Capacitors connected to each other

  • Thread starter magevivi
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  • #1
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Homework Statement


Two capacitors, one that has a capacitance of 4.30 µF and one that has a capacitance of 12.1 µF, are connected in parallel. The parallel combination is then connected across the terminals of a 9.0-V battery. Next, they are carefully disconneted so that tehy are not discharged. They are then reconnected to each other--the positive plate of each capacitor connected to the negative plate of the other. Find the potential difference across each capacitor after they are reconnected.

C1 = 4.30µF
C2= 12.1µF
V = 9v

Homework Equations



C = Q/V

The Attempt at a Solution



Before they are reconnected each capacitor is charged to the following:
C1(V) = Q1 --> (4.3µF)(9v) = 38.7µC
C2(v) = Q2 --> (12.1µF)(9v) =108.9µC

But I lose my way after the reconnection. I understand that the voltage will change are charge will rearrange but when it comes to the specifics im not entirely sure. Any help would be appreciated.
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Welcome to PF.

What happens to the positive charge of the one capacitor and the negative charge of the other when they are connected?

Did I hear you say they cancel? Good.

So if they cancel what is the net charge across the 2 capacitors?

And since you are already armed with the equivalent capacitance ... and you know the charge ... hmm I wonder how to find the voltage?
 

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