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ritwik06
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Homework Statement
Image URL:
http://img13.imageshack.us/img13/1935/capacitor.png
Before the switch was closed, Capacitor 1 had a charge of magnitude CE/3 on its plates and Capacitor 2 had a charge of magnitude CE/6. The third capacitor was initially neutral.
Find the charge flown through battery when switch 'S' is made to close where E is the emf of the battery.
The Attempt at a Solution
Firstly, I solve this problem using the conventional method. I assume that a charge 3q flows out from the positive end of the battery. It divides itself in the ration 2:1 between capacitors 1 and 2, in order to keep the potential across them equal, since they are connected in parallel. Therefore the net final charges on the capacitors are:
1== (CE/3)+2q
2== (CE/6)+q
3== 3q
From potential balancing in the circuit when steady state is achieved, I get:
(((CE/3)+2q)/C)+(3q/C)=E
q=2CE/15
therefore charge flown through battery=3q=2CE/5
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Till now, I had no problem. But after this method, an idea struck me:
I calculated the initial total +VE charge=(CE/3)+(CE/6)=CE/2
Now, I calculated the final total +VE charge.
C(equivalent for the circuit)=3C/5
The net positive charge on the equivalent single capacitor would have been 3CE/5.This extra positive charge must have come from the charge flown through the battery.
Charge flown through battery=3CE/5 - CE/2= CE/10 (which is different from the one calculated previously)
This method worked right until now and I thought it was another version of charge conservation. But this problem really shocked me literally. Please explain the anomaly with the second method.
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