An accelerated electron produces ...

$A$ parallel plate capacitor consists of two circular plates each of radius $2 \,cm$, separated by a distance of $0.1 \,mm$. If the potential difference across the plates is varying at the rate of $5 \times 10^6 \,Vs^{-1}$, then the value of displacement current is

$A$ magnetic field can be produced by

Match List-$I$ with List-$II$:

List-$I$	List-$II$
$A$. Gauss's Law in Electrostatics	$I$. $\oint \vec{E} \cdot d \vec{l} = -\frac{d \phi_B}{d t}$
$B$. Faraday's Law	$II$. $\oint \vec{B} \cdot d \vec{A} = 0$
$C$. Gauss's Law in Magnetism	$III$. $\oint \vec{B} \cdot d \vec{l} = \mu_0 i_C + \mu_0 \epsilon_0 \frac{d \phi_E}{d t}$
$D$. Ampere-Maxwell Law	$IV$. $\oint \vec{E} \cdot d \vec{s} = \frac{q}{\epsilon_0}$

Choose the correct answer from the options given below:

"Changing electric field produces magnetic field". Explain the importance of the given statement.

Consider a parallel plate capacitor which is maintained at a potential of $200 \, V$. The separation distance between the plates of the capacitor and the area of the plates are $1 \, mm$ and $20 \, cm^2$, respectively. Calculate the displacement current in $1 \, \mu s$. (in $ \, mA$)