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Re: capacitance of a disk



According to the formula derived by David's a circular
plate whose diameter is 12" (R=0.1524 m) has C=10.79 pF.
The value of C=10.7 pF, for the plate of the same size
was posted by JohnM. On that basis I decided to use this
diameter as a benchmark for my code.

Calculating E as dV/dl with dl=1 cell_size makes my
C too small (factor of more than 2). That is why I had
"a good idea" of reducing dl by the factor of two. This
brought me to a better agreement with the expected
value (8.4 pF instead of 10.8 pF). But I agree with
JohnD that this is not correct; dl=1 cell size is the
way to go.

So why do I have a discrepancy? I do not think it is
because my object is a disk (aspect ratio 2*R/h=25)
rather than a flat plate. I also think that my cell sizes
are sufficiently small (see below) for at least 10%
agreement. Was somebody else more successful
in trying to match a theoretical expectation with the
numerical approach?

By the way, the shape of my radial distribution of
sigma over the flat surface is nearly the same as that
plotted according to David's formula. It is nearly
flat up to r=R/3 and rises quickly above r>2*r/3.

If somebody else wants to compare numerical and
analytical approaches may I suggest that you use
the same geometry as I do. This would allow us
to compare numbers and possibly discover bugs.
Here is the description of my geometry.

1) Height of the universe =200 cells. Let me clarify this.
My universe is cylindrical. The outer cells, the boundary
of the universe (where V=0=const), does NOT belong
to the universe. It surrounds it.

2) Radius of the universe = 200 cells. (again without
counting the cell of r=201 which belongs to the boundary).
Note that the 2D iteration (in the r,z grid) involves only
the right side of my disk. To account for the presence
of the left side of the disk one column of cells had to be
added on the left side of my grid. The potentials in this
column were forced to be (during consecutive iterations)
mirror images of potentials in the r=1 column. Without
this "symmetry trick" cells at r=1 would have no neighbors
and I would not be able to relax them correctly.

3) height of the disk = 2 cells (two layers, one above and
one below the equator of the universe.)

4) Radius of the disk = 25 cells. The disk radius of 0.1524 m
makes the value of cell size equal to 0.1524/25=0.0061 m
or 6.1 mm.

Note that this choice makes the height of my universe
equal to 122 cm and its radius 122 cm. The disk is
in the middle; its diameter is 30.48 cm and its height
is 1.2 cm. I wish I could make the universe ten times
larger, but this is not practical. I am using cylidrical
coordinates; the cell averaging is according to the
method described by JohnM.

My I suggest that those who are interested contact me
in private; others may not be interested in too many
details. (We will let them know what we did in one
message at the end.) That does not mean that
discussion of general principles should be cut.

Ludwik Kowalski kowalskiL@mail.montclair.edu