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*From*: Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>*Date*: Sat, 3 Feb 2001 11:27:14 -0500

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

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