If the distance between the plates of a capacitor having capacity $C$ and charge $Q$ is doubled,then the work done will be:

The energy of a charged capacitor resides in

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.

$A$ 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:

$A$ capacitor of capacitance $C$ has stored energy $W$ and charge $Q$. If the charge is increased to $2Q$,what will be the new stored energy?

The energy stored in a capacitor is in the form of