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
Concept. For a capacitor filled with a leaky dielectric, $R=\dfrac{\rho d}{A}$ and $C=\dfrac{K\varepsilon_0A}{d}$, so $RC=\rho K\varepsilon_0$ whatever the shape.
Step 1. Solve. $K=\dfrac{RC}{\rho\varepsilon_0}=\dfrac{17.7\times10^{14}\times10^{-6}}{10^{13}\times8.85\times10^{-12}}=\dfrac{1.77\times10^9}{88.5}=2\times10^7$.
Answer. $\alpha=2$.
Solution by Sreeraj P, M.Sc Physics
