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Re: [Phys-L] charge distribution leading up to a capacitor



I had asked:
Let r be the *ratio* of negative to positive particles on the
negative plate. What is the ratio on the wire leading up to that
negative plate: one, less than r, equal to r, or greater than r?

To which John Denker replied on Friday, July 29, 2016 4:36 PM

I don't understand the question. I don't see how to attach physical
significance to the question or to any of the proposed answers.

In an ordinary capacitor, the number of charged particles is on the
order of a mole, i.e 6e23 particles. The amount of «net» charge on
either plate is on the order of a microcoulomb or less, i.e. at most
6e12 elementary charges. So to an exceedingly good approximation, r
is equal to 1 on the capacitor plates, on the wires, and everywhere
else.

I understand and agree that the ratio is essentially 1.

However, it is not exactly 1 on the plate. If it was, it would be neutral.

All I want to know is if there is a similar imbalance in the wires.

I know that such an imbalance, if it existed, wouldn't produce a significant charge.

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Robert A. Cohen, Department of Physics, East Stroudsburg University
East Stroudsburg, PA 18301