The energy required to charge a capacitor of $5\,\mu F$ by connecting a $d.c.$ source of $20\,kV$ is......$kJ$.

$A$ 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:

The total energy stored in the capacitor system shown in the figure will be (in $\mu J$)

The plates of a parallel plate capacitor have an area of $90 \,cm^{2}$ each and are separated by $2.5 \,mm$. The capacitor is charged by connecting it to a $400 \,V$ supply.
$(a)$ How much electrostatic energy is stored by the capacitor?
$(b)$ View this energy as stored in the electrostatic field between the plates,and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

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 $C$)

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