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(a) What are the change on the potential difference across each capacitor.

This is easy.

C(total) = (C1)(C2) / (C1 + C2) = 1.6uF

Q = C(total) * (V) = 4.8 x 10^-4 C This is the charge on C1 and C2.

V1 = Q/C1 = 240V

V2 = Q/C2 = 60V.

(b) The charged capacitors are disconnected from each other and from the battery. They are then reconnected, positive plate to positive plate and negative plate to negative plate, with no external voltage being applied. What are the charge and the potential difference for each capacitor now?

I'm not so sure about my answers to this one.

Because the capacitors are charged, connected positve to positve, and negative to negative, there would be no current flowing through the circuit. By charge conservation, Q(before) = Q(after), so in the series circuit Q = 4.8x10^-4C, so

V = Q/C(total) = Q/ (C1 + C2) = 48V across C1 and C2.

Therefore

Q1 = C1 * V = 2uF * 48V = 9.6x10^-5C

Q2 = C2 * V = 8uF * 48V = 3.84x10^-5C

(C)Suppose the charged capacitors in (a) were reconnected with plates of opposite sign together. What then would be the steady-state charge and potential difference for each capacitor?

Same as before Q = 4.8x10^-4C. But this time, there would be current flowing. So the charge is being distributed.

Q1 + Q2 = 4.8x10^-4C

The potential for the two capacitors must be the same, so

Q1/C1 = Q2/C2.

Solve the simultaneous equation for the answer.

Is this the right answer for (b) and (c)? I'm not so sure about the value for the conserved charge...