In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

Two capacitors of capacitance $2C$ and $C$ are joined in parallel and charged to potential $V$. The battery is now removed and the capacitor $C$ is filled with a medium of dielectric constant $K$. The potential difference across each capacitor will be

The capacitance of a capacitor becomes $\frac{7}{6}$ times the original value if a dielectric slab of thickness $t = \frac{2d}{3}$ is introduced between the plates, where $d$ is the distance of separation between the plates. The dielectric constant of the slab is

$A$ parallel plate capacitor of capacitance $C$ has spacing $d$ between two plates having area $A$. The region between the plates is filled with $N$ dielectric layers,parallel to its plates,each with thickness $\delta = \frac{d}{N}$. The dielectric constant of the $m^{\text{th}}$ layer is $K_m = K(1 + \frac{m}{N})$. For a very large $N (> 10^3)$,the capacitance $C$ is $\alpha \left( \frac{K \varepsilon_0 A}{d \ln 2} \right)$. The value of $\alpha$ will be. . . . . . . .
[$\varepsilon_0$ is the permittivity of free space]

$A$ parallel plate capacitor is charged by a battery and the battery remains connected. $A$ dielectric slab of constant $K$ is inserted between the plates and then taken out. What happens to the electric field between the plates?

$A$ $60 \ \mu F$ parallel plate capacitor whose plates are separated by $6 \ mm$ is charged to $250 \ V$,and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 \ mm$ is placed between the plates,find the change in the potential difference across the capacitor (in $V$)?