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Re: Capacitor charge, were is it?
Tesla List wrote:
>
> >From lod-at-pacbell-dot-netSun Nov 3 22:01:23 1996
> Date: Thu, 02 Nov 1995 18:52:00 +0000
> From: GE Leyh <lod-at-pacbell-dot-net>
> To: tesla-at-pupman-dot-com
> Subject: Re: Capacitor charge, were is it?
>
> Tesla List wrote:
> big snip
>
> Again, it's quite confusing to use the words 'energy' and 'charge'
> interchangeably!! They really are very different things.
> May I suggest that a section of above rebuttal should read:
>
> _____________________________________________________________________________
> "You have hit on what I have been saying all along about the dielectric/
> metal interface being the point of charge separation... IN A CAPACITOR
> AND A CAPACITOR ONLY! IF A CHARGED CAPACITOR IS DISMANTLED, THE _energy_ IS
> ONLY FOUND IN THE DIELECTRIC AND ZERO _energy_ IS RETAINED BY THE PLATES.
>
> Thus, you are right and wrong at the same time!
>
> Only a dielctric can store (electrical)_ENERGY_! In a capacitor the..."
> _____________________________________________________________________________
>
> This makes a lot more sense to me, since _any_ isolated object with a
> different number of positive and negative charges has a net charge,
> regardless of its conductivity.
> The electric field that emanates from any charged object contains a certain
> amount of energy (not charge), depending the the dielectric constant of
> the medium (1.00 for vacuum) and the total potential drop.
>
> -GL
I'll agree for the sake of argument. But where is the charge? I would
think that it is where the energy is!! Energy, in this discussion is
just differential, scalar charge doing work! The electric field which so
many of the physics types use, is just another word for the potential
ability to do work. In this case, in space or air, by coulombic forces
and in closed metallic circuits by electrodynamic means. A differeential
field, a differential charge, stored energy, it is just the same glove
turned inside out depending on our viewpoint. A naked charge has no
energy, a fixed electrical field has no energy. But with metallic
conductors, work can be done with both.
Richard Hull, TCBOR
rom hullr-at-whitlock-dot-com Mon Nov 4 18:32:35 1996
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Date: Mon, 04 Nov 1996 18:46:18 -0800
From: hullr-at-whitlock-dot-com (Richard Hull)
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To: tesla-at-pupman-dot-com
Subject: Re: 50%
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Tesla List wrote:
>
> >From MALCOLM-at-directorate.wnp.ac.nzMon Nov 4 06:45:29 1996
> Date: Mon, 4 Nov 1996 20:09:43 +1200
> From: Malcolm Watts <MALCOLM-at-directorate.wnp.ac.nz>
> To: tesla-at-pupman-dot-com
> Subject: Re: 50%
>
> Wow! I just have to respond to this post on the 2 cap problem.....
>
> > Energy is conserved. Mistakes in capacitor circuit theory and
> > algebra are the problem. That is why you did not remember it from
> > college. The voltage across the two capacitors after the
> > reconnection is not V/2. It is V/sqrt2. This can be easily checked
> > by test and correct algebra.
> >
> > The voltage will be a little less than V/sqrt2 because there is
> > some spark loss in the reconnection. In the test that I made the
> > voltage across the two capacitors was slightly less than V/sqrt2
> > because of losses but was still much more than V/2. Using V/sqrt2
> > in the equation will give a total of 100% energy (50% in each) for
> > the two capacitors if the algebra is done correctly.
>
> So we now have Q = 2C x V/sqrt2 when we started with Q = CV. I think
> your result begs an explanation. If no energy is expended in
> connecting the two caps, how did the extra charge separation come
> about?
> I have just this very minute done this experiment with two high
> quality capacitors and the voltage comes to V/2 near as.
>
> Malcolm
Good boy Malcolm,
Don't let 'em talk yo out of the fact of the matter.
I did the experiment a year ago for the first time in my life with my own
hands or I wouldn't have gone on so long over this topic! v/2 it is and
always will be.
Richard Hull, TCBOR