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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).
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. 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. 8?/

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.
Analogies welcome
Dave Huffman