Re: question about series capacitors

[brian whatcott <inet@intellisys.net>]

At 15:34 9/10/97 -0500, you wrote:
> Dear PhysLers: I have a question about the usual formula for capacitors in
> series, which is different than Ludwik's business about the leakage which
> was discussed a few years ago.
>
> Suppose I connect two capacitors in series across a battery. Label the four
> capacitor plates from left to right as A, B, C, and D. Okay, suppose A is
> connected to the positive terminal of the battery, so out goes charge +Q to
> it, and compensating charge off D, leaving -Q on it. My question is: why
> does the charge on B and C have to be -Q and +Q, respectively?
>
> If the isolated circuit consisting of plates B and C and the wire between
> them is initially uncharged, then the sum of the charges presumably has to
> remain zero. But why couldn't I get -1.1Q and +1.1Q on these two plates,
> say? Why *exactly* -/+Q? Is it always exact: what if plates A and B have
> different shapes? Or what if I imagine distorting the wire between plates B
> and C, so that B and C are both portions of some larger object, say the two
> ends of a solid rectangular block, or even a sphere? Surely at some point
> the answer will no longer be -/+Q. At what point - in other words, what
> assumptions go into the usual derivation? I've looked in several textbooks
> and it's presented as though -/+Q is patently obvious. Carl
>
> ------------------------------------------------------------------------
> Dr. Carl Mungan

This is an engineering-style response, so it may lack rigor in 
mathematical physics terms.

For two caps in series, the charge is carried by the conduction current
in the conductors, and the displacement current through the caps.
 The series connection necessitates that the current is the same
through the whole loop.

 There is no reason why the caps cannot have a prior charge where
the 'battery' ends ( A,D) are at zero potential, but the common capacitor 
point (B,C) has introduced charge to both caps at some previous time.
(In this case, the charges in the two caps may be quite different,
depending on the capacitance values; the charging path being in 
parallel for this case, i.e B or C to A and B or C to D)

  Even so, the extra series current provided by the change of voltage
applied across the ends (A,D) is the same through both caps, and for the
same time, and so it follows that the EXTRA charge is always equal
at the two caps.

Regards
brian whatcott <inet@intellisys.net>
Altus OK