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JEE Main 2021 (26 Feb, Shift 1) · Electric Field & Field Lines

Electric Field & Field Lines question from JEE Main 2021 (26 Feb, Shift 1)

Q34JEE Main 2021 · 26 FEB · SHIFT 1Medium

Electric Field & Field Lines

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 wire is a = $\frac{\sqrt{3}}{2}L$ .

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Answer: (c)

On the perpendicular bisector of a rod: $E = \dfrac{kQ}{a\sqrt{a^2 + L^2/4}}$ .
With $a = \dfrac{\sqrt3}{2}L$ : $a^2 + \dfrac{L^2}{4} = L^2$ , so $E = \dfrac{kQ}{(\sqrt3 L/2)L}$ .
$E = \dfrac{2Q}{4\pi\varepsilon_0\sqrt3 L^2} = \dfrac{Q}{2\sqrt3\pi\varepsilon_0 L^2}$
So the answer is (c).

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