Re: changing dielectrics in a capacitor (fwd)

[Steven Roys <sroys@xxxxxxxx>]

---------- Forwarded message ----------
Date: Mon, 19 Jan 2004 21:43:27 -0200
From: Antonio Carlos M. de Queiroz <acmq@xxxxxxxxxxxxxxxx>
To: High Voltage list <hvlist@xxxxxxxxxx>
Subject: Re: changing dielectrics in a capacitor (fwd)

High Voltage list wrote:

> From: Alfred Erpel <alfred@xxxxxxxxx>

>     Imagine a parallel plate capacitor where all metal plates of the
> capacitor are fixed, and you can rotate the dielectric material with a k=10
> from full coverage between the metal plates to no coverage where the new
> dielectric is air k=1.
>     Lets say we start with the air as dielectric and the capacitor is rated
> at .001µF and charged to 10,000 volts.
>     When the k=10 dielectric material is rotated totally into place to
> replace the k=1 air, what happens?
> I will speculate:
> Q=C*V wants to stay the same so as the capacitance increases to .01µF the
> voltage drops to 1000 volts.

Ok.

> The capacitor now has 1/10 of its original energy.  If my speculations are
> correct, what happened to the 9/10 of the energy that went away?  Does the
> new dielectric want to get "sucked" in?

Yes. The energy will be spend accelerating the dielectric block, or
dissipated by you trying to impede it to accelerate.

An electrostatic machine of a kind that I have never seen made can be
built, apparently, using this principle. The conductors would all be
fixed, and only a dielectric block would move. It's just necessary to
impede charge deposition over the dielectric, otherwise the phenomenon
of the "dissectable Leyden jar" would impede it to work. (In your
example, when the dielectric is between the plates, the charges in
the plates can spark to the dielectric surface and remain there. If
the dielectric is then removed, the plates would not receive their
charges back and the voltage would not increase again). This can be
avoided by always keeping a guard distance between the plates and the
dielectric.

Antonio Carlos M. de Queiroz