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

$A$ parallel plate capacitor has a potential of $20 \, kV$ and a capacitance of $2 \times 10^{-4} \, \mu F$. If the area of the plates is $0.01 \, m^2$ and the distance between the plates is $2 \, mm$,find the energy stored in the capacitor.

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

The capacitance of a capacitor is $4 \times 10^{-6} \ F$ and it is charged to a potential difference of $100 \ V$. The energy required to completely discharge it is . . . . . . $J$.

$A$ 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 (where $\epsilon_{0}$ is the permittivity of free space).

Four capacitors each of capacity $2 \mu F$ are connected as shown in the figure. If $V_{A}-V_{B}=10 \,V$,the energy stored in the system is: