[Prev][Next][Index][Thread]

Re: Capacitor Charge-Were is it?

Tesla List wrote:
> 
> >From huffman-at-fnal.govTue Oct 29 22:47:16 1996
> Date: Tue, 29 Oct 1996 11:03:10 -0600
> From: huffman <huffman-at-fnal.gov>
> To: List Tesla <tesla-at-pupman-dot-com>
> Subject: Capacitor Charge-Were is it?
> 
>     [The following text is in the "ISO-8859-1" character set]
>     [Your display is set for the "US-ASCII" character set]
>     [Some characters may be displayed incorrectly]
> 
> Hi All,
> I want to thank everyone for the replies. For me this has been a very
> interesting discussion although I still am having problems with it. It
> seems like the chicken and the egg question to me. You can't have a
> capacitor without a conductive plate/s and a dielectric. If I say the
> charge is on the plates, I probably should say the charge is on the surface
> of the plate/s, since that would put it in the dielectric. If you dump
> charge on the inside of a conductor it moves to the surface, and I
> understand the charge cannot be in the metal plate. I still envision all
> those electrons crowding around on the metal plate of a capacitor.
> Dielectrics are strong, dielectrics are impressive, but it's conductors
> that do all the work. (sorry Mr. Clemens)
> 
> In a realm we cannot see, can we really know?
> Sockit to me, just don't kick me off the list.
> Dave Huffman
> 
> P.S. The two capacitor problem...
> The lost energy goes into making heat, light, EM radiation, sound, etc. Why
> 1/2?
> I know it's not the same situation, but does half the energy also get lost
> going from the Cp to Cs?
> Can I transfer energy from one cap to another with minimal loss?

Dave,

1. 'Cause that's the way it is. :^)
  Seriously, I really don't have a GOOD explanation other than this is
the way the world seems to work when you've got losses in the circuit. 
  
2. You're right its definately not the same situation, and I certainly
don't think the theoretical losses are 50%! See my post on "50%"
yesterday.

3. Yes! Same answer as 2. In your PSPICE model, try setting the
resistance in the primary circuit to a relatively low value and then
compute the maximum energy in the secondary versus your initial energy
in the primary. You can even leave the secondary "lossy" and not change
the results by much at all..


-- Bert --