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Re: funny capacitor



John said that the ambiguous situation (more than one
solutions) was created by asking a "wrong question". He
also showed a trick which resolves the ambiguity. Remove
any one row and any one column from the singular matrix
and the unique answer, presumably the same no matter
which row is eliminated, will result from the inversion.

I think I know where the error was; it was in the way in
which Cij were determined. But first let me explain why
I think that the way of "asking the question" was correct.
In using the term "potential" (for V1, V2, V3 and V4) I
was referring to a difference of potentials with respect to
the object # 4, the enclosure where V4 was declared to
be zero. The purpose of adding the enclosure (walls
called "infinity") was to make this crystal-clear.

We all know that the term potential, like the term elevation,
has no unique meaning unless a reference level is defined.
This is embodied in the unit of potential (volt=joule/coulomb)
where joule is for the work needed to bring a small probe
charge (q coulombs) from a reference level to a given
location. In trying to solve the V(Q) problem, after the
Cij coefficients have been identified, we faced this set of
equations:

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

where the matrix of coefficients, Cij, is singular. Clearly
something was wrong in the previous step; that is when
the Laplace method was used to solve the Q(V) problem
four times. So let us go back and reexamine the way in
which the Cij coefficients were determined.

To find Ci,j for the first column, we declared that V1=1 and
V2=V3=V4=0. Dr. Laplas (number crunching) provided us
with the values of Q1, Q2, Q3 and Q4. We identify these
values with Cij from the first column (because V1=1). Then
we do the same for the 2nd column (declaring V2=1 and
V1=V3=V4=0), for the 3rd column, and, finally, for the
4th column. I suspect that the 4th column put us in trouble.
We have no right to declare that V4=1 because the enclosure
is our "sea level reference"; it must remain zero no matter
what else we are doing. Right?

We need a method of calculating Cij which does not introduce
the ambiguity. Do you agree? By using this method we would
not be forced to use tricks to solve V(Q) problems. What is the
correct way of calculating nine (not sixteen) Cij coefficients
for the three equations with three unknowns? I suppose the
3 by 3 matrix of such coefficients would not be singular.

Q1=C11*V1 + C12*V2 + C13*V3
Q2=C21*V1 + C22*V2 + C23*V3
Q3=C31*V1 + C32*V2 + C33*V3

Ludwik Kowalski wrote:

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). .....