Question 96 of 180

The material filled between the plates of a parallel plate capacitor has resistivity $200\Omega m$. The value of capacitance of the capacitor is $2pF$. If a potential difference of $40V$ 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 permittivity of material is $50$)

A$0.9mA$B$9.0mA$C$9.0\mu A$D$0.9\mu A$

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Answer: (A) $0.9mA$

$\rho =200\Omega m$
$K=50$
$C=\dfrac{{ϵ}_{0}KA}{d}=2\times {10}^{-12}\text{F}$
$R=\dfrac{\rho d}{A}$
$I=\dfrac{V}{R}=\dfrac{AV}{\rho d}=\dfrac{V}{\rho }\cdot \dfrac{C}{{ϵ}_{0}K}=\dfrac{40}{200}\times \dfrac{2\times {10}^{-12}}{8.85\times {10}^{-12}\times 50}=0.9mA$