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Re: [Phys-l] Capacitor problem reprise



My take is this:

Once the capacitor plate closer to the battery is disconnected, there's no way (short of ionization leakage or something not in the problem) to change the charge upon it. The capacitor prepared with the EMF>0 keeps whatever (+) charge was deposited upon it; that prepared with EMF<0 keeps that (-) charge.
The other plates of these capacitors stay at ground potential, whether or not they're connected to ground, because their charges are defined by their counterpart plates, once things are at equilibrium. They can be connected to ground as often as one likes, with no net charge flow to/from ground potential. Disconnecting the capacitors from ground allows no rearrangement of energies or charges in the system, because C is fixed, as are the charges on the non-grounded plates.
One *can* tell in which way the respective capacitors are prepared -- hook a current meter between each of the (formerly) battery-adjacent plates and ground, and see which way current flows to discharge the plates. Equivalently, measure the voltage between the (plus-prepared and minus-prepared charged) plates. As long as the counterpart plates are at the same (ground) potential, the voltage will tell one how the capacitors were prepared.

One final thought (which may be pedagogically useful for instructors): Imagine two disconnected plates, lying fixed in the same plane. Then connect them via a battery with some EMF. All one can say is that one plate becomes positive relative to the other. Now bring in an infinite "ground" plane parallel to the plates' plane. Does the total energy of the battery-connected plates change? One approach says "no" -- there is no change of potentials between the plates, and no net charge flow. However, there will obviously be a rearrangement of the electrostatic field lines as the ground plane comes closer. There may also be a net torque on the ground plane due to the unequal potentials through which it moves.




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Down with categorical imperative!
flutzpah@yahoo.com
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----- Original Message ----
From: "mrmeyer@mtu.edu" <mrmeyer@mtu.edu>
To: phys-l@carnot.physics.buffalo.edu
Sent: Friday, March 28, 2008 12:05:29 PM
Subject: [Phys-l] Capacitor problem reprise

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




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