Re: capacitance of a disk

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

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.

I suppose the quantity S in your derivation has nothing to
do with the S defined above. Is this correct? I know that
many people are able to read physics from your equations.
Unfortunately, I am not one of them. The best I can do is
to appreciate the ease with which you do mathematics.

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.
Ludwik Kowalski


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)))) .
>
> 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.