Why does Maxwell law consider displacement current and actual current to be same?

The aim of using displacement current was to allow currrent ot flow inspite of a change in the field. This was not possible in the case of actual current since it lead to some complex situations, due to which scientists had to find a new solution for generating an electric field in such cases.

The key difference between these two terms is that displacement current is deprived of electrons which flow in the case of an actual current. It's role is to help generate magnetic field which is produced when the medium changes between two different bodies.

According to definition of displacement current, we can write

$I_d={\epsilon }_0\frac{d?}{dt}={\epsilon }_0\frac{d\left(ES\right)}{dt}={\epsilon }_0\frac{d\left(\frac{V}{\mathcal{l}}S\right)}{dt}=\frac{{\epsilon }_{0}S}{\mathcal{l}}\left(\frac{dV}{dt}\right)$  

$\Rightarrow \mathcal{l}=\frac{8.85\times 10^{-12}\times 40\times 10^{-4}\times 10^6}{4.425\times 10^{-6}}8\times 10^{-3}m$  

$\mathrm{q}=\mathrm{C}\mathrm{V}$

^{$\frac{\mathrm{d}\mathrm{q}}{\mathrm{d}\mathrm{t}}=\frac{\mathrm{C}\mathrm{d}\mathrm{V}}{\mathrm{d}\mathrm{t}}$}

$I_d=C\left(V_0\omega \mathrm{c}\mathrm{o}\mathrm{s}?\omega t\right)$

$={\mathrm{V}}_0\omega \mathrm{C}\mathrm{o}\mathrm{s}\omega \mathrm{t}$