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## Homework Statement

Initially, a 120V emf is connected to C1 by a switch. When the switch is flipped C1 is connected to C2 and C3 in series. What are the charge and potential difference across each capacitor?

C1 = 15 uF

C2 = 16 uF

C3 = 30 uF

## Homework Equations

C= q/V

1/Ceq = 1/C1 + 1/C2 + ... + 1/Cn

## The Attempt at a Solution

Initially with C1 is only connected to the 120V source it will charge to 120V. Once the switch is flipped C1 will become the emf source. C1 will discharge while C2 and C3 will charge.

The initial charge on C1 I found to be Qi=CV=(15uF)(120V)= 1800 uC

Since C1, C2 and C3 are in series, the charge on the capacitors will be the same (Q1=Q2=Q3). I assume that Qi = Q1 + Q2 + Q3, as in the charge will distribute evenly within the new circuit. So Q1=600 uC. So the charge on each capacitor is Q=600 uC. But this isn't right.

The potential difference across each capacitor would then be V=Q/C.

So V1=Q/C1, V2 = Q/C2, V3=Q/C3.