Re: Capacitor energy experiment

[David Rutherford <drutherford@SOFTCOM.NET>]

"Kilmer, Skip" wrote:
> Hey, David
> I think I'm missing the point as well. Are you implying that the firs=
> t charges arriving at the uncharged capacitor have CV^2 of energy? Si=
> nce there is no potential difference across the capacitor at that tim=
> e, it requires essentially zero work (in the absence of dissipative c=
> oncerns) to put them on the plates. As time goes on the capacitor's p=
> otential difference rises to some final value V determined by the cha=
> racteristics 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 store=
> d energy, equal to the work done in charging, it will be CV*V/2, or 1=
> /2CV^2. Where's the missing energy?

I can only assume that you are playing games, now. Nobody can be THIS
stupid in real life.

--
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