Question

The potential difference between the plates of a parallel plate capacitor is changing at the rate of $10^{6}V/s$. If the capacitance is $2\mu F$, the displacement current in the dielectric of the capacitor will be :

• A. $1A$
• B. $2A$
• C. $3A$
• D. $4A$

Answer 1

Answer: (B)

Displacement current is given by
$i_d={\epsilon }_0\frac{\partial {\varphi }_E}{\partial t}$
where ${\varphi }_E$ is the electric flux given as ${\varphi }_E=$ EA where $A$ is area of the capacitor plates and $E$ is the electric field between the plates.
So, $i_d={\epsilon }_0\frac{\partial (EA)}{\partial t}={\epsilon }_{0}A\frac{\partial E}{\partial t}$
But $E=V/d$ where $V$ is the potential difference between the capacitor plates separated at d distant apart
So, $i_d=\frac{{\epsilon }_{0}A}{d}\frac{\partial V}{\partial t}=C\frac{\partial V}{\partial t}$
where $C$ is the capacity of the capacitor.
$\therefore i_d=2\times 10^{-6}\times 10^6=2A$