The material filled between the plates of a parallel plate capacitor has resistivity \(200 \Omega \, {m}\). The value of capacitance of the capacitor is \(2\, {pF}\). If a potential difference of \(40 \,{V}\) is applied across the plates of the capacitor, then the value of leakage current flowing out of the capacitor is (given the value of relative permitivity of material is \(50\) )

\({C}=2 \times 10^{-12}\, {F}\)

\({V}=40\, {V}\)

\({K}=56\)

\(C =\frac{ K \varepsilon_0 A }{ d }\) (\(K\) is the dielectric constant or relative permittivity) and

\(R =\frac{\rho d }{ A }\)

Now, charge will be discharged through the resistance between the plates.

Now, time constant ( \(( T )\) of discharging,

\(\tau= RC =\frac{\rho d }{ A } \times \frac{ K \varepsilon_0 A }{ d }\)

\(\tau=\rho K \varepsilon_0\)

For a given R-C circuit, the discharged current is given by

\(i =\frac{ Q }{ RC } e ^{-\frac{ t }{ RC }}\)

\(i =\frac{ Q }{ pK \varepsilon_0} e ^{-\frac{ t }{ pK \varepsilon_0}}\)

The above discharge current is the leakage current,

\(i _{\text {leakage }}=\frac{ Q }{\rho K \varepsilon_0} e ^{-\frac{ t }{\rho K \varepsilon_0}}\)

Maximum leakage current,

\(\left( i _0\right)_{\text {leakage }} =\frac{ Q }{\rho K \varepsilon_0}=\frac{ CV }{\rho K \varepsilon_0}\)

\(=\frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}\)

\(=903 \mu A =0.9 mA\)

\(\left( i _0\right)_{\text {leakage }} =0.9 \;mA\)