Re: Capacitor energy experiment

[Robert Cohen <Robert.Cohen@PO-BOX.ESU.EDU>]

Once again, I'm lost.

It seems to me that the real issue is how we define work.
Now we've gotten into a discussion about where the energy
goes when we charge a capacitor.  These are two separate
issues, it seems to me.

Let's suppose that the energy associated with the capacitor
is QV (or Q^2/C, from the definition of C=Q/V).  Capacitor
A has charge Q.  It is then connected to capacitor B with
the same capacitance C.  The charge Q is split between
the two capacitors.  What is the energy associated with
capacitors A and B after being connected?  Feel free to
assume steady-state, oscillations, etc., whatever you'd like.

Is the energy after being connected the same as that before
being connected?  Does it depend upon the assumption of
steady-state, oscillations, etc.?

Finally, do we get a different answer if we assume the
energy associated with the capacitor is 0.5QV instead
of QV?

____________________________________________
Robert Cohen; rcohen@po-box.esu.edu; 570-422-3428; http://www.esu.edu/~bbq
Physics, East Stroudsburg Univ., E. Stroudsburg, PA 18301

> -----Original Message-----
> From: David Rutherford [mailto:drutherford@SOFTCOM.NET]
> Sent: Wednesday, February 05, 2003 6:40 PM
> To: PHYS-L@lists.nau.edu
> Subject: Re: Capacitor energy experiment
>
>
> Ludwik Kowalski wrote:
> > My description of the experiment with the air capacitor
> > was not complete. There should be an electronic switch
> > so that charging is done in a different loop that discharging.
> > The electronic switch would perform the same function as
> > a mechanical switch but much more rapidly, for example,
> > every 0.1 ms. Then the calorimeter would measure the
> > "amount of heat" generated in N discharges, each
> > delivering only 0.5*C*V^2 of thermal energy.
>
> The point of my experiment is to show that for Ec ~ 0, that is for the
> energy Ec << 1/2 CV^2 dissipated while charging the capacitor, the
> thermal energy dissipated while discharging the capacitor will be Ed ~
> CV^2. For Ec = 0, I predict that you would get Ed = CV^2, not 1/2
> CV^2, as predicted conventionally, therefore the energy stored on the
> capacitor after charging and before discharging would be CV^2, not 1/2
> CV^2, as predicted conventionally. That would disprove one of the most
> basic laws of physics - the energy of a distribution of charges.
>
> Unless I'm mistaken, the experiment you're describing is old hat.
> There's nothing new to be discovered doing the experiment your way. If
> I'm incorrect in assuming that your way is the same old thing, please
> explain to me why it's not.