A charge Q is placed on a capacitor of capacitance C. The capacitor is connected

Question

A charge Q is placed on a capacitor of capacitance C. The capacitor is connected into the circuit as shown in the figure below, with an open switch, a resistor, and an initially uncharged capacitor of capacitance 3C. The switch is then closed and the circuit comes to an equilibrium. In terms of Q and C, find the final potential difference between the plates of each capacitor. smaller capacitor larger capacitor Find the charge on each capacitor. smaller capacitor larger capacitor Find the final energy stored in each capacitor, smaller capacitor larger capacitor Find the internal energy appearing in the resistor.

Explanation / Answer

a.)Potential difference is same for both capacitors.

Q = (C+3C)*V

So V = Q/4C for both capacitors

b.) Charge on smaller capacitor = C*(Q/4C) = Q/4

Charge on larger capacitor = 3C*(Q/4C) = 3Q/4

c.) Final energy stored in smaller capacitor = (0.5)*C*(Q/4C)^2 = Q^2/32C

Final energy stored in larger capacitor = (0.5)*3C*(Q/4C)^2 = 3Q^2/32C

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