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Re: Capacitor problem



It is emitted as E-M radiation, since there is no other place for it to
have gone. If there are normal wires, then it is quoted in Halliday and
Resnick that the loss is via Joule heating in the wires.


On Fri, 28 Mar 1997, Donald E. Simanek wrote:


Since this group gets charged up over electrostatics, here's an old
problem which many of you may have seen before. If so, you can restrain
yourself and let the others have a crack at it.

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.

-- Donald

.....................................................................
Dr. Donald E. Simanek Office: 717-893-2079
Prof. of Physics Internet: dsimanek@eagle.lhup.edu
Lock Haven University, Lock Haven, PA. 17745 CIS: 73147,2166
Home page: http://www.lhup.edu/~dsimanek FAX: 717-893-2047
.....................................................................




----------
Tom K. McCarthy Email:mcca6300@spacelink.msfc.nasa.gov