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

The mean electrical energy density between the plates of a charged air capacitor is (where $q=$ charge on capacitor,$A=$ area of capacitor plate):

The energy required to charge a parallel plate capacitor with plate separation $d$ and plate area $A$ such that the uniform electric field between the plates is $E$,is

The charge $Q$ on a capacitor varies with voltage $V$ as shown in the figure,where $Q$ is taken along the $X$-axis and $V$ along the $Y$-axis. The area of triangle $OAB$ represents

$A$ 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:

The work done in placing a charge of $8 \times 10^{-18} \, C$ on a condenser of capacity $100 \, \mu F$ is: