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Re: Capacitor energy experiment



Hey, David
I think I'm missing the point as well. Are you implying that the first charges arriving at the uncharged capacitor have CV^2 of energy? Since there is no potential difference across the capacitor at that time, it requires essentially zero work (in the absence of dissipative concerns) to put them on the plates. As time goes on the capacitor's potential difference rises to some final value V determined by the characteristics of the device (C) and the amount of charge stored on its plates. Since the charge of the "full" capacitor will equal CV, and the average potential difference across it will be (V+0)/2, the stored energy, equal to the work done in charging, it will be CV*V/2, or 1/2CV^2. Where's the missing energy?
skip

-----Original Message-----
From: David Rutherford [mailto:drutherford@SOFTCOM.NET]
Sent: Thursday, February 13, 2003 11:56
To: PHYS-L@lists.nau.edu
Subject: Re: Capacitor energy experiment


Michael Burns-Kaurin wrote:

I will try asking Bob's question in a different way. Suppose I have two 1
F capacitors, each with potential difference 1 V. What physical difference
between the capacitors makes one of them "store" 1 J while the other
"stores" 0.5 J?

You are missing my point, completely. Suppose, for the sake of argument,
that it is _impossible_ to charge a capacitor without 1/2 CV^2 of energy
being dissipated during charging. Also suppose that the energy stored on
the capacitor is fixed at 1/2 CV^2. I'm not saying that I believe that
this is the only possibility; I'm only trying to show that, if this was
the only possibility, the energy dissipated during charging, should be
included in the total energy of the charges. Otherwise, if you don't
include it, the experiment doesn't provide valid evidence for the work
done on distribution of charges (or the energy of the field) since,
theoretically, the work done to assemble a distribution of charges
doesn't consider dissipation of energy during the assembling of the
charges. The perfect experiment would be to charge the capacitor in the
absence of dissipation. But since that is probably not going to happen,
I'm trying to convince you that, in the conventional experiment (the one
where 1/2 CV^2 is dissipated), 1/2 of the energy of the charges - the
energy dissipated during charging - that should be counted, is not being
counted. That energy may no longer be associated with the charges stored
on the capacitor after the capacitor is charged, but it was at one time
associated with those charges and is still somewhere in the field. Work
can be derived from it and it should be counted.

--
Dave Rutherford
"New Transformation Equations and the Electric Field Four-vector"
http://www.softcom.net/users/der555/newtransform.pdf

Applications:
"4/3 Problem Resolution"
http://www.softcom.net/users/der555/elecmass.pdf
"Action-reaction Paradox Resolution"
http://www.softcom.net/users/der555/actreact.pdf
"Energy Density Correction"
http://www.softcom.net/users/der555/enerdens.pdf
"Proposed Quantum Mechanical Connection"
http://www.softcom.net/users/der555/quantum.pdf