In the given figure,a capacitor is formed by placing a compound dielectric between the plates of a parallel plate capacitor. The expression for the capacity of the said capacitor will be (Given area of plate $= A$):

$A$ capacitor $C$ is charged to a potential difference $V$ and the battery is disconnected. Now,if the capacitor plates are brought closer slowly by some distance,then:

Three different dielectrics are filled in a parallel plate capacitor as shown. What should be the dielectric constant of a material,which when fully filled between the plates produces the same capacitance?

$A$ 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]

Consider a parallel plate capacitor of $10\,\mu F$ with air filled in the gap between the plates. Now,one half of the space between the plates is filled with a dielectric of dielectric constant $K = 4$,as shown in the figure. The capacity of the capacitor changes to.......$\mu F$.

The area of the plates of a parallel plate capacitor is $A$ and the distance between the plates is $10\,mm$. There are two dielectric sheets in it,one of dielectric constant $10$ and thickness $6\,mm$ and the other of dielectric constant $5$ and thickness $4\,mm$. The capacity of the capacitor is