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ibc said:I just said that it's problematic to tak about these plates potentials, because we talk about uniform field that only exists inside the capacitor, and that it's problematic to choose a general point of perspective for both capacitors, to know the potential on each plate.
Yes, that was puzzling to me too. The equilibrium distribution is dependent on the initial charge distribution, but fundamentally, it seemed to me that equilibrium for charged nodes could still be handled by the condition of equilibrium, equipotentiality of connected plates, charge conservation at each node, and the definition of capacitance for each capacitor in series (which if you read below may not be quite right). But the equations I got were unduly messy for such a simple situation.
schroder said:When you connect two caps in series, the plates that are connected via the conducting leads become One plate, electrically speaking, and that is the language we are speaking here today.
schroder said:You can extend this model to include any number of caps you wish keeping in mind that where the leads are connected together, there is effectively only one plate in play.
And schroder's comments also indicated a good trick for the equilibrium distribution, but I had trouble getting the trick right with more capacitors
Then I found this interesting comment on Wikipedia: C=Q/V does not apply when there are more than two charged plates, or when the net charge on the two plates is non-zero. To handle this case, Maxwell introduced his "coefficients of potential". If three plates are given charges Q1,Q2,Q3, then the voltage of plate 1 is given by V1 = p11Q1 + p12Q2 + p13Q3, and similarly for the other voltages.
http://en.wikipedia.org/wiki/Capacitance
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