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# Re: capacitance of a disk

• From: "John S. Denker" <jsd@MONMOUTH.COM>
• Date: Fri, 16 Feb 2001 10:46:10 -0500

At 10:15 AM 2/16/01 -0500, Prof. John P. Ertel wrote:
This thread is so long running that I'm no longer sure that I'm

A number of very good questions have been asked, and a number of very good

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).

In this context, that is an inappropriately narrow definition of
"capacitor". If we restrict ourselves to the subset of the world where all
capacitors have two plates, and if we restrict ourselves to totalling the
charge on the two plates, that's fine. But the question that started this
extremely reasonable and well-posed question.

This violates Kirchoff's "laws", but that's not uncommon in the real world.

In particular, for N objects, including but not limited to N=2, there is an
N-by-N capacitance matrix. You can ask about the charge on object i
induced by the voltage on object j. We know that each row of the matrix
sums to zero, but for N>2 that does not imply any simple relationship
between any two of the objects.

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.

No, it's not nearly so interesting. There's no physics in this factor of
2. It just has to do with a totally arbitrary decision of how to account
for the area of a disk as it passes from nonzero thickness to zero
thickness. A thick disk has two distinct faces, while the zero-thickness
disk arguably has two faces, but arguably they are not distinct because
they are in the same place.

For consistancy, we require the the integral of the charge
density over the surface yield q (either +q or -q depending on which plate).

Of course, but that doesn't resolve the aforementioned arbitrariness as to
what is "the" surface.