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

Answer 1

Answer: (B)

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