A capacitor of capacity $C$ is connected with a battery of potential $V$ in parallel. The distance between its plates is reduced to half at once, assuming that the charge remains the same. Then to charge the capacitance upto the potential $V$ again, the energy given by the battery will be

  • A

    $C{V^2}/4$

  • B

    $C{V^2}/2$

  • C

    $3C{V^2}/4$

  • D

    $C{V^2}$

Similar Questions

If $E$ is the electric field intensity of an electrostatic field, then the electrostatic energy density is proportional to

A parallel plate capacitor whose capacitance $C$ is $14\, pF$ is charged by a battery to a potential difference $V =12\, V$ between its plates. The charging battery is now disconnected and a porcelin plate with $k =7$ is inserted between the plates, then the plate would oscillate back and forth between the plates with a constant mechanical energy of $..........pJ$. (Assume no friction)

  • [JEE MAIN 2021]

A $2\,\mu F$ capacitor is charged to $100$ $volt$ and then its plates are connected by a conducting wire. The heat produced is........$J$

Obtain the expression for the energy stored per unit volume in a charged capacitor.

A condenser of capacity ${C_1}$ is charged to a potential ${V_0}$. The electrostatic energy stored in it is ${U_0}$. It is connected to another uncharged condenser of capacity ${C_2}$ in parallel. The energy dissipated in the process is