Re: non-dissipative circuitry

[Chuck Britton <britton@NCSSM.EDU>]

At 12:53 PM -0500 2/4/03, John S. Denker wrote:

> Chuck Britton wrote:
>
> >  It seems to ME that the METHOD of connecting the capacitors is
> >  absolutely NO effect on the final state of the two capacitors.
> >
> >  Connect them with and inductor, connect them with a resistor. Once
> >  things calm down, the charge will have redistributed itself in a
> >  totally predictable way. Energy has been 'dissipated'. Exactly the
> >  SAME amount of energy, regardless of the connection method. nicht
> >  wahr???
>
> Ganz nicht.  Method matters.

ok, I understand your mechanical non-dissipative examples BUT:

       Back to the twin capacitor test.

Conservation of charge seems to ME to be powerful enough to define
the final state of the system.

Final state requires that the two voltage differences are equal AND
the total amount of separated charge remains unchanged.

Where is my thinking going astray????


e.g.    two identical capacitors of capacitance C.
`       one charged to a voltage Vi, the other uncharged.
       charge on the charged one = CVi

       After they are coupled together - spaaacks or whatever -

       The final capacitance is now 2C, holding the original charge = CVi
       at a final voltage of Vf = (CVi) / 2C = 1/2 Vi

       Initial electrostatic energy = 1/2 CVi^2
       Final electrostatic energy = 1/2 (2C) Vf ^2 =
               1/2 (initial electrostatic E)   `regardless of method !?

Find my error, before I jump over to the Rutherford side!  ;-)