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Re: Capacitor charge, were is it?



On 10/25/96 22:25:22 you 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?
>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).

Since a vacuum is a conductor (i.e vacuum tube), you cannot have a potential 
difference (charge) in a pure vacuum.  This concept is theoretical.

>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?

Wasn't work done transferring the energy?

>
>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.

Indeed, this should be the case.  I designed a device to measure pressure in 
water (k=80) using a parallel plate capacitor.  When the pressure of the 
water increased (actually a high pressure ultrasonic shockwave) between the 
plates, there was a charge transfer on the capacitor.  The voltage on the 
capacitor was held constant using an op amp configured as a current to 
voltage converter.  A measurable voltage change was produced at the op 
amp output when the shockwave was focussed between the plates of the 
capacitor.  This device was termed a capacitance hydrophone.

In this example, V was held constant by the op amp.  Since Q=CV, when the 
dielectric constant of the water changed under pressure, C changed producing 
a current flow, I=dQ/dT.  The op amp configured as a I/V converter produced 
a voltage change.

>Analogies welcome
>Dave Huffman
>
>