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Re: a funny capacitor.



Following JohnD's prescription (see details below) I now
can calculate the Cij coefficients for the "funny capacitor".
The system consists of four objects because the infinity is
now considered as object #4. The other three objects are
the plates (left, right and bottom).

* *
* *
1 * * 2
* *
* *

*********
3

The following matrix shows the Cij values for my funny capacitor.

I +2.86, -0.460, -1.15, -1.25 I
I -0.460, +2.86, -1.15, -1.25 I
I -1.15, -1.15, +3.26, -0.965 I
I -1.25, -1.25, -0.965, +3.46 I

The left plate is object #1, the right plate is object #2, the bottom
plate is object #3, and the infinity is object #4. Recall that Cij are
defined by the following equations:

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

What does it mean that C13=-1.15 F? I suppose it means that
that positive net charge on object 1 (left plate) decreases when
the positive potential on object 3 (bottom plate) decreases. Is
this a correct intepretation? For my "funny capacitor"

Q1=+2.86*V1 - 0.46*V2 - 1.15*V3 - 1.25*V4
Q2=-0.46*V1 + 2.86*V2 - 1.15*V3 - 1.25*V4 equation (2)
Q3=-1.15*V1 - 1.15*V2 + 3.26*V3 - 0.965*V4
Q4=-1.25*V1 - 1.25*V2 - 0.965*V3 + 3.46*V4

By choosing V1=-50, V2=+50, V3=20 and V4=0 the net charges
received by the four objects were calculated from the above as
-189, +143, +65.2 and -19.3 units, respectively. Therefore,
I expect to recover the imposed potentials by solving the
following matrix equation:

+2.860*V1 - 0.465*V2 - 1.147*V3 - 1.246*V4 = -189
-0.465*V1 + 2.860*V2 - 1.147*V3 - 1.246*V4 = +143
-1.147*V1 - 1.147*V2 + 3.263*V3 - 0.965*V4 = +65.2
-1.246*V1 - 1.246*V2 - 0.965*V3 + 3.460*V4 = -19.3

No singularity was encountered and the values of potentials
were recovered, as expected. I am emphasizing the success of
the inversion because the determinant of the Cij matrix which
John posted here three days ago was zero and the inversion
was no possible. Once again I would like to thank John for
the help he provided.

Note that in each row (of equation 2) the term on the diagonal
is larger than other terms. This is not surprising; a change in
Q due to a change of V on the object itself is larger than the
identical change of V on another component of my funny
capacitor.

HOW WERE THE VALUES OF Cij DETERMINED?

Recall that the capacitance coefficients Cij are defined by

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

Following John's prescription I "solved the Laplace equation"
by assuming that V1=-50 volts, while other three potentials
are kept at zero. The C11, C21, C31 and C41 were calculated
by dividing these charges by V1. Be careful, it must be V1
and not the absolute value of V1. {I made this mistake at one
point but John corrected me in a private message].

The values of C21, C22, C23 and C24 were obtained in the
same way but imposing V2=+50 volts and keeping the other
four potentials at zero. The coefficients in the third and fourth
columns were found in the same way. After that the "Laplace
equation was solved" for the fourth time but this time the
boundary condition was V1=-50, V2=+50, V3=+20 and
V4=0. Not surprisingly the values of Q were found to
consistent with the superposition principle.

Note that by "solving Laplace equation" I am referring to the
brute force number crunching; I would not dare to touch the
problem without going back to school to learn calculus.
Ludwik Kowalski