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Parallel Capacitors connected to Battery then D/C 
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#1
Feb1311, 03:46 PM

P: 4

1. The problem statement, all variables and given/known data
Capacitors of 19.1 uF and 2.1 uF are charged as a parallel combination across a 229 V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Find the resulting charge on the ﬁrst capacitor. 2. Relevant equations Q = CV 3. The attempt at a solution Since they are connected in parallel, I found the charge for each individual capacitor. Q1 = C1V Q2 = C2V Then I know when reconnected, the net charge has to equal the new charge. So: q1+q2 = Q1Q2 Then the voltage across them has to be the same. So: V' = q1/c1 = q2/c2 Then I solved them sim. and got: q1 + (c2/c1)q1 = Q1  Q2 But when I plug in my numbers, I get the wrong answer. What am I doing wrong? 


#2
Feb1311, 04:04 PM

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P: 11,689

You might find it convenient to assign charge (coulombs) to each plate before the reconnection, then determine what will remain after the cancellation happens when opposite charges are "introduced" to each other via the new connections. 


#3
Feb1311, 04:09 PM

P: 4

Edited for more that I've done to solve the problem.
So the charges essentially cancel out? 


#4
Feb1311, 04:49 PM

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Parallel Capacitors connected to Battery then D/C



#5
Feb1311, 04:55 PM

P: 4

So would it be Q1Q2 (The initial charges of the capacitor) = q1 + q2 (The charges after reconnection)?



#6
Feb1311, 05:00 PM

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#7
Feb1311, 05:06 PM

P: 4

So because they are parallel, the voltage across the plate have to be the same, so I have:
V' = q1/c1 = q2/c2 and solving simulatenous, I get: q1 + (c2/c1)q1 = Q1  Q2 Plug in the numbers, and I get ~3507 which is still the wrong answer... 


#8
Feb1311, 05:10 PM

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P: 11,689

Ah, but 3507 what? And how do your significant figures compare with those given as original information?



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