The source of displacement current is . . . . . . .

What contradiction is found by using Ampere's circuital law to obtain the magnetic field during the charging of a capacitor?

Match the following Maxwell’s equations: (The symbols used here have their usual meanings)

List-$I$	List-$II$
$(a)$ Gauss’ law for electrostatics	$(i)$ $\oint \vec{E} \cdot d\vec{A} = \frac{Q}{\epsilon_0}$
$(b)$ Gauss’ law for magnetism	(ii) $\oint \vec{B} \cdot d\vec{l} = \mu_0 [i_c + \epsilon_0 \frac{d\phi_E}{dt}]$
$(c)$ Faraday’s law	(iii) $\oint \vec{B} \cdot d\vec{A} = 0$
$(d)$ Ampere-Maxwell’s law	(iv) $\oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$

The source of an electromagnetic wave can be a charge

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:

$A$ parallel plate capacitor is charged to $60\,\mu C$. Due to a radioactive source,the plate loses charge at the rate of $1.8 \times 10^{-8}\, C/s$. The magnitude of the displacement current is: