The work done in placing a charge of $8 \times 10^{-18} \ C$ on a capacitor of $800 \ \mu F$ is:

$A$ battery is used to charge a parallel plate capacitor until the potential difference between the plates becomes equal to the e.m.f. of the battery. The ratio of the energy stored in the capacitor to the work done by the battery will be

Energy is stored in what form in a capacitor?

$A$ cylindrical capacitor has charge $Q$ and length $L$. If both the length and the charge are doubled (keeping other parameters constant),the energy stored in the capacitor will:

$A$ 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 $A,$ the energy stored in the capacitor 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: