# Finding the charge of a capacitor

1. Jul 20, 2008

### Indeed

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

Problem 62. Capacitors C1 (4 uF) and C2 (2 uF) are charged as a series combination across a 100 V battery. The two capacitors are disconnected from the battery and from each other. They are then connected positive plate to positive plate, and negative plate to negative plate. Calculate the resulting charge on each capacitor.

2. Relevant equations

(Charge) = (voltage) or (potential difference) x (capacitance)
(Q)=(V)(C)

3. The attempt at a solution

I added both capacitors to get a net capacitance, C12 (1.33 uF). Then I used 100V and 1.33 uF to get the charge, Q12 (133). I though that series capacitors had the same charge, but 133 is not the answer.

I don't understand the bit about connecting positive plates to positive plates, what is that?

2. Jul 20, 2008

### LowlyPion

They are asking that if you disconnected both capacitors while still charged from the circuit and then connected them in parallel in isolation from the battery, how much charge would there then be on the new combination. (Ignore the practicality of ever actually doing it.)