If the charge on a capacitor is increased by $2 \ C$,the energy stored in it increases by $44 \%$. The original charge on the capacitor is (in $C$):

$A$ parallel plate capacitor having plate area $A$ and separation $d$ is charged to a potential difference $V$. The charging battery is disconnected and the plates are pulled apart to four times the initial separation. The work required to increase the distance between the plates is ($\epsilon_0 =$ permittivity of free space).

$A$ 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.

$A$ parallel plate capacitor has a uniform electric field $E \ (V/m)$ in the space between the plates. If the distance between the plates is $d \ (m)$ and the area of each plate is $A \ (m^2)$,then the energy (joules) stored in the capacitor is:

Work done in placing a charge of $8 \times 10^{-18} \text{ C}$ on a capacitor of capacitance $800 \mu\text{F}$ is . . . . . . .

Consider a parallel plate capacitor of area $A$ (of each plate) and separation $d$ between the plates. If $E$ is the electric field and $\varepsilon_0$ is the permittivity of free space between the plates,then the potential energy stored in the capacitor is $:-$