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Re: a question from a student



I guess that I follow the "simplest is best" philosophy in this (and in
most) cases.

First: I'll consider a simple SERIES circuit made up of a switch, battery,
a resistor (to keep the current finite), and a capacitor; and consider
the capacitor to be unenergized* initially with the switch open. Both
plates of the capacitor have charges on board, they are just in equal
amounts so that the net charge on either plate is zero.

Second: I'll close the switch and the battery will pull the "no net
charge" on one plate apart, pumping the positive charge around circuit
(if you want to use the "positive conventional current" description) to
the other plate of the capacitor.

Third: If I simply accept the postulate that we can neither create nor
distroy charge** then we have the fact that Q1=-Q2 by inspection.

* I don't like the phrase "charge a capacitor" since there is never any
net charge deposited on a capacitor used in any circuit. We could imagine
the deposition of a net excess of static charge on an isolated capacitor
(on either plate or even both plates). But this what we never do in any
circuit. We do store energy in the E-field between the plates of a
capacitor; and so I would rather use the phrase "energize a capacitor".
This is the description that I regularly use with my students and they
seem to understand the important distinction.

** Electron/positron pair production still follows this postulate in that
no net charge is created.

On Sun, 6 Apr 1997, LUDWIK KOWALSKI wrote:

OOP, an older file was mailed a minute ago by mistake. It could be worse !

Last Thursday a student asked me why Q1 and Q2 are equal on a capacitor
connected to a battery. Would this also be true if the area of one plate
were twice as large as the other? How should I answer? Keep in mind this
is a non-calculus course and Gauss's law is not even mentioned in our text,
except in the appendix (at the end of the book).
Ludwik Kowalski



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