Re: Capacitor problem

[Herb Schulz <herbs@interaccess.com>]

> ...
> 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?
> ...

Howdy,

You actually have the same problem in just charging a capacitor with a
battery in a circuit with negligible resistance.  The energy supplied to
the charge by the battery is QV while the final energy stored in the
capacitor is QV/2.

Good Luck,

Herb Schulz
(herbs@interaccess.com)