# The material filled between the plates of a parallel plate capacitor has resistivity

Question:

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 )

1. $9.0 \mu \mathrm{A}$

2. $9.0 \mathrm{~mA}$

3. $0.9 \mathrm{~mA}$

4. $0.9 \mu \mathrm{A}$

Correct Option: , 3

Solution:

$\rho=200 \Omega \mathrm{m}$

$\mathrm{C}=2 \times 10^{-12} \mathrm{~F}$

$\mathrm{V}=40 \mathrm{~V}$

$\mathrm{K}=56$

$\mathrm{i}=\frac{\mathrm{q}}{\rho \mathrm{k} \varepsilon_{0}}=\frac{\mathrm{q}_{0}}{\rho \mathrm{k} \varepsilon_{0}} \mathrm{e}^{-\frac{\mathrm{t}}{\rho \mathrm{k} \varepsilon_{0}}}$

$\mathrm{i}_{\max }=\frac{2 \times 10^{-12} \times 40}{200 \times 50 \times 8.85 \times 10^{-12}}$

$=\frac{80}{10^{4} \times 8.85}=903 \mu \mathrm{A}=0.9 \mathrm{~mA}$

Option (3)