Question

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 permitivity of material is $50$ )

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

Answer 1

Answer: (C)

$\rho =200\Omega m$
$C=2\times 10^{-12}F$
$V=40V$
$K=56$
$i=\frac{q}{\rho k{\epsilon }_0}=\frac{q_0}{\rho k{\epsilon }_0}{e}^{-\frac{t}{\rho k{\epsilon }_0}}$
$i_{\max }=\frac{2\times 10^{-12}\times 40}{200\times 50\times 8.85\times 10^{-12}}$
$=\frac{80}{10^4\times 8.85}=903\mu A=0.9mA$