Re: Electric Field Question

[Bernard Cleyet <anngeorg@PACBELL.NET>]

since the process (of discharge) will involve current it may!! be detectable.
unfortunately any radiation will be shielded if the freq. is high (skin effect)  if
the discharge is done with a resistor a mag. field detector might detect it.

a capacitor is any conductor or combination of -- some  one else (JD) has discussed
(only skimmed, so could be wrong) a matrix (simultaneous eqs.).   Harnwell begins
with V's = pQ's      V1 = p11Q1 +p12Q2 .....; V2=p21Q1 + p22Q2 .....etc.

the p's are coefficients of potential; to get the coefficients of CAPACITANCE, one
manipulates the matrix using the determinant of the coefficients and cofactors:

qsub i =  SUM (1 to n) c sub ij * Vsub j ;   c sub jj are the coefficients of
capacity.   He works the example of three spheres one outside and one inside the
third.  He writes:  the combination such as the one inside the third is said to
form a condenser.  The capacitance is, in this case, the coefficient of cap. of the
inner conductor.  If one conductor is merely in the presence of others it's
coefficient of cap. is frequently referred to as a stray capacity.  Two pages later
he finds the cap. of an isolated sphere -- my justification for any conductor is a
cap (or has stray capacity.  (the other conductor is at infinity (the room and it's
objects -:)     ).

Sum:  I haven't found that caps. require equal and opposite charges,

Kowalski wrote:

> P.S.
> It seems that two concentric shells with net charges +Q and +Q
> (or -Q an -Q) represent a charged capacitor. Why not? There
> is an electric field between the two equally charged electrodes.
>
> Will the system be discharged by touching the electrodes
> with a conducting rod, for example, by somebody sitting on the
> surface of the smaller shell? It depends on what is meant by
> "discharged." The rod will allow the positive charge to move
> from the the smaller shell to the outer surface of the larger
> shell. The field between the electrodes will disappear. The
> inner shell will become neutral but the charge on the outer
> shell will double. And somebody outside the system will continue
> observing a constant E = k*(2*Q)/r^2. How would he know
> that these processes occured inside?
>
> A spherical capacitor with two equal charges of the same sign?
> It would be confusing to call such setup a capacitor. What would
> be a better name?" I would say that the word "capcitor" is
> reserved for a setup with two equal charges of opposite sign.
> Ludwik Kowalski

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