Calculate Relative Permittivity of Insulating Material in Capacitor

To find the relative permittivity $\varepsilon_r$ of the insulating material, we use the standard formulas for the capacitance and resistance of a parallel plate system filled with a material of resistivity $\rho$ and permittivity $\varepsilon = \varepsilon_r \varepsilon_0$.

1. Formula for Capacitance ($C$): $C = \frac{\varepsilon_r \varepsilon_0 A}{d}$ where:

   • $A$ is the area of the capacitor plates,
   • $d$ is the separation between the plates,
   • $\varepsilon_0$ is the permittivity of free space,
   • $\varepsilon_r$ is the relative permittivity of the material.

2. Formula for Resistance ($R$): $R = \rho \frac{d}{A}$ where $\rho$ is the resistivity of the insulating material.

3. Product of Resistance and Capacitance ($RC$): Multiplying the expressions for $R$ and $C$ eliminates the geometric terms $A$ and $d$: $RC = \left(\rho \frac{d}{A}\right) \left(\frac{\varepsilon_r \varepsilon_0 A}{d}\right) = \rho \varepsilon_r \varepsilon_0$

4. Solving for Relative Permittivity ($\varepsilon_r$): $\varepsilon_r = \frac{R C}{\rho \varepsilon_0}$

5. Substituting the given values:

   • $d = 0.885 \text{ mm} = 0.885 \times 10^{-3} \text{ m}$
   • $C = 1 \ \mu\text{F} = 10^{-6} \text{ F}$
   • $\rho = 1 \times 10^{13} \ \Omega\text{m}$
   • $R = 17.7 \times 10^{14} \ \Omega$
   • $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$

   $\varepsilon_r = \frac{(17.7 \times 10^{14} \ \Omega) \times (10^{-6} \text{ F})}{(1 \times 10^{13} \ \Omega\text{m}) \times (8.85 \times 10^{-12} \text{ F/m})}$

   $\varepsilon_r = \frac{17.7 \times 10^8}{88.5} = 0.2 \times 10^8 = 2 \times 10^7$

Comparing $\varepsilon_r = 2 \times 10^7$ with the given form $\alpha \times 10^7$, we get: $\alpha = 2$