$A$ capacitor is charged to $200\,V$. It has $0.1\,C$ of charge. When it is discharged,the energy released will be $J$.

Two capacitors,one of capacity $C$ and other of capacity $C/2$ are connected to a $V$ volt battery,as shown in the figure. The work done in charging fully both the capacitors is

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

$A$ 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:

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 $6 \ \mu F$ is charged from $10 \ V$ to $20 \ V$. The increase in its stored energy is: