Q 12-01-118JEE MainJEE Main 2021 (26 Feb, Shift 1)Medium
Find the electric field at point $P$ (as shown in figure) on the perpendicular bisector of a uniformly charged thin wire of length $L$ carrying a charge $Q$. The distance of the point $P$ from the centre of the rod is $a = \frac{\sqrt{3}}{2}L$
Answer: (B) $\frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$
For a uniformly charged rod of length $L$, the field on its perpendicular bisector at distance $a$ is
$$E = \frac{1}{4\pi\varepsilon_0}\frac{Q}{a\sqrt{a^2 + L^2/4}}$$
With $a = \frac{\sqrt{3}}{2}L$: $a^2 + \frac{L^2}{4} = L^2$.
$$E = \frac{Q}{4\pi\varepsilon_0\cdot\frac{\sqrt{3}}{2}L\cdot L} = \frac{Q}{2\sqrt{3}\pi\varepsilon_0 L^2}$$
Solution by Sreeraj P, M.Sc Physics