Question

A uniform line charge is moving with a constant velocity $v=10m/s$ along its axis. It produces an electrostatic field of $9\times {10}^{4}V/m$ at a perpendicular distance of $1m$ from its axis. What will be the current (in $\mu A$) due to movement of line charge? Assume that Gauss' Law holds true.

Answer 1

Electric field due to a line charge having linear charge density $\lambda$ at a point at perpendicular distance $r$ from the charge,
$E=\frac{2k\lambda }{r}$
where $k=\frac{1}{4\pi {ϵ}_o}=9\times {10}^{9}N{m}^2{C}^{-2}$
$9\times {10}^4=\frac{2k\lambda }{r}$
$\Rightarrow \lambda =\frac{9\times {10}^4}{2\times 9\times {10}^9}=\frac{1}{2}\times {10}^{-5}C/m$
Since current is rate of flow of charge,
Current, $i=\frac{dq}{dt}=\frac{dq}{dx}\frac{dx}{dt}$
where $\frac{dq}{dx}=\lambda$ and $\frac{dx}{dt}=v=10m/s$
$i=\frac{1}{2}\times {10}^{-5}\times 10=\frac{1}{2}\times {10}^{-4}A=50\mu A$