capacitor of 1E8 ohms.

I can see that when equilibrium is reached, the voltage at either
end of the test capacitor is equally disposed about 'ground'.

The potential across the cap is ostensibly the same as when it
was charged, but it slowly decays through the internal and
external leakage paths.

In the case of the capacitor charged negative with respect to
ground, it is convenient to retain the significance of the red
marked end - as positive with respect to the other end.
But we will reverse the position of the cap so the red spot is
now at the other end as viewed from the constant viewpoint.

This time, the negative end leaks up to a half potential
difference below 'ground', while the 'grounded' end floats
slowly up to a half p.d. above 'ground'.
Then the PD leaks down with time
The capacitor charging source polarity is easily distinguished.
There is no mysterious effect at all, that I can see.....



Brian Whatcott Altus OK Eureka!