Q 12-03-211JEE MainJEE Main 2021 (26 Aug, Shift 1)Medium
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 permittivity of material is $50$)
Answer: (A) $0.9$ mA
$C = \dfrac{K\varepsilon_0A}{d}$ and $R = \dfrac{\rho d}{A}$, so $RC = \rho K\varepsilon_0$.
$$I = \frac{V}{R} = \frac{VC}{\rho K\varepsilon_0} = \frac{40\times2\times10^{-12}}{200\times50\times8.85\times10^{-12}} \approx 9.0\times10^{-4}\ \text{A} = 0.9\ \text{mA}$$
Solution by Sreeraj P, M.Sc Physics