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Re: [Phys-l] Capacitance problem



On 03/25/2008 01:30 PM, mrmeyer@mtu.edu wrote:
John -

I don't disagree with anything you said - for the first time through the
sequence when the capacitors are uncharged. I get stuck starting on the
2nd iteration...because at that point, the top plate of the top capacitor
has a charge different than the bottom plate.

It's a *linear* system.

Linearity implies superposition.

This is a principle of physics, indeed a principle of mathematics,
that is absolutely priceless. Linearity implies superposition.

Just before the start of the second cycle, the 2-capacitor system,
taken has a whole, has some overall black-box capacitance and some
overall black-box voltage. You know the capacitance and you know
the voltage, and you don't need to know anything else at the moment.
(S1 is connected to the top terminal of the black box; S2 is inside
the black box.) Linearity comes in as follows: The black-box
capacitance is *independent* of the value of V2. (If the capacitors
were nonlinear this would not be true.)

Ask how much charge must flow into the terminal of the black box in
order to charge it up when switch S1 closes.

After that amount of charge has flowed through the system, you can
then peek inside the black box and see what that charge did to the
V2 node. Then dump that charge to ground via S2.