Re: electrostatic shielding

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Leigh Palmer wrote:

> > 1) Let me add the obvious; the value of G changes
> > gradually from 1/(2*Pi) to 1/(4*Pi*) when the distance
> > is increasing. The first value is for a very large flat
> > metallic disk, where sigma=Q/(2*Pi*R^2). The second
> > is for a point charge (the disk seen from "infinity").
>
> That is certainly not obvious to me. The value of G cannot be
> as large as 1/(2*pi) because the charge on the disk will flow
> preferentially to the edge; the surface charge density at the
> center must be less than Q/(2*pi)R^2

That is correct. I was thinking (not realistically) about a disk of
uniform S(r) distribution; Q/2 on each side.

> > The sharpness of the disk edges will probably play a
> > very significant role in the S(r) distribution.
>
> I think that is not the case if the disc is thin (thickness t
> of disc is much less than the radius R).

Intuitively I would expect the opposite. For a thinner disk
more of the total Q will be near the edge. I am assuming
that the edge is a circle (radius=0.5*thickness). For a very
thin disk the E near the edges will result in a corona flow,
like from a needle. Is this wrong?
Ludwik Kowalski