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JEE Main 26 February 2021, Shift 1, Physics Q12

Question 12 of 90 in this shift, Physics question 12 of 30, Section A.

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}\mathrm{L}

The figure, in words

Drawn: a vertical thin rod (drawn as a narrow rectangle) with a double-headed vertical arrow on its left labelled L and the label Q at its lower end. The midpoint of the rod is labelled O (a small right-angle mark at O). A dotted horizontal line runs from O to a point P on the right. A horizontal arrow above it, marked a, spans O to P and ends in an arrowhead labelled E.

(1) $\frac{\mathrm{Q}}{3\pi\varepsilon_0\mathrm{L}^{2}}$
(2) $\frac{\mathrm{Q}}{4\pi\varepsilon_0\mathrm{L}^{2}}$
(3) $\frac{\sqrt{3}\mathrm{Q}}{4\pi\varepsilon_0\mathrm{L}^{2}}$
(4) $\frac{\mathrm{Q}}{2\sqrt{3}\pi\varepsilon_0\mathrm{L}^{2}}$ Official answer

Official answer

Option 4

NTA final key.

Chapter: Electric Charges and Fields
Topic: Electric Field
Idea tested: Finite/Infinite Line Field

JEE Main 26 February 2021, Shift 1, Physics Q12

Same idea in other shifts