A parallel plate capacitor is having separation between plates 0.885 mm. It has a capacitance of 1 $$\mu$$F when the space between the plates is filled with an insulating material of resistivity $$1 \times 10^{13}$$ $$\Omega$$m and resistance $$17.7 \times 10^{14}$$ $$\Omega$$. Relative permittivity of the insulating material is $$a \times 10^7$$. The value of $$a$$ is __________. (Take permittivity of free space $$= 8.85 \times 10^{-12}$$ F/m)

Solution :

For a capacitor filled with dielectric :

$$C = \frac{\epsilon_0K A}{d}$$

Resistance of dielectric slab :

$$R = \frac{\rho d}{A}$$

Eliminating $$A$$ :

$$A = \frac{\rho d}{R}$$

Substituting in capacitance formula :

$$C=\frac{\epsilon_0K}{d}\times\frac{\rho d}{R}$$

$$C = \frac{\epsilon_0K\rho}{R}$$

Therefore,

$$K = \frac{CR}{\epsilon_0\rho}$$

Given :

$$C = 1\mu F = 10^{-6}\text{ F}$$

$$R = 17.7 \times 10^{14}\Omega$$

$$\rho = 10^{13}\Omega\text{m}$$

$$\epsilon_0 = 8.85 \times 10^{-12}\text{ F/m}$$

Substituting :

$$K=\frac{(10^{-6})(17.7 \times 10^{14})}{(8.85 \times 10^{-12})(10^{13})}$$

$$=\frac{17.7 \times 10^8}{8.85 \times 10^1}$$

$$=2 \times 10^7$$

Given,

$$K = a \times 10^7$$

Therefore,

$$a = 2$$

Final Answer :

$$2$$