Displacement current is given 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:

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

Match List-$I$ with List-$II$ and choose the correct answer from the options given below:
| List-$I$ | List-$II$ |
| :--- | :--- |
| $A. \oint \vec{B} \cdot d\vec{l} = \mu_0 i_c + \mu_0 \varepsilon_0 \frac{d\phi_E}{dt}$ | $I. \text{Gauss' law for electricity}$ |
| $B. \oint \vec{E} \cdot d\vec{l} = -\frac{d\phi_B}{dt}$ | $II. \text{Gauss' law for magnetism}$ |
| $C. \oint \vec{E} \cdot d\vec{A} = \frac{Q}{\varepsilon_0}$ | $III. \text{Faraday law}$ |
| $D. \oint \vec{B} \cdot d\vec{A} = 0$ | $IV. \text{Ampere-Maxwell law}$ |

$A$ parallel-plate capacitor with circular plates is being discharged. The radius of the circular plates is $10 \ cm$. $A$ circular loop of radius $20 \ cm$ is concentric with the capacitor and located halfway between the plates. If the electric field between the plates is changing at the rate of $3.6 \times 10^{12} \ V/(m \cdot s)$,then the displacement current through the loop is (Assume $\frac{1}{4 \pi \varepsilon_0} = 9 \times 10^9 \ N \cdot m^2/C^2$) (in $A$)

Write inferences from Maxwell's equations.