The capacitance of a capacitor is $1\,pF$ when there is air between its plates. If the distance between the plates is doubled and the space between them is filled with wax,the new capacitance becomes $2\,pF$. What is the dielectric constant of the wax?

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

$A$ 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 $x = \frac{d}{3}$ and $x = \frac{2d}{3}$,respectively. If $C_1 = 2 \mu F$,the value of $C_2$ is $........... \mu F$.

$A$ 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 having capacity $C_0$ is charged to $V_0$. With the battery disconnected,if the separation between the plates is doubled,then the energy stored in it is $E_1$. Instead,if the separation between the plates is doubled with the battery in connection,the energy stored in it is $E_2$. Then the value of $\frac{E_2}{E_1}$ is

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. Which graph shows the variation of electric field $(E)$ with distance $(x)$ from the left plate?