What is the difference between actual current and displacement current?

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.

Maxwell simply wanted to generate a magnetic field using electric current, which displacement current had the capability to offer. Since both current flows were able to produce an electric field, both of them could be simultaneously used for using in the equation.

According to definition of displacement current, we can write

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

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

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

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

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

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

Velocity in medium v