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)
Solution
$\rho$ = 200 $\Omega$m<br><br>C = 2 $\times$ 10<sup>$-$12</sup> F<br><br>V = 40 V<br><br>K = 56<br><br>$$i = {q \over {\rho k{\varepsilon _0}}} = {{{q_0}} \over {\rho k{\varepsilon _0}}}{e^{ - {t \over {\rho k{\varepsilon _0}}}}}$$<br><br>$${i_{\max }} = {{2 \times {{10}^{ - 12}} \times 40} \over {200 \times 50 \times 8.85 \times {{10}^{ - 12}}}}$$<br><br>$= {{80} \over {{{10}^4} \times 8.85}}$ = 903 $\mu$A = 0.9 mA
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Coulomb's Law
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