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# Re: capacitance of a disk

John's observation points again to a chink in the development of
a coherent description of the charge on one plate of a capacitor.
It pleases me to think of this as "the Physicist's Complaint"

Put it this way: if one insists on treating one plate of a capacitor
in a stand-alone manner, and supposing that the balancing charge
which exists at infinite regress, exists at zero potential then
the opposing charge which one realises is equal in
magnitude and opposite in sign, is supposed to be zero.

Or so it seems to me.
:-)

Brian

At 10:15 2/16/01 -0500, John P Ertel wrote:
/// 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.
////
Bob Sciamanda

----- Original Message -----
From: "David Bowman" <David_Bowman@GEORGETOWNCOLLEGE.EDU>
To: <PHYS-L@lists.nau.edu>
Sent: Sunday, February 11, 2001 01:44 PM
Subject: Re: capacitance of a disk

/// 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))
. . .

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