From: Jim Choate <ravage@einstein.ssz.com>
Date: Mon, 23 Nov 1998 03:46:43 +0800
To: cypherpunks@einstein.ssz.com (Cypherpunks Distributed Remailer)
Subject: Re: Oh, a warning about playing with glass & charge.... (fwd)
Message-ID: <199811221916.NAA18764@einstein.ssz.com>
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Forwarded message:

> Date: Sun, 22 Nov 1998 10:05:41 -0800
> From: Tim May <tcmay@got.net>
> Subject: Re: Oh, a warning about playing with glass & charge....

> And when you do your "...and then take away one of the copper plates"
> experiment, you'll find that force is needed. Force times distance is
> energy. The energy to pull one of the plates away equals the 0.5CV^2 energy
> no longer stored in the capacitor.

This is true ONLY if the dielectric doesn't touch either plate and is simply
used to increase the storage capability (via Gauss' Law) of the capacitor
over an air gap type of the same physical structure. The reason you have
this force is the induced charge on the surface of the dielectric and the
electrostatic attraction this causes with the charge on the plate. If the
plate ACTUALY TOUCHES there is NO INDUCED CHARGE via Gauss's Law, it's a
REAL charge stored on the surface of the insulating dielectric and the plate
remains NEUTRALY charged.

Why is there no induced charged? Because as you place charge on the
conducting plate it induces a charge via Gauss' Law on the dielctric. Since
the metal has outer orbital electrons available (which an insulator doesn't)
the charge in the plate gets transported to the surface of the insulator and
bonds there (remember unlike charges attract) with the induced charge.

Why does the charge storage increase because of the dielectric? The
electrical moment of the atoms is greater than that of the air. This is
another reason the charge once deposited on the dielectric surface will
stay there when the plate is removed.

If you go back and look at the examples in your physics books that your
basing your position on you'll find that in EVERY case there is an air gap
between the plates and the dielectric. Remove the air gap and the behaviour
of the charge and hence it's explanation changes accordingly.

> The energy stored is not stored either on the plate or on the insulator. It
> is in the electric field.

Q: Where does the electric field come from Tim?

A: The charges (ie electrons and protons) that are sitting on the surface of
   the dielectric (where the di- comes from). The electric field strength is
   a measure of the wavelength of the photons those two contrary sets of
   particles are bouncing back and forth between themselves.

Q: Explain why the charge sitting on the surface of that dielectric will
   stay on the metal plate when the metal plate is removed. The metal plate
   is a conductor so as it is removed it effectively carries the charge of
   the part of the dielectric that it still retains contact with. Why would
   that e- choose to move TOWARD a ne- charge when it's sitting there with
   an opposite charge already in a minimum energy configuration?

A: It won't. To do so violates the minimal energy requirements of the
   system as well as the standard behaviour of charges (ie like repel,
   opposite attract).

Another experiment....

Try removing the metal foil when there is a charge and when there isn't.

You won't find an appreciable difference.

As to the energy required to remove the foil, yes force times distance is
energy and it's the energy requisite to overcome the inertia of the metal
foil. There is also a small additional force caused by Gauss' Law because the
the distance between the e- (for example) stored on the dielectric and the
metal plate will form a 'virtual' charge on the surface of the plate. However,
since the plate itself is neutral (by definition) there will be a contrary
charge on the opposite side. Hence, you won't loose a single coulomb of charge
to the metal plate. You can check this by using either an electronic charge
meter (1-800-OMEGA if you're interested) or you  can use a gold leaf
electrometer. If the charge varies as Tim claims then the angle between the
leaves should change (ie get smaller). It won't.

As to the force, the maximal force will occur when the plate is removed as a
single piece perpendicular to the surface of the dielectric. If you roll or
peel it off it will for all intents and purposes come free effortlessy
because you won't have to fight the electrostatic force.

                --------------------------------------  < Cu +
              ------------------------------------------
                --------------------------------------  < Cu -
              ------------------------------------------
                --------------------------------------  < Cu +
              ------------------------------------------
                --------------------------------------  < Cu -

NOTE: there is NO air gap between the glass and the Cu sheeting. I use
      normal old cinter blocks to press them together as tightly as I
      can.

Another experinment:

Build the stack as shown above and remove each later one at at time with
and without charge. Note the behaviour of the metal plate. It will tend to
go with the glass plate that is being removed. Because of induced charges
via Gauss' Law the foil will tend to go with the glass plate as it's
removed.

            ---------------------------------------------
             -------------------------------------------  Cu +
            ---------------------------------------------

As the top plate is removed the air gap will force a virtual charge to build
up on the surface of the foil. As a consequence an opposite and equal charge
will build up on the other side of the foil and it will be + and in
opposition to the + charge already sitting on the bottem glass plate. The
induced charge on the top of the Cu will be - and will be attracted to the +
charge on the bottem of the top plate.

This of course assumes you're using very thin foil. If it's very thick at
all you won't see this because of gravity. I tend to use very heavy foil so
that I don't get shocks from these various induced fields when I'm handling
the plates. Be sure to use heavy rubber gloves. Rubbermaid kitchen gloves
intended for industrial washing are sufficient unless you're going to use a
high voltage and charge.

Anybody who wants to understand the theory (despite Tim's claims to the
contrary) is welcome to check out:

Physics
Jay Orear
ISBN 0-02-389460-1
Chpt. 16, Electrostatics
pp. 308 - 338
in particular,
pp. 327, 16-6 Dielectrics


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