$A$ capacitor of capacity $50\,\mu F$ is charged to $10\;V$. Its energy is equal to:

$A$ 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:

$A$ $10 \, pF$ capacitor is connected to a $50 \, V$ battery. How much electrostatic energy is stored in the capacitor?

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

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:

$A$ parallel plate capacitor has a plate separation $d$ and plate area $A$. If it is charged to a potential $V$ and then disconnected from the battery,calculate the work done in increasing the separation between the plates to $2d$.