Re: capacitance except on a set of measure zero

["John S. Denker" <jsd@MONMOUTH.COM>]

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.