The energy of a charged capacitor is given by the expression ($q$ = charge on the conductor and $C$ = its capacity).

$A$ parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates,the time rate of change of electrostatic energy of the capacitor is proportional to

$A$ capacitor is charged by a battery and the energy stored in it is $U$. The battery is now removed and the separation distance between the plates is doubled. The energy stored now is:

If the charge on a capacitor is increased by $2 \ C$,the energy stored in it increases by $21 \%$. The original charge on the capacitor is (in $C$)

Three identical capacitors are combined in different ways. For the same voltage applied to each combination,the one that stores the greatest energy is:

$A$ 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