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Re: Capacitor charge, were is it?
Tesla List wrote:
>
> >From huffman-at-fnal.govFri Oct 25 21:56:12 1996
> Date: Fri, 25 Oct 1996 10:33:24 -0500
> From: huffman <huffman-at-fnal.gov>
> To: List Tesla <tesla-at-pupman-dot-com>
> Subject: Capacitor charge, were is it?
>
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> Group,
> I'm having trouble with the idea of charge being stored in the dielectric.
> This may not be totally Tesla related but I would like some comments,
> stones, etc.
>
> R. Hull post - All of the charge is held in the dielectric of the secondary
> and not the
> metallic components. This is the case in all capacitors. The plates can
> never store charge!.. Only conduct it to a point where work can be done
> electrodynamically.
>
> If this is true we could not have a capacitor with a charge that has no
> dielectric (vacuum).
Richard Hull comments:
AHA! But the vacuum is a dielectric!! yessiree... Space (absolute
vacuum is endowed with both a dielectric constant (permittivity) and an
inductive nature (permeability) Eo and Uo.
The multiplacation of the permittivity and permiability of free space
yields the reciporical of the square of the speed of light. This is one
of those classic mathematical identities which is true, but has no real
physical or causal meaning. Just a fun identity.
For a fantastic read on just this subject, check out the last issue of
Electric Spacecraft Journal. (issue #18?) Leslie Kulba did the piece and
very insightful, it was. She has taught mathematics and really knows her
stuff. We had a pleasant visit and discussion session over last
Christmas holidays. But I digress.
Vacuum capacitors are real and are used extensively in radio work. Space
is a dielectric and can store charge. Metal plates and often a closed
metallic circuit are needed to commute the charge to do real work. These
are subtilties which are at the very core of our understanding of what
really goes on in electrical exchanges. They are often quickly glossed
over in freshamn physics in engineering along with the prayer from the
teacher that smart kids such as yourself won't dig too deep or really
think about the deeper issues so that they can get on with stuffing your
head with more glossed over issues to get you out the door and working
in the drone like work-a-day world.
However did we come think that the charge was held on the plates when the
most no brainer of simple experiments would show it to be otherwise!!!!
Richard Hull
> We can, however, and the charge must be held on the plates. The energy is
> stored in the electric field which can only be there if the plates have a
> different charge from each other.
They only have a differing charge due to the work that was performed on
the dielectric! No real work was ever performed on the plates. The
dielectric holds 100% of the charge. R. Hull
I recall a post of an experiment in which
> a capacitor is charged, then carefully dismantled. The two metal plates
> were handled, shorted together, then reassembled getting the charged
> capacitor which can be shorted out yielding a large spark.
> There must be something going on here that is not obvious.
Yeah, real emperical experiment from which truths can be unearthed at a
core level not often considered in a world where we are told that the
"plates" are the thing. Actually, a good physics school will make all
this manifest to advancing students. They often, upon matriculating,
revert back to the plate concept of charge storage, however. Often, only
upon being pressed, will they reflect back to the more theoretical
aspects of charge storage. Richard Hull
>
> There is a classic problem in which two identical capacitors are connected
> with a switch. Before the switch is thrown one capacitor has a certain
> voltage, the other none. If we give numbers, let's say C=1uF and V=1000V on
> the first capacitor. The energy is (cv^2)/2 = 0.5 joules. Now after the
> switch is closed both capacitors will have v/2 = 500V across them. This can
> actually be done. But now the energy in the system is half the original
> value.
> (1uf*500^2)/2 = 0.125 joules/cap times two caps = 0.250 joules. Were is the
> other half?
> The classic things I've been taught have me confused with the actual
> workings of nature. If two charged plates (air dielectric) have a certain
> charge Q and then a dielectric is inserted between them with a K>1, the
> voltage should decrease since the charge hasn't changed but the capacitance
> has increased.
You can't have two plates charged, as you say.... You have charged the
dielectric between them. Remove that and you remove all the charge, so
you can't change the dielectric in a charged capacitor and have the new
dielectric contain any charge. (the old or original plates where never
charged, and the new dielectric has had no work performed within it.)
You will definitely change the capacitance, oh yes, there will be a
change in capacitance but that has nothing to do with a charge or the
original charge, which is now gone. It is just the new capacity to store
charge which would have to be forthcoming. This charge must be
delivered to the dielectric via a closed metallic circuit. By the way,
if you think about it, metal plates can never escape a dielectric, as
space is omnipresent.
All the foregoing has assumed you didn't leave the power supply
connected and active, but removed it immediately after charging the
original capacitor.
The post about the dissectable capacitor was mine. I have it all on a
recent video report in color for all to see. Anyone can do the
experiment with a large sheet of mylar and a couple of aluminum plates
and about a 8KV DC supply to see for themselves. Beware of the minor
problem of killing yourself, instantly. Insulate well and think before
you touch! R. Hull
> Analogies welcome
> Dave Huffman