1. Jan 7, 2009

jakey

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

so, we have a capacitor which has capacitance C1 and charge Q0. This capacitor is connected to a second one (capacitance C2) with initial charge Q=0. The capacitors are in a series fashion with no voltage source. What charge does each carry now? Also find the potential difference.

2. Relevant equations

I think we can use PE = Q^2/2C and Q=CV

3. The attempt at a solution

I thought of tackling the problem by using the fact that potential energy is constant. I know that some of the charge from the first resistor will transfer to the second.
So, because Q0 is the initial charge of the first capacitor, PE = Q0^2/2C1.
After some time, let Q2 be the charge which leaves the first capacitor and enters the uncharged capacitor. PE= (Q0-Q2)^2/C1 + Q2^2/C2.

I then equated the two of them. It seems that Q2= 2Q0/(C1+C2) but the correct answer should be the one without a 2. Does anyone know how to tackle this problem? thanks

2. Jan 7, 2009

jakey

okay, so now, i kinda got a solution from a person. he got the right answer. you may disregard what ive written above. but why is the voltage eqaul??? thanks

3. Jan 7, 2009

Pumblechook

If they are in series with no other component or voltage source then they are also in parallel so the (final) voltages are the same.

The ratio of the final charges will be the same as the ratio of the capacitance values.