JEE Main · 2023 · 10 Apr 2023 · Shift 2 · Electrostatics

The electric field of an infinite line charge at perpendicular distance $r$ is $E=\frac{2k\lambda }{r}$. The electron moves in a circle, so the attractive electrostatic force supplies the centripetal force, $eE=\frac{m{v}^2}{r}$, which gives $e\cdot \frac{2k\lambda }{r}=\frac{m{v}^2}{r}$. The radius cancels on both sides, leaving $v=\sqrt{\frac{2ek\lambda }{m}}$. Substituting the data, $v=\sqrt{\frac{2\times 1.6\times 10^{-19}\times 9\times 10^9\times 2\times 10^{-8}}{9\times 10^{-31}}}=\sqrt{6.4\times 10^{13}}$. The electron revolves with a speed of $8\times 10^6{\text{m s}}^{-1}$.