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Re: Capacitor problem, another question

I should think that the "missing" energy has been radiated. As the
charges re-distribute themselves from the originally "fully-charged"
capacitor into the formally un-charged capacitor, they briefly form a
current. If there is resistance in the connecting wires, it is easy to
see that the energy would be dissipated as heat in the wires. If there
is no resistance in the wires, then generated magnetic fields created
from the current would become non-negligible. The wires have some
inductance, hence the current builds "momentum" and we would have an LC
oscillator. My thoughts obviously evolve as I write this. The energy
has gone into LC oscillations. If the capacitors and and connecting
wires are superconductors, then the LC oscillations should last forever,
eventually radiating E-M waves.

Donald F. Collins
~~~~~~~~~~~~~~~~~
Physics Department Warren Wilson College Asheville, NC 28815
(704) 298-3325 dcollins@warren-wilson.edu

On Tue, 1 Apr 1997, LUDWIK KOWALSKI wrote:


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.