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*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.

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