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

$A$ parallel plate capacitor having a plate separation of $2\, mm$ is charged by connecting it to a $300\, V$ supply. The energy density is.....$J/m^3$

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 $:-$

If the energy density in a parallel plate capacitor is given as $1.8 \times 10^{-9} \ J/m^3$,then the value of the electric field in the region between the plates is ....... $NC^{-1}$. (Given: $\epsilon_0 = 9 \times 10^{-12} \ C^2/N \cdot m^2$)

$A$ 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 plate area $A$ and separation $d$. It is charged to a potential difference $V_0$. The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is