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.

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

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

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

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

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

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

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