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Re: Capacitors problem

On Fri, 28 Mar 1997, in responding to Donald's problem (see below)
Mark Shapiro wrote:

The "lost" energy goes to electromagnetic radiation. The system oscillates!

Are you answering in this way because of:

1. energy must be conserved.
2. where else can "the lost energy" go under specified conditions?
3. this is a reasonable extrapolation of what we know. The final L of
wires is responsible for an oscillation of charges. Large LC systems
oscillate and, in classical physics, this produces e.m. waves.

I did discover this "paradox" five years ago and I liked it very much.
I agree with your answer but I would be much more satisfied if I knew
that somebody really measured the amount of energy radiated and showed
that it is equal to what is missing (after accounting for the heat
generated in wires). Let me add another question. What happens when, in
addition to R-->0, we also impose L-->0 (infinitely short wires, etc.)?

Contrary to what Don said (an earlier message) I think that "gedanken-ing"
is not as convincing as real experimental data. It is a great tool in the
hands of great minds when nothing else is available. (Did somebody answer
the "how many volts" question experimentally? Yes, this is a digression.)

Mark, did you find the solution entirely by yourself? I needed a hint to
realize that emission of radiation is involved in this "electrostatic"
problem.

A good "paradox" for an advanced course but not for an introductory course
when "electricity at rest" is presented before "electricity in motion".
I tentatively plan to revers this order next year.
Ludwik Kowalski
**********************************************************************

Date: Fri, 28 Mar 1997 21:46:06
From: "Donald E. Simanek" <dsimanek@eagle.lhup.edu>

Two identical parallel plate capacitors are used. One is charged with
charge Q, so the plates have Q and -Q respectively. The other capacitor
is still uncharged. The stored energy of the charged capacitor is CV^2/2,
or Q^2/2C and the stored energy of the other one is zero.

Now perfectly *resistanceless* wires are used to connect the capacitors
as shown, being careful not to add or remove any charge from the system.

-------------
| |
------- -------
------- -------
| |
-------------

The charges redistribute, so that each capacitor has charge Q/2. The
total stored energy (both capacitors) is now

2 2
(Q/2) 1 Q
2 ------ = - --
C 4 C

This is half as much stored energy as before. Where did the rest of the
energy go?

The charge redistribution can be accomplished by having the wires in
place and a remote-controlled pair of switches complete the connection.
No mechanical work need be done by the agent performing this operation.
**************************************************************************