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. `3/2CV^(2)`
C. `25/6CV^(2)`
D. `9/2CV^(2)`

Answer

B

Explanation

The digramatic repensentation of given problem is shown in figure .
The net charge shared between the two capacitors is
`Q '=Q_2-Q_1=4CV-CV=3CV`
The two capacitor will have the same potential say V'
The net capacitance of the parallel combinations of the two capacitors will be
`C' =C_1+C_2=C+ 2C=3C`
The potential difference across the capacitors will be
`V '(Q ')/(C ')=(3 CV)/(3C)=V`
The electrostatic energy of the capacitor will be
`U '=1/2C ' V '=1/2(3C) V^(2)=3/2CV^(2)`