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Re: funny capacitor



Last week the following problem was formulated.
Given several conducting objects which are not
connected to anything (electrically floating) find
their potentials (V1, V2, V3, etc.) when the net
charges on the objects are given (Q1, Q2, Q3, etc.).

I believe that mother nature has one, and only one,
reproducible solution. In trying to find that solution,
for a particular example (funny capacitor), two steps
were clearly identified. The first step consists of
calculating the influence coefficients, Cij, defined by:

Q1=C11*V1 + C12*V2 + C13*V3 + C14*V4
Q2=C21*V1 + C22*V2 + C23*V3 + C24*V4 Eqns (2)
Q3=C31*V1 + C32*V2 + C33*V3 + C34*V4
Q4=C41*V1 + C42*V2 + C43*V3 + C44*V4

In this example objects 1, 2 and 3 were metallic plates
while object 4 was the enclosure (infinity). For the chosen
geometrical arrangement the Cij coefficients, calculated
according to JohnD's prescription, turned out to be:

+2.86, -0.46, -1.15, -1.25
-0.46, +2.86, -1.15, -1.25
-1.15, -1.15, +3.26, -0.96
-1.25, -1.25, -0.96, +3.46

Knowing these coefficients we can calculate charges for
any given set of potentials. But that is not our problem.
We want to calculate potentials when charges are known.
This is step 2. Many textbooks tell us that this simply a
matter of solving n equations with n unknowns, for example,

+2.86*V1 - 0.46*V2 - 1.15*V3 - 1.25*V4 = -189
-0.46*V1 + 2.86*V2 - 1.15*V3 - 1.25*V4 = +143
-1.15*V1 - 1.15*V2 + 3.26*V3 - 0.96*V4 = +65.2
-1.25*V1 - 1.25*V2 - 0.96*V3 + 3.46*V4 = -19.3

where numbers on the right side are charges Q1,
Q2, Q3 and Q4 (in arbitrary units) calculated at
the same time as Cij coefficients. These are charges
to produce: V1=-50, V2=+50, V3=+20 and V4=0.
John convinced us that the above equation can not
be solved because the determinant of the matrix of
coefficients is always zero (due to gauge invariance
and to the law of charge conservation).

Trying to solve the above equation is like trying
to find out the intersection point of two parallel
lines, wrote John. The two parallel lines do not
intersect. Does it it mean our problem does not have
a single unique solution? That is where I am lost
again. John eliminated one row and one column
from the Cij matrix. This produced a matrix

2.86 -0.46 -1.25
-1.15 -1.15 -0.96
-1.25 -1.25 3.46

whose inverse is:

0.30 -0.18 0.06
-0.30 -0.48 -0.24
0.00 -0.24 0.22

My question is this. How can we use the elements of
this 3 by 3 matrix to solve the problem? Presumably,
the one, and only one, solution is:

V1=B11*Q1 + B12*Q2 + B13*Q3 + B14*Q4
V2=B21*Q1 + B22*Q2 + B23*Q3 + B24*Q4 Eqns (2)
V3=B31*Q1 + B32*Q2 + B33*Q3 + B34*Q4
V4=B41*Q1 + B42*Q2 + B43*Q3 + B44*Q4

We need sixteen coefficient Bij to calculate V's
from Q's. What are the values of these coefficients
for our specific example?
Ludwik Kowalski