# Capacitors Connected, Disconnected, and Connected Again

1. Mar 21, 2009

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1. The problem statement, all variables and given/known data

Capacitors 1,3, and 3 have capacitances equal to 2.0uF, 4.0 uF, and 6.0uF, respectively. The capacitors are connected in parallel, and the parallel combination is connected across the terminals of a 200-V source. The capacitors are then disconnected from both the voltage source and each other, and are connected to three switches as shown in the image. a) What is the potential difference across each capacitor when switches S1 and S2 are closed but switch S3 remains open? b) After switch S3 is closed, what is the final charge on the lefmost plate of each capacitor? c) Give the final potential difference across each capacitor after switch S3 is closed.
2. Relevant equations

C = Q/V

Vsys=Cequiv*Qtotal

Series Connection: 1/Cequiv = 1/C1 + 1/C2 + .... + 1/Cn
Parallel Connection: Cequiv = C1 + C2 +....+ Cn
n= number of capacitors under consideration.

3. The attempt at a solution
For a) the potentials are the same, but I think I am not justifying it by the right reasons. My argument is that since all the capacitors were at the same potential when connected in parallel, as they are connected as asked by a there can be no direct change in potential because none of the capacitors are directly attached to one another.
b) I really don't know exactly, and trust me, I'm not trying to slack off but I'm just clueless as to how to approach it. My thoughts is that C1 and C3 are connected in parallel but I don't know how to account for C2 so everything remains unsolved. If they were all in series then they would all end up with the same charge, but that's not the case so I am not sure what to do. Any help is greatly appreciated.

Last edited: Mar 21, 2009
2. Mar 21, 2009

### LowlyPion

Welcome to PF.

When they are in || across the voltage source they are endowed with specific charges from the supply. Q = V*C

What they want you to consider then is what happens when you take these charged capacitors and now place them in a series configuration.