If the distance between the plates of parallel plate capacitor is halved and the dielectric constant of dielectric is doubled, then its capacity will

A parallel plate capacitor of capacitance $12.5 \mathrm{pF}$ is charged by a battery connected between its plates to potential difference of $12.0 \mathrm{~V}$. The battery is now disconnected and a dielectric slab $\left(\epsilon_{\mathrm{r}}=6\right)$ is inserted between the plates. The change in its potential energy after inserting the dielectric slab is_______.$\times 10^{-12} \mathrm{~J}$.

In a parallel plate capacitor with air between the plates, each plate has an area of $6 \times 10^{-3}\; m ^{2}$ and the distance between the plates is $3 \;mm$ the capacitance of the capacitor is $17.71 \;pF$. If this capacitor is connected to a $100\; V$ supply, $3\; mm$ thick mica sheet (of dielectric constant $=6$ ) were inserted between the plates,

$(a)$ while the voltage supply remained connected.

$(b)$ after the supply was disconnected.

A capacitor of capacitance $9 n F$ having dielectric slab of $\varepsilon_{ r }=2.4$ dielectric strength $20\, MV / m$ and $P.D. =20 \,V$ then area of plates is ....... $\times 10^{-4}\, m ^{2}$

When a slab of dielectric material is introduced between the parallel plates of a capacitor which remains connected to a battery, then charge on plates relative to earlier charge

In a parallel plate capacitor the separation between the plates is $3\,mm$ with air between them. Now a $1\,mm$ thick layer of a material of dielectric constant $2$ is introduced between the plates due to which the capacity increases. In order to bring its capacity to the original value the separation between the plates must be made......$mm$