(પદાર્થની સાપેક્ષ પરમીએબિલિટી $50$ છે)
\({C}=2 \times 10^{-12}\, {F}\)
\({V}=40\, {V}\)
\({K}=56\)
\(C =\frac{ K \varepsilon_0 A }{ d }\) (\(K\) 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,
\(\left( i _0\right)_{\text {leakage }} =\frac{ Q }{\rho K \varepsilon_0}=\frac{ CV }{\rho K \varepsilon_0}\)
\(=\frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}\)
\(=903 \mu A =0.9 mA\)
\(\left( i _0\right)_{\text {leakage }} =0.9 \;mA\)