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

$A$ 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:

$A$ capacitor of capacitance $5 \mu F$ is charged by a battery of emf $10 \text{ V}$. At an instant of time,the potential difference across the capacitor is $4 \text{ V}$ and the time rate of change of potential difference across the capacitor is $0.6 \text{ Vs}^{-1}$. Then,the time rate at which energy is stored in the capacitor at that instant is:

$A$ parallel plate capacitor has a uniform electric field '$E$' in the space between the plates. If the distance between the plates is '$d$' and the area of each plate is '$A$', what is the energy stored in the capacitor?

The energy stored in a capacitor of capacitance $10 \mu F$ when charged to a potential of $6 \text{ kV}$ is: (in $\text{ J}$)

If the energy density in a parallel plate capacitor is given as $1.8 \times 10^{-9} \ J/m^3$,then the value of the electric field in the region between the plates is ....... $NC^{-1}$. (Given: $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$)