|Chronology||Current Month||Current Thread||Current Date|
|[Year List] [Month List (current year)]||[Date Index] [Thread Index]||[Thread Prev] [Thread Next]||[Date Prev] [Date Next]|
/// the total charge on anyother).
capacitor in any normal circuit is ZERO (+q on one plate and -q on the
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.
----- Original Message -----
From: "David Bowman" <David_Bowman@GEORGETOWNCOLLEGE.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))
. . .