$A$ parallel plate capacitor is charged by a battery until the potential difference between the plates equals the electromotive force of the battery. What is the ratio of the energy stored in the capacitor to the work done by the battery?

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

$A$ parallel plate capacitor has plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is

In the figure,the charge on the capacitor is plotted against the potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively:

$A$ $700\,pF$ capacitor is charged by a $50\,V$ battery. The electrostatic energy stored by it is

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