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Re: capacitance of a disk

This thread is so long running that I'm no longer sure that I'm
answerign the right question.

But, if I do understand, then of course the total charge on any
capacitor in any normal circuit is ZERO (+q on one plate and -q on the other).

It seems to me that the "twice q" as compared to Smythe is due simply to
the question of how much charge it takes to produce a certain charge
density. For consistancy, we require the the integral of the charge
density over the surface yield q (either +q or -q depending on which plate).


Bob Sciamanda wrote:

This surface charge density result is just twice Smythe's result: pg 114;
eq (3).
Perhaps there is a discrepancy in what "Q" stands for; in Smythe it is the
total charge, added up over all disck surfaces.

David, I have been able to spend only sporadic moments on these things -
but I will "shortly" e-mail you some scanned pages of Smythe in TIFF


Bob Sciamanda
Physics, Edinboro Univ of PA (em)
----- Original Message -----
From: "David Bowman" <David_Bowman@GEORGETOWNCOLLEGE.EDU>
To: <>
Sent: Sunday, February 11, 2001 01:44 PM
Subject: Re: capacitance of a disk

Regarding Ludwik's question:
. . .
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))
. . .

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