$A$ variable condenser is permanently connected to a $100 \ V$ battery. If the capacity is changed from $2 \ \mu F$ to $10 \ \mu F$,then the change in energy is equal to:

$A$ parallel plate capacitor has a plate separation $d$ and plate area $A$. If it is charged to a potential $V$ and then disconnected from the battery,calculate the work done in increasing the separation between the plates to $2d$.

$A$ 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 capacity $C$ has charge $Q$ and stored energy is $W$. If the charge is increased to $2Q$,the stored energy will be

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.

The energy stored in a capacitor of capacitance $C$ charged to a potential $V$ is: