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 ( $\varepsilon_0=$ permittivity of free space)

A parallel plate capacitor has plate area $50 \mathrm{~cm}^2$ and plate separation 3 mm . The space between the plates is filled with a dielectric medium of thickness 1 mm and dielectric constant 4 . The capacitance becomes ( $\varepsilon_0=$ permittivity of free space)

The equivalent capacitance between plates ' A ' and ' B ' ( A -area of each plate, d-separation between plates) ( $\varepsilon_0$ - permittivity of free space) is

The voltage between the plates of a parallel plate capacitor of capacity $1 \mu \mathrm{~F}$ is changing at the rate of $4 \mathrm{~V} / \mathrm{s}$. the displacement current in the capacitor is