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Re: Electrostatics Problem



Bob's message (see below) triggered a though. It has to do with the
Faraday cup. We all know that when an electrified object, say a tiny
metallic sphere with ten zillion of unbalanced electrons on its surface,

is inserted into the cup the field inside is zero. To simplify, assume
that the cup is spherical and that the inserted object is exactly in the
center.

We bring a probe charge, say a single electron, into the cavity, and the
net
force acting on it is zero everywhere, even very close to the center. I
am
ignoring the tiny attractive force due to induction. The radius of the
cup
shell, R, has no effect. Now start making this radius bigger and bigger.

We are gedankening, the cup is located in empty space many light years
from the nearest galaxy. At what R does the presence of the outer shell
become irrelevant? The intuition tells me that for very large R the
local
force should be consistent with Coulomb's law. But this is not
consistent
with the Faraday cup model. What is wrong with my intuition?

And for very large R the shape of the cup can be changed without any
effect on the situation near the center. The large sphere can become a
box,
even a relatively flat one. The side walls can be removed and replaced
by
only a wire which keeps the two "plates" at the same potential.

Ludwik Kowalski
................................................................................

Bob Sciamanda wrote:

Sorry, I am repeating my last post with some typos corrected:
( I also add the solutions to the 4 simultaneous equations):

Assumptions, previously proved theorems, and definitions:

The two parallel conducting plates are infinite in length and width
(neglect end effects). Once electrostatic equilibrium is achieved, the

E field inside each plate =0

Once electrostatic equilibrium is achieved, non zero net charge
resides only as sheets of surface charge on the 4 plate surfaces.
A sheet of surface charge of density s(Coulombs/m^2) produces
everywhere an E field of s/(2*epsilon) - directed away from the
sheet if s is positive and toward the sheet if s is negative.
S1 = the net charge density on the left plate (the net plate charge
divided by the area of one side);
S2 = the net charge density on the right plate (the net plate charge
divided by the area of one side);
s1 = the net charge density on the left surface of the left plate;
s2 = the net charge density on the right surface of the left plate;
s3 = the net charge density on the left surface of the right plate;
s4 = the net charge density on the right surface of the right plate.

The problem:
Given S1 and S2, calculate s1, s2, s3 and s4

We need 4 equations to determine the 4 unknowns:
By definition:
(1): s1 + s2 = S1
(2): s3 + s4 = S2

The E field inside the left plate (due to the 4 sheets of charge) =0:
(3): (s1 - s2 -s3 -s4 )/(2*epsilon) = 0

The E field inside the right plate (due to the 4 sheets of charge) =
0:
(4): (s1 +s2 +s3 - s4 )/(2*epsilon) =0

The above 4 numbered equations determine s1, s2, s3 and s4,
given S1 and S2. The solutions are:


s1 = s4 = (S1 + S2)/2

s2 = - s3 = (S1 - S2 )/2