Re: Missing Energy

["JACK L. URETSKY (C)1998; HEP DIVISION, ARGONNE NATIONAL LAB ARGONNE, IL 60439" <JLU@HEP.ANL.GOV>]

Reply to Mark Sylvester and Herb Schulz:
*********************************************88*************
At 12:22 PM -0500 on 5/20/99, C wrote:
>        So do you conclude, as I do, that there is no issue of
> "missing energy" because the closed system in which energy
> is conserved consists of the two capacitors and the person
> (or device) that brings the two capacitors together?  The
> person must exert a force to combine the two plus leads and
> the two minus leads.
>        Thank you.
>                        Regards,
>                                Jack

Howdy,

But there is less stored energy afterward than before, not more!?  Or do I
have this wrong?  I always thought that the energy was radiated away (and
also went into thermal energy if the resistance in the wire was taken into
account.

Good Luck,

Herb Schulz
(herbs@interaccess.com)
********************************************
        You read my posting too narrowly.  The point is that an external
agent must manipulate the wires and exert whatever forces are necesssary
to bring them together.  The agent, in exerting the forces, does either
positive or negative work on the system.
        Thermal energy cannot account for the difference, because you'd
calculate the same energy difference with lossless wires.  Also, the energy
change in the capacitors cannot be accounted for by radiation, because the
change is independent of the size of the system and the speed with which
the connection is made.
        I'll illustrate the last point by changing the problem a bit.
Connect the negative ends of the capacitor through a resistor having
resistance R, and the positive ends through a switch that is initially
open.  Close the switch at t=0 and calculate the charges on the capacitor
plates as a function of time.  At the end of the day the energy stored in
the capacitors is the same as in the original problem, but now we have
had current flow through the resistor.  I think that you will find that
in this new setup, the joule heating in the resistor just accounts for
the energy loss in the capacitors.  We can neglect the miniscule amount
co energy required to close the switch.
                                        Regards,
                                                Jack

        "I scored the next great triumph for science myself,
to wit, how the milk gets into the cow.  Both of us
had marveled over that mystery a long time.  We had
followed the cows around for years - that is, in the
daytime - but had never caught them drinking fluid of
that color."
                Mark Twain, Extract from Eve's
                Autobiography