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Re: [Phys-l] RC Discharge



Michael Edmiston wrote:

But we can also remove, say, negative charge from each plate, such that
both plates end up with positive net charge. If we remove different
magnitudes of negative charge from the two plates, both plates have net
positive charge, but there is a potential difference between the plates.
One plate would be at higher potential than the other even though both
are positive. I would designate the higher-potential plate with a plus
and the lower-potential plate with a minus where these signs are used to
indicate which plate is at higher/lower potential.

Now... If deltaV = q/C, which plate has charge q? I would say neither.
I would say q in the capacitor equation represents half the difference
in the net charges on the plates. I would say deltaV = q/C with q = (Q+
minus Q-)/2.

Am I missing something here?

It isn't this simple. For instance, imagine two capacitors--a symmetric parallel plate capacitor and a cylindrical coaxial capacitor--BOTH of which have +1 units of charge on one conductor and +5 units on the other. Now use the ideas of Gauss' law to analyze the different possible cases.


Case 1: Symmetric parallel plate capacitor. Both conductors end up with +3 units of charge on their outer surfaces. One gets -2 units on its inner surface (the one facing the other conductor's inner surface) and the other gets +2 units on its inner surface. The "charge on the capacitor" is 2 units in keeping, I think, with Michael's suggestion.


Case 2: Cylindrical coaxial capacitor.

Subcase a: If the outer conductor is the one with +1 unit of charge, then IT ends up with +6 units of charge on its outer surface and -5 units on its inner surface (the one facing the inner conductor's outer surface. The inner conductor ends up with no charge on its inner surface and +5 units on its outer surface. (Draw a picture!) The "charge on the capacitor" is 5 units in contradicting Michael's suggestion.

Subcase b: If the outer conductor is the one with +5 units of charge, then it ends up with +6 units of charge on its outer surface and -1 unit on its inner surface (the one facing the inner conductor's outer surface. The inner conductor ends up with no charge on its inner surface and +1 unit on its outer surface. Now the "charge on the capacitor" is 1 unit AND the less positively charged conductor is the "positive plate."


Two points:

1) When we talk about "the charge" on a capacitor, we are referring to the magnitude of the NECESSARILY equal and opposite charges on the facing surfaces. Any excess charge that ends up on the nonfacing surfaces is irrelevant to the fundamental operation of the capacitor.

2) The amount of excess charge that ends up on the nonfacing surfaces depends on the geometric details of the capacitor's construction.

Disclaimer: I know that one can nit-pick the above with regard to its neglect of the effects of fringing fields. My intent is to get at the fundamental physics hinted at by Michael's post.

--
John "Slo" Mallinckrodt

Professor of Physics, Cal Poly Pomona
<http://www.csupomona.edu/~ajm>

and

Lead Guitarist, Out-Laws of Physics
<http://www.csupomona.edu/~hsleff/OoPs.html>