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# Re: capacitance except on a set of measure zero

• From: "John S. Denker" <jsd@MONMOUTH.COM>
• Date: Tue, 6 Feb 2001 20:17:03 -0500

At 04:33 PM 2/6/01 -0500, Ludwik Kowalski wrote:
My brain is not comfortable with the term "mutual capacitance"
in the context of two metallic objects carrying charges of identical
sign.

Well, let's see. Let's start by taking C=Q/V as our provisional definition
of capacitance. That definition works fine for any value of the voltage
except for V=0 in which case Q=0 and C=0/0=indeterminate. Our provisional
definition is good everywhere except on a set of measure zero. So what
next? We have no reason to believe the capacitance has a singularity
(indeed in the systems under discussion it is CONSTANT in the neighborhood
of V=0) so why not take a leap of faith and assign that one last point the
obvious value?

In particular, if we take C() and Q() to be functions of voltage, then
C(V) = lim Q(v)/v as v -> V for any V including V=0 (eq. 1)

Things look even simpler if we consider the "small-signal" capacitance
C = dQ/dV (eq. 2)
which is well defined for all V ... and for linear systems is identical to
the "chord" capacitance defined by eq. 1.