Question
A point charge is moving along a straight line with a constant velocity u. Consider a small area A perpendicular to the direction of motion of the charge (El). Calculate the displacement current through the area when its distance from the charge is x. The value of x is not large so that the electric field at any instant is essentially given by Coulomb's law.

Answer

$\text{E}=\frac{\text{Kq}}{\text{x}^2}$ [from coulomb's law]$\phi_\text{E}=\text{EA}=\frac{\text{KqA}}{\text{x}^2}$
$\text{l}_\text{d}=\epsilon_0\frac{\text{d}\phi\text{E}}{\text{dt}}=\epsilon_0\frac{\text{d}}{\text{dt}}\frac{\text{kqA}}{\text{x}^2}=\epsilon_0\text{KqA}=\frac{\text{d}}{\text{dt}}\text{x}^{-2}$
$=\epsilon_0\times\frac{1}{4\pi\epsilon_0}\times\text{q}\times\text{A}\times-2\times\text{x}^{-3}\times\frac{\text{dx}}{\text{dt}}=\frac{\text{qAv}}{2\pi\text{x}^3}$

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