$A$ $12\,pF$ capacitor is connected to a $50\,V$ battery. How much electrostatic energy is stored in the capacitor?

The energy of a parallel plate capacitor when connected to a battery is $E$. With the battery still in connection,if the plates of the capacitor are separated so that the distance between them is twice the original distance,then the electrostatic energy becomes

If potential difference across a capacitor is changed from $15 \ V$ to $30 \ V,$ work done is $W.$ The work done,when potential difference is changed from $30 \ V$ to $60 \ V,$ will be $.......W$

$A$ $6\,\mu F$ capacitor is charged from $10\,V$ to $20\,V$. The increase in energy will be:

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between plates is:

The charge $Q$ on a capacitor varies with voltage $V$ as shown in the figure,where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents