In your spherical capacitor example, you have to distinguish between the charge on the inner surface of the outer sphere, and the charge on the outer surface. The charge on the inner surface of the outer sphere will adjust itself to be equal to -Qa, since there can be no electrostatic field within the metal of the outer sphere. The capacitance between the inner surface of the outer sphere and the inner sphere, which is a purely geometric parameter, won't change.
But, since in this case the outer sphere is "floating" and has an excess charge on its outer surface, there also will be a capacitance between the outer surface of the outer sphere and infinity.
Of course in "real life" we usually don't put arbitrary amounts of static charge on capacitor plates, but instead use a battery or other power supply to maintain a known potential difference between the plates.
-----Original Message-----
From: phys-l-bounces@carnot.physics.buffalo.edu [mailto:phys-l-bounces@carnot.physics.buffalo.edu] On Behalf Of Carl Mungan
Sent: Wednesday, February 06, 2008 1:44 PM
To: phys-l@carnot.physics.buffalo.edu
Subject: [Phys-l] defn of capacitance
Some of us (locally) have been discussing the definition of
capacitance (restricted for simplicity to the case of two isolated
conductors only) and we seem to have run across a problem with how to
define it if the charges on the two conductors are NOT (necessarily)
equal and opposite. For simplicity let's say that we only have vacuum
outside of both conductors.
My initial stab at a definition was the following: Capacitance is
uniquely defined if I specify the geometry of the two conductors
(shapes, sizes, separations, etc) as delta(V)/Q where delta(V) is the
absolute value of the line integral of E between the two conductors
and Q is the absolute value of the average difference in charges of
the two plates (ie. half of the difference in charges on the two
plates).
Here's an example for the case of two parallel plates, charge Qa on
the left plate and Qb on the right plate each of area A. Say Qa>Qb
for specificity.
between the plates E = Qa/2*A*e - Qb/2*A*e where e=1/4*pi*k
so delta(V) = (Qa-Qb)*d/2*A*e where d=distance between plates
but Q=(Qa-Qb)/2
conclude C=Q/delta(V)=A*e/d as usual
Note the answer doesn't matter what charges we put on the two plates.
But now here's a case where the answer DOES depend on them, which
seems to contradict the usual idea of capacitance being independent
of charge:
Choose two concentric spheres, an inner one of radius a and an outer
one of radius b. Place charge Qa on the inner one and Qb on the outer
one. Again say Qa>Qb for specificity.
between the spheres E = k*Qa/r^2 independent of Qb
so delta(V) = k*Qa*(b-a)/a*b
but Q=(Qa-Qb)/2
conclude C depends on the charges!
Please fix my definition of C for the case of two isolated conductors
in vacuum carrying charges which are not necessarily equal and
opposite. As a test, demonstrate that your fixed definition works for
both of the examples listed above. -Carl
--
Carl E Mungan, Assoc Prof of Physics 410-293-6680 (O) -3729 (F)
Naval Academy Stop 9c, 572C Holloway Rd, Annapolis MD 21402-5002 mailto:mungan@usna.eduhttp://usna.edu/Users/physics/mungan/
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