A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is ismilarly charged to a potential difference 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the poistive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

Options

A. zero
B. `(3)/(2)CV^(2)`
C. `(35)/(6)CV^(2)`
D. `(9)/(2)CV^(2)`

Answer

B

Explanation

Net charge `Q=Q_(2)-Q_(1)` potential is `V_(1)`
`therefore V_(1)=((C_(0))/(C+C_(0)))V_(0)`
Similarly after nth operation , `E=1//2C^(1)V^(2)`