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……$mm$

A capacitor is connected to a battery of voltage $V$. Now a di electric slab of dielectric constant $k$ is completely inserted between the plates, then the final charge on the capacitor will be

(If initial charge is $q_{0}$ )

A parallel plate capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with the capacitor are given by $Q_0, V_0, E_0$ and $U_0$ respectively. A dielectric slab is introduced between plates of capacitor but battery is still in connection. The corresponding quantities now given by $Q, V, E$ and $U$ related to previous ones are

A parallel plate capacitor with air between the plates has a capacitance of $8 \;pF \left(1 \;pF =10^{-12} \;F \right) .$ What will be the capacitance (in $pF$) if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant $6 ?$

A parallel plate capacitor with plate area $A$ and plate separation $d$ is filled with a dielectric material of dielectric constant $K =4$. The thickness of the dielectric material is $x$, where $x < d$.

Let $C_1$ and $C_2$ be the capacitance of the system for $x =\frac{1}{3} d$ and $x =\frac{2 d }{3}$, respectively. If $C _1=2 \mu F$ the value of $C _2$ is $……….. \mu F$