Finding Charge and Potential Difference in Capacitor Series | Homework Help

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In a series configuration of two capacitors, one initially charged and the other uncharged, charge will redistribute between them until they reach the same voltage. The final charge on each capacitor can be determined using the relationship Q = CV, where the voltage across both capacitors will be equal due to the absence of a voltage source. The charge on the initially charged capacitor (C1) will decrease while the uncharged capacitor (C2) will gain charge. The correct final charge on capacitor C2 is derived from the capacitance values, leading to the conclusion that the ratio of the final charges corresponds to the ratio of their capacitances. Understanding this charge distribution is key to solving problems involving capacitors in series.
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Homework Statement



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.

Homework Equations



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


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