The capacitance of an air-filled parallel plate capacitor is $10 \, pF$. The separation between the plates is doubled and the space between the plates is then filled with wax,giving the capacitance a new value of $40 \times 10^{-12} \, F$. The dielectric constant of wax 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?

$A$ parallel plate capacitor has a capacitance of $C_0$. If a dielectric slab of relative permittivity $\varepsilon_r$ and thickness equal to one-fourth of the distance between the plates is inserted,the new capacitance is $C$. Then,the ratio $\frac{C}{C_0}$ is:

What will be the capacity of a parallel-plate capacitor when the half of the parallel space between the plates is filled by a material of dielectric constant $\varepsilon_r$? Assume that the capacity of the capacitor in air is $C$.

When a dielectric material is introduced between the plates of a charged capacitor,the electric field between the plates:

If a slab of insulating material $4 \times 10^{-3} \ m$ thick is introduced between the plates of a parallel plate capacitor,the separation between the plates has to be increased by $3.5 \times 10^{-3} \ m$ to restore the capacity to its original value. The dielectric constant of the material will be: