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

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

$A$ 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:

$A$ parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If $x$ is the separation between the plates,the time rate of change of electrostatic energy of the capacitor is proportional to