$A$ 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?

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

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 having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

$A$ 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: