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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).
> 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

Dave, 

These are GREAT questions!

Your "posers" remind me of a crotchety old Power Systems professor who
used to teach Power Engineering at U of Illinois - a recognized expert
in power systems distribution and stability. When the poor Power
Engineering students took his tests, their average grade would be 15%.
Fortunately, he graded on a curve...

Each test consisted of exactly three problems:
  - The first problem, any average genius could solve. 
  - The second problem, only God and Dr. Fett could solve. 
  - God would consult with Dr. Fett to solve the third one!

Anyways... back to your questions, last to first:

The answer to your last question is YES. For a constant charge, Q,
anything you do which changes the capacitance changes the voltage. If
you took the same air-dielectric capacitor and decreased the plate
separation, the voltage would also decline. Suppose you seperated the
plates by a very large distance. In this case, theory predicts that the
voltage across the cap would approach infinity. In practice, leakage
currents and parasitic capacitance to _everything else in the world_
prevents this infinate voltage rise. 

Your second question is a classic! Another way I've seen this one is
equal caps at opposite voltages suddenly shorted. You end up with 0
volts on both. Where'd all the energy go? There's two answers, neither
of which are particularly satisfying.

A. From a purely theoretical standpoint, you've got an irrestistable
force and immovable object type of problem. Caps can't instantaneously
change voltage, yet they MUST if we have perfect caps, no inductance, no
resistance, and a perfect switch. We get an infinate current, flowing
for an infinately short period of time - an impulse function. All bets
are off when we let currents go to infinity. "After the smoke clears"
there can be _whatever_ energy is necessary to match the steady state
result that theory predicts! I know this isn't very satisfying, but its
the best explanation I could come up with...

B. From a "real world" standpoint, we really have series resistance and
inductance, and a switch that can't instantaneously close. In this case,
the difference between the energy we started with, and what we ended up
with is lost in heat and electromagnetic radiation (light, RF, etc...).
That "bang" you hear also disrupts a good chunk of the EM spectrum... 

Now for your first (and toughest) question. Although a conductor CAN
collect charge, it cannot stored energy as an electrostatic field inside
itself - the mobility of the electrons prevents this from happening.
However, good dielectric materials are also insulators: dielectrics
_will_ support internal electric fields, and they can store energy in
this field independent of whether we have metal plates attached or not.
You and Richard agree here - there's never energy stored _inside_ the
metal plates, only in the electrostatic field between the plates.

If you (carefully!) disassemble a charged cap, most of the stored energy
_stays_ inside the dielectric, and shows up as surface charges on the
outside surface of the dielectric. Simply removing the outside plates
doesn't change this significantly. As long as you don't permit any
current to flow from one plate to the other while removing it, the total
charge in the system (Q) is virtually unaffected! The dielectic material
still stores all the original energy in slightly-displaced bound charges
INSIDE the dielectric material. Even though you've removed the outer
plates, the "effective" capacitance of the system didn't change! Put the
plates back on and you can collect these charges, and have your charged
cap back again. Not at all obvious, is it!

HOWEVER, suppose you tried this experiment in the vacuum of space with a
vacuum capacitor. In this case, you could never truly remove the
capacitor plate from the system. As you moved one plate away from the
other, you'd decreases the capacitance of the system and boost the
effective voltage between the plates. Since there's nothing but
free-space to store the field, this situation is analagous to the
air-dielectric cap above. However, if you didn't permit any charge to
leak away from the plates, the experiment would still work! 

If you want a better explanation, please contact Dr. Fett...

I think I'll go rest for a while now - you've triggered 30 year-old
flashbacks!  Safe coilin' to ya' Dave! 

PS.. Let me know how the neons worked out for you!

-- Bert --