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# Re: electrostatic shielding

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").

2) The formula for S(r) is intuitively acceptable but
how do you justify it formally, Leigh? There may be
many even functions with a minimum at r=0. The
sharpness of the disk edges will probably play a
very significant role in the S(r) distribution.
Ludwik Kowalski

Leigh Palmer wrote:

Rephrasing Fred Lemmerhirt's question:

Suppose an isolated conducting disc of radius R carries an electric
charge Q. Will the charge be uniformly distributed on both
surfaces, producing a uniform field pointing straight toward the
sheet on both sides? If a positive point charge is brought near one
side of the sheet, will the charge only on that side of the sheet
be rearranged and the field on that side "distorted", while the
charge and field on the opposite side remain uniform?

The electric field due to an isolated conducting disc will be
approximately uniform in the region near its center. The scale size
in this statement is the radius of the disc, and the term "near"
means at distances r from the center for which r<<R, for example,
inside the concentric spherical volume of radius R/10. At distances
r>>R the field will be approximately that of a point charge having
the same value as the total charge on the disc, say outside the
concentric spherical volume of space of radius 10 R. In between the
these regions the field is more complicated, but it retains the
azimuthal symmetry of the problem. I don't know what the magnitude
of the electric field near the center will be, but it will be of the
form

Q
E = G ---------
2
eps0 R

where G is a constant. (I can't calculate its value, but I would be
very surprised if the number were not transcendental.) The charge
density, S(r), on the surface near the center will be approximately
uniform. It has a minimum value S(0) at the center, and it varies in
distance from the center from that value in a second order manner or
greater, i.e.:

/ 2 4 \
| / r \ / r \ |
S(r) = S(0) | 1 + a | --- | + b | --- | + ... |
| \ R / \ R / |
\ /

where a, b, ... are constants of order unity or less. The charge
density will be the same on both sides, from symmetry
considerations.

The second part of Fred's question is best answered by considering
the influence of a point charge on an isolated neutral conducting
disc, calculating the field, and then producing a solution by the
principle of superposition. A positive point charge q is placed at a
point on axis a distance d/2 from the center of the disc. The
resulting field is approximately that of a dipole of magnitude qd
within the hemisphere for which 0 < r < R/10. The solution of the
problem as stated is the vector sum of these two results. To answer
the remainder of Fred's question we must again consider the
appropriate regime. I will approach this for the case of a small
point charge (q << Q). In this case there will be a small effect
near the center on the other side of the disc. A positive surface
charge will be induced to compensate for the negative charge induced
on the top, and there is a nearly uniform field *away from* the
surface due to the positive charge density.

Superposition of the solutions will show that there is a relatively
large increase in the surface charge density at the center under the
positive charge, and hence an increase in the magnitude of the
electric field at the center of the disc. On the opposite side the
electric field intensity is diminished somewhat by the contribution
from the second part of the superposition.

I hope that qualitative discussion helps. I hope I got it more or
less right, and clear enough to do some good.

Leigh