Question

A potential difference V = 4.200×10²V is applied to two capacitors connected in series. One capacitor, C1, is 4.60 ?F and the other, C2, is 7.50 ?F.

The charged capacitors are disconnected carefully 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 is the charge on the positive plate of C1?

What is the potential difference across C1?

What is the potential difference across C2?

Suppose the charged capacitors had been reconnected with plates of opposite sign together. What then would be the steady-state potential difference for C1 ? (Do not try this at home!)

Answer 1

When connected in series,

Ceq = (4.60 x 7.50) / (4.6 + 7.5)

Ceq = 2.85 uF

charge on each capacitor, Q = V Ceq = 420 x 2.85 uF

     = 1197.5 uC

     total charge = 2 C = 2395 uC

charge will be conserved.

suppose new potential diff across them is V.

now total charge = (C1 + C2)V = 2395

V = 198 Volt ...Across both C1 and C2.

charge on C1 = 198 x 4.60 uF = 910.5 uC before and after