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 not large so that the electric field at any instant is essentially given by Coulomb's law.

• A. $-\frac{qAv}{4\pi x^3}$
• B. $\frac{qAv}{2x^3}$
• C. $-\frac{qAv}{2\pi x^3}$
• D. $\frac{qAv}{4x^3}$

Answer 1

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

Displacement current$\left(i_d\right)$
$i_d={\epsilon }_0\frac{d{\varphi }_E}{dt}\text{Now we know that}{\varphi }_E=E.A$
Since
$E=\frac{kq}{x^2}$
$\Rightarrow {\varphi }_E=\frac{kq}{x^2}A=kqA{x}^{-2}\Rightarrow \frac{d{\varphi }_E}{dt}=-2kqAv{x}^{-3}\text{Multiplying both side by}{\epsilon }_0\Rightarrow {\epsilon }_0\frac{d{\varphi }_E}{dt}=-2{\epsilon }_0\frac{1}{4\pi {\epsilon }_0}qAv{x}^{-3}\Rightarrow i_d=-\frac{qAv}{2\pi x^3}$