# Problem on Capacitor

Hyperreality
A potential difference of 300V is applied to a series connection of two capacitors, of capacitance C1 = 2uF and capacitance C2 = 8uF.

(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?

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...

Homework Helper
Hyperreality said:
A potential difference of 300V is applied to a series connection of two capacitors, of capacitance C1 = 2uF and capacitance C2 = 8uF.

(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?

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...
You need to look again at b) and c). There will be a charge flow when the capacitors are connected in both cases. Charge will flow as long as there is a potential difference between the two capacitors. You are correct about charge conservation; the implication is that whatever charge flows from one capacitor will be transferred to the second capacitor. The sum of the charges on the two capacitors cannot change. Case b) is probably a bit easier to think about because the charges on the connected plates have the same sign. See if you can determine how the total charge will redistribute so that the final potential across each capacitor will be the same. It will have to be a potential somewhere between the intial potentials on each capacitor. Once you get b), the same thinking applies to c). The difference is that the initial charges on the connected plates will have opposite sign resulting in a lower final charge and lower potential than case b).

Hyperreality
See if you can determine how the total charge will redistribute so that the final potential across each capacitor will be the same.

For (b) Okay, so V1 = V2 acros both capacitors, because the sign between the connected plates are the same. So

(Q - q)/C1 = (Q + q)/C2,

q = Q(C2 - C1) / (C1 + C2) = 2.88 x 10^-4 C, so

Q1 = Q - q = 1.92 x 10^-4C
Q2 = Q + q = 7.68 x 10^-4C
V1 = Q1 / C1 = 96V = V2.

Is (b) right now??