A parallel plate capacitor is having separation between plates $0.885$ mm. It has a capacitance of $1\,\mu\text{F}$ 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 $\alpha \times 10^7$. The value of $\alpha$ is ______.
(Take permittivity of free space $= 8.85 \times 10^{-12}$ F/m)
Numerical value type. Enter your answer.
Answer: 2
For the same slab between the plates, $C = \dfrac{\epsilon_r\epsilon_0A}{d}$ and $R = \dfrac{\rho d}{A}$, so
$$RC = \rho\,\epsilon_r\epsilon_0 \quad (\text{independent of } A \text{ and } d)$$
$$\epsilon_r = \frac{RC}{\rho\epsilon_0} = \frac{17.7 \times 10^{14} \times 10^{-6}}{10^{13} \times 8.85 \times 10^{-12}} = \frac{17.7 \times 10^8}{88.5} = 2 \times 10^7$$
$\alpha = 2$.
Solution by Sreeraj P, M.Sc Physics