There is an air-filled $1\,pF$ parallel plate capacitor. When the plate separation is doubled and the space is filled with wax,the capacitance increases to $2\,pF$. The dielectric constant of wax is

Explain the effect of a dielectric on the capacitance of a parallel plate capacitor and obtain the formula for the dielectric constant.

As shown in the figure,a dielectric slab of thickness $d/2$ and dielectric constant $K$ is placed between the plates of a parallel plate capacitor of plate separation $d$. The capacitor is charged to a potential $V$ using a battery. If the dielectric slab is pulled out after disconnecting the battery from the capacitor,the final potential difference across the plates of the capacitor is

$A$ capacitor stores $60\,\mu C$ charge when connected across a battery. When the gap between the plates is filled with a dielectric,an additional charge of $120\,\mu C$ flows through the battery. The dielectric constant of the dielectric inserted is

$A$ parallel plate capacitor has two layers of dielectric as shown in the figure. This capacitor is connected across a battery. The graph between electric field $(E)$ and distance $(x)$ from the left plate will be:

The capacity of an air-filled parallel plate capacitor is $C_0$. One-half of the space between the plates is filled with a dielectric of constant $K$ as shown in the figure. The new capacity becomes $C_n$. The ratio of $C_n$ to $C_0$ is: