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 \(\alpha\times 10^7\). The value of \(\alpha\) is ________. (Take permittivity of free space \(=8.85\times 10^{-12}\) F/m)

The resistance of the insulating material is given by \(R = \dfrac{\rho d}{A}\).

The capacitance of the parallel plate capacitor is given by \(C = \dfrac{\epsilon_0 \epsilon_r A}{d}\).

Multiplying the expressions for \(R\) and \(C\), we get:

\(RC = \rho \epsilon_0 \epsilon_r\)

Rearranging for relative permittivity \(\epsilon_r\):

\(\epsilon_r = \dfrac{RC}{\rho \epsilon_0}\)

Substituting the given values:

\(\epsilon_r = \dfrac{17.7 \times 10^{14} \times 10^{-6}}{10^{13} \times 8.85 \times 10^{-12}}\)

\(\epsilon_r = \dfrac{17.7 \times 10^8}{8.85 \times 10^1}\)

\(\epsilon_r = 2 \times 10^7\)

Comparing this with the given expression \(\epsilon_r = \alpha \times 10^7\), we get:

\(\alpha = 2\)

Answer: \(2\)