A potential difference V = 1.520×102 V is applied to two capacitors connected in

Question

A potential difference V = 1.520×102 V is applied to two capacitors connected in series. One capacitor, C1, is 4.20 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 charge on the positive plate of C2?

What is the potential difference across C2?

Explanation / Answer

Earlier the capacitors are connected in sereis,

Ceq = 4.2 x 7.5 / (4.2 + 7.5) = 2.69 uF

Q = CV = 2.69 x 1.52 x 10^2 = 408.88 uC

Now when they are connected in parallel,

Ceq = 4.2 + 7.5 = 11.7 uF

Charge is conserved.

Q' = 2Q = 2 x 408.88 = 817.76 uC

Q1 + Q2 = 817.76 uC

Capacitors in parallel have same voltage across them

V1 = V2 = 817.76/11.7 = 69.89 V

Q1 = C1V1 = 4.2 x 69.89 = 293.54 C

Q2 = C2V2 = 7.5 x 69.89 = 524.18 C

Hence, Q1 = 293.54 uC ; Q2 = 524.18 uC and V2 = 69.89 Volts.

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