Question

The material filled between the plates of a parallel plate capacitor has resistivity $200\Omega \text{m}$. The value of capacitance of the capacitor is $2pF$. If a potential difference of $40\text{V}$ 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 $(k=50)$$]$

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

Answer 1

The correct option is C $0.9\text{mA}$

Given, resistivity of the filled material
$\rho =200\Omega \text{m}$
Capacitance of capacitor $C=2pF=2\times {10}^{-12}F$
$k=50$
Potential difference applied across the capacitor, $V=40\text{V}$
The setup can be shown below as
Here, $d$ is the length of the plate and $A$ is the surface area of the plate.
So, the equivalent resistance is given by
$R=\frac{\rho d}{A}$
Also, we know that
$C=\frac{k{ϵ}_{0}A}{d}$
or, $\frac{d}{A}=\frac{k{ϵ}_0}{C}=\frac{50\times (8.85\times {10}^{-12})}{2\times {10}^{-12}}\Rightarrow \frac{d}{A}=221.25{\text{m}}^{-1}$
So, we have
$R=\frac{\rho d}{A}=200\times 221.25\Omega =44250\Omega$
Thus, the leakage current is given as
$i_l=\frac{40}{R}=\frac{40}{44250}=0.9\text{mA}$
Hence, option (b) is correct.