Question

A point charge is moving along a straight line with a constant velocity $v$. Consider a small area $A$ perpendicular to the direction of motion of the charge. Calculate the displacement current through the area when its distance from the charge is $x$. The value of $x$ is too large so that the electric field at any instant is essentially given by Coulomb's law.

• A. $\frac{qAv}{2\pi x^3}$
• B. $\frac{qAv}{\pi x^3}$
• C. $\frac{2qAv}{\pi x^3}$
• D. $\frac{qAv}{x^3}$

Answer 1

The correct option is A $\frac{qAv}{2\pi x^3}$

The electric field due to the point charge $q$ at a distance $x$ is given by,

$E=\frac{Kq}{x^2}$

Since, the area $A$ is small we can assume that the electric field is constant and perpendicular to the plane of area $A$ at every point.

Therefore, the electric flux due to this field is given by,

${\varphi }_E=EA=\frac{KqA}{x^2}$

Hence, the displacement current through area $A$ is given by

$I_d={\epsilon }_0\frac{d{\varphi }_E}{dt}={\epsilon }_0\frac{d}{dt}\left(\frac{KqA}{x^2}\right)$

$={\epsilon }_{0}KqA\frac{d{x}^{-2}}{dt}$

$={\epsilon }_0\times \frac{1}{4\pi {\epsilon }_0}\times q\times A\times (-2x^{-3})\frac{dx}{dt}[\because K=\frac{1}{4\pi {\epsilon }_0}]$

$=\frac{-qAv}{2\pi x^3}$ $[\because v=\frac{dx}{dt}]$

$\therefore |I_d|=\frac{qAv}{2\pi x^3}$

Hence, option $\left(A\right)$ is the correct answer.