Re: capacitance of a disk

[brian whatcott <inet@INTELLISYS.NET>]

At 16:41 2/8/01 -0500, Carl Mungan wrote:
> ...start with two concentric spheres, work out the capacitance,
> then let the radius of the outer sphere go to infinity to get the
> capacitance of an isolated sphere.
>
> If you try the same thing for a parallel-plate capacitor C = A*e0/d
> and let d go to infinity, you get zero which is obviously wrong. I
> assume this is because we neglected the fringing field, which is only
> valid if d << sqrt(A).
>
> So what is the capacitance of an isolated disk? ...

> Carl E. Mungan

Assume that the expression I mentioned earlier for the capacitance
of concentric spheres has been experimentally verified to
the accuracy given,
i.e Capacitance = R1R2/(R1+R2) 111.27 pF/m
for
R1 radius of inner sphere meters
R2 radius of outer sphere  meters

As R2 grows very large, its capacitance tends towards
    R1 111.27 pF/m

As R2 grows very large, the influence of asperities
on the inner shape declines,
so that its projected area dominates the expression.

In terms of its area, the capacitance of a sphere is
4 pi R1^2 m^2 := R1 111.27 pF/m

The area of a disk of radius R is given by
2 pi R^2 + 2 pi R t = 2 pi R (R + t)
(t thickness in meters)

Where R2 is very large, its capacitance is in the ratio

(2 pi R)(R + t)/(4 pi R1^2)
to that of the sphere.

So the capacitance of the isolated disk is

(R + t)/2R1  R1 111.27 pF/m = (R + t) 55.64 pF/m


(This is a plausibility argument, one step better
I'm sure you will agree, than demonstration by
loud assertion, and two steps better than proof
by arm-waving...)


brian whatcott <inet@intellisys.net>  Altus OK
                Eureka!