Calculating Capacitor Charge and Potential Difference in Series

[Soaring Crane]

A capacitor C1 carries a charge Q0. Is is then connected directly to a second, uncharged, capacitor C2.

What charge will each carry now? Q1 = Q0C1/(C1 + C2); Q2 = Q0C2/(C1 + C2)


What will be the potential difference across each?
V = Q0 / (C1 + C2)

-------||-------||--------
|*****C1*****C2******|
|********************|
|********************|
|********************|
|_______________________|

I know that the formula Q = CV is used, but how do you manipulate it (other than V = Q/C) to get the answers in bold?

Thanks.

 

[StatusX]

The voltage around the loop will be 0 in equilibrium, and whatever charge the second capacitor gets, the first one must lose. Does that help at all?

 

[Doc Al]

I know that the formula Q = CV is used, but how do you manipulate it (other than V = Q/C) to get the answers in bold?

Since the capacitors are connected in parallel, they have the same voltage (V). So the charges on the capacitors are: $Q_1 = C_1V$ and $Q_2 = C_2V$. The total charge remains the same, so $Q_0 = Q_1 + Q_2$. Solve for V.

 

[Soaring Crane]

I think I understand what you're saying, but I don't know how to apply it to the questions.

 

[Soaring Crane]

Oh, OK. Let me try it.