Re: There's work, and then there's work

[David Rutherford <drutherford@SOFTCOM.NET>]

John Mallinckrodt wrote:
> The discharging case is similarly well-modeled as an (initially
> charged) capacitor in series with an (initially open) switch and a
> resistor.  After the switch is closed, the capacitor discharges
> through the resistor.  The charge Q on the capacitor flows as the
> voltage across both the capacitor and resistor falls from V to 0.  It
> is simple to perform the required integrals that show that a total
> energy of QV/2 is removed from the capacitor (which had stored it in
> the form of electrostatic energy) and a total energy of QV/2 is
> delivered to the resistor (which dissipates it as heat.)  Pleasingly
> (and completely unsurprisingly), the energy supplied by the capacitor
> is completely accounted for by the energy delivered to the resistor.

Then how do the charges continue on to the other plate of the capacitor,
in order to equalize the potentials of the plates, if they've expended
all of their kinetic energy in the resistor? If what you say is true,
you could install a galvanometer downstream from the resistor and it
would show zero current during the discharging of the capacitor. I
assume that's not what's observed.

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

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