$A$ quantity $X$ is given by $\varepsilon_0 L \frac{\Delta V}{\Delta t}$, where $\varepsilon_0$ is the permittivity of free space, $L$ is the length, $\Delta V$ is a potential difference, and $\Delta t$ is a time interval. The dimension of $X$ is the same as that of

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 potential difference between the two plates of a parallel plate capacitor changes at a rate of $10^6 \; V/s$. If the capacitance of the capacitor is $2 \; \mu F$, what is the displacement current in the dielectric of the capacitor?

Suppose that the electric flux inside a parallel plate capacitor changes at a rate of $7 \times 10^{14} \text{ V} \cdot \text{m/s}$. If the area of the plates is $1 \text{ m}^2$,calculate the magnetic field $B$ at a distance $r = 0.1 \text{ m}$ from the axis of the capacitor. (Given: $\varepsilon_0 = 8.85 \times 10^{-12} \text{ F/m}$,$\mu_0 = 4\pi \times 10^{-7} \text{ T} \cdot \text{m/A}$)

$A$ parallel plate capacitor with plate area $A$ and plate separation $d$ is charged by a constant current $I$. $A$ plane surface of area $A/2$,parallel to the plates,is drawn between the plates. The displacement current through this area is

An accelerated electric charge emits