Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: capacitance of a disk

Regarding Ludwik's question:

Can the distribution of relative sigma, S, (the surface charge
density at a given r with respect to that in the center) be
extracted from your mathematical formalism, David? I am
referring to the flat disk, whose radius, R, is given. You
posted the solution in terms of the total capacitance last
night. I assume you had to deal with the distribution of
surface densities to calculate C.

Yes. In my most recent post I showed how to calculate the surface
charge density for this limiting flat disk case as well as for the
generic spheroid conductor case. To recap the result for the disk
the surface charge density is:

[sigma](r) = Q/(2*[pi]*R^2*sqrt(1 - (r/R)^2))

where r ranges over the interior of the disk, i.e. 0 <= r <= R . Note
the singularly divergent charge density on the outer edge of the disk.

I suppose the quantity S in your derivation has nothing to
do with the S defined above. Is this correct?

Yes. you are correct. The quantity S in my formula for the potential
below has *nothing* to do with the surface charge density. In fact
S has no physical meaning *at all*. I only introduced it as a way of
breaking up a very complicated formula that would have been spread out
over two lines into two parts which has each part more easily read.

I know that
many people are able to read physics from your equations.
Unfortunately, I am not one of them.

I'm sorry about the cumbersome ASCII notation. That makes the reading
and comprehending job much more difficult than with real mathematical

The best I can do is
to appreciate the ease with which you do mathematics.

What about the understandability of my picturesque descriptions of the
confocal "conicoidal" coordinate system?

I would like to compare your formula for S(r/R) with
what my code is producing numerically for a disk
whose aspect ratio (R/h) is very large.

What you call S(r/R) I called [sigma](r) above (i.e. it's the surface
charge density per actual area, not per reduced area in units of R^2).

David Bowman wrote (in part):

V(r_vec) = Q*S/(4*[pi]*[epsilon]_0*R)

where we have defined the quantity S by

S = arcsin(sqrt(2/((r/R)^2 + 1 + sqrt(((r/R)^2 - 1)^2 + (2*z/R)^2)))) .

Note that these two formulae are really just one formula that has been
broken up into two pieces so each piece would have fit onto one line.

When V(r_vec) is evaluated on the charged disk we have u --> R and
S --> [pi]/2 since the arcsin(1) = [pi]/2. If we let V be the
potential on the charged conducting disk we get:

V = Q/(8*[epsilon]_0*R) = Q/C .

Thus the capacitance is C = 8*[epsilon]_o*R.

David Bowman