Question

A point charge is moving along a straight line with a constant velocity v. Consider a small area A perpendicular to the motion of the charge. 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 1

From Coulomb's law:
Electric field strength,
$E\mathit{=}\mathit{}\frac{kq}{x^{\mathit{2}}}$
Electric flux,
$$\begin{gathered} {\varphi }_{\mathrm{E}}=EA \\ {\varphi }_{\mathrm{E}}=\frac{kqA}{x^2} \end{gathered}$$
Displacement current = I_d
$$\begin{gathered} I_{\mathrm{d}}=\left|{\in }_0\frac{d{\varphi }_{\mathrm{E}}}{dt}\right| \\ {I}_{\mathrm{d}}=\left|{\in }_0\frac{d}{dt}\left(\frac{kqA}{x^2}\right)\right| \\ {I}_{\mathrm{d}}=\left|{\in }_{0}kqA\frac{d}{dt}{x}^{-2}\right| \\ {I}_{\mathrm{d}}=\left|{\mathit{\in }}_{\mathit{0}}\frac{\mathit{1}}{\mathit{4}\mathit{\pi }{\mathit{\in }}_{\mathit{0}}}\mathit{\times }\mathit{q}\mathit{\times }\mathit{A}\mathit{\times }\mathit{(}\mathit{-}\mathit{2}\mathit{)}{\mathit{x}}^{\mathit{-}\mathit{3}}\mathit{\times }\frac{\mathit{d}\mathit{x}}{\mathit{d}\mathit{t}}\right| \\ {I}_{\mathrm{d}}=\left|\frac{qAv}{2\mathrm{\pi }{x}^3}\right| \end{gathered}$$