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Re: capacitance of a disk/Roundup



At 16:41 2/8/01 -0500, Carl Mungan wrote:
The discussion of capacitance and of the charge distribution on a
disk seems to have ripened to the point that a question I brought up
in class the other day can perhaps be answered. (I certainly don't
know the answer.)

You can 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? Feel free to make
whatever simplifications or approximations you want and otherwise
change the problem to make it more tractable or meaningful. Carl
--
Carl E. Mungan, Asst. Prof. of Physics 410-293-6680 (O) -3729 (F)
U.S. Naval Academy, Stop 9C, Annapolis, MD 21402-5026
mungan@usna.edu http://physics.usna.edu/physics/faculty/mungan/


Well, Carl, I hope you took as much pleasure in the various responses
which actually spoke to your question as I did.
I tabulated the capacitance values offered, for a constant diameter
disk of 1 meter diameter to ease comparison.
They all agree that the disk's capacitance is a linear function of
diameter not area (though at least one respondent wavered a little
with areal comparisons of capacitance for different shapes...)

Here they are:
John Mallinckrodt hereafter called JM at 8 Feb 15:08 + 8 hrs GMT
32 pF +- 1 pF (footnote 1)

John Denker 8 Feb 19:28 + 5 hrs GMT
28 pF +28 -17 pF (footnote 2)

Brian Whatcott * Feb 20:28 + 6hrs GMT
28 pF
(Not quite sure why John put up such a critique of my estimate:
it seems to have been identical to his best guess)

JM 8 Feb 23:24 + 8hrs GMT
39 pF +- 1 pF (footnote 3)

JM 9 Feb 17:52 +8 hrs GMT After consulting an engineering reference
35 pF

It looks like our communal best guess at the 1 meter diam disk is
35 pF then....
32 + 28 + 28 + 39 + 35 / 5 = 32 pF


Footnotes:
1) JM: C = 2 pi Epsilon_0 R X 1.15 +- 5%

Epsilon-0 is 8.854E-12 F/m free space
(Science Data Book, Tennent, Open Univ)

2) JD: C = 4 pi Epsilon_0 R X 0.5 + 100% - 62%

3) JM: C = 2 pi Epsilon_0 R X 1.4 +- 3%

(Sure hope I didn't misrepresent any of the worthy responders.
I'm sure I'll hear about it if I did! :-)


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