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(C) Given: Resistivity $ho = 200 \, \Omega \, m$,Capacitance $C = 2 	imes 10^{-12} \, F$,Potential difference $V = 40 \, V$,Relative permittivity $

Vedclass · Accepted Answer

$0.9 \, mA$

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(C) Given: Resistivity $ho = 200 \, \Omega \, m$,Capacitance $C = 2 	imes 10^{-12} \, F$,Potential difference $V = 40 \, V$,Relative permittivity $K = 50$.The resistance $R$ of the dielectric material between the plates is given by $R = \frac{ho d}{A}$,and the capacitance is $C = \frac{K \varepsilon_0 A}{d}$.Multiplying these,we get the time constant $	au = RC = \left( \frac{ho d}{A} ight) \left( \frac{K \varepsilon_0 A}{d} ight) = ho K \varepsilon_0$.The leakage current $I$ is given by Ohm's Law: $I = \frac{V}{R}$.Substituting $R = \frac{ho d}{A}$ and $C = \frac{K \varepsilon_0 A}{d} \implies \frac{A}{d} = \frac{C}{K \varepsilon_0}$,we get $R = \frac{ho K \varepsilon_0}{C}$.Thus,$I = \frac{V}{R} = \frac{VC}{ho K \varepsilon_0}$.Substituting the values: $I = \frac{40 	imes 2 	imes 10^{-12}}{200 	imes 50 	imes 8.854 	imes 10^{-12}}$.$I = \frac{80 	imes 10^{-12}}{10000 	imes 8.854 	imes 10^{-12}} = \frac{80}{88540} \approx 0.000903 \, A = 0.903 \, mA$.Rounding to the nearest option,the leakage current is $0.9 \, mA$.

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