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The material filled between the plates of a parallel plate capacitor has resistivity $200 \Omega \mathrm{m}$. The value of capacitance of the capacitor is $2 \mathrm{pF}$. If a potential difference of $40 \mathrm{~V}$ 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 )
Correct Option: , 3
$\rho=200 \Omega \mathrm{m}$
$\mathrm{C}=2 \times 10^{-12} \mathrm{~F}$
$\mathrm{V}=40 \mathrm{~V}$
$\mathrm{K}=56$
$\mathrm{i}=\frac{\mathrm{q}}{\rho \mathrm{k} \varepsilon_{0}}=\frac{\mathrm{q}_{0}}{\rho \mathrm{k} \varepsilon_{0}} \mathrm{e}^{-\frac{\mathrm{t}}{\rho \mathrm{k} \varepsilon_{0}}}$
$\mathrm{i}_{\max }=\frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}$
$=\frac{80}{10^{4} \times 8.85}=903 \mu \mathrm{A}=0.9 \mathrm{~mA}$
Option (3)
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