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JEE Main 27 August 2021, Shift 2, Physics Q10

Question 10 of 90 in this shift, Physics question 10 of 30, Section A.

Figure shows a rod AB, which is bent in a

120^\circ

circular arc of radius R. A charge

(-Q)

is uniformly distributed over rod AB. What is the electric field

\vec{E}

at the centre of curvature O ?

The figure, in words

Axes y (up) and x (right) with origin O. A circular arc AB of radius R, centred at O, spans 120 degrees, with A above the x-axis and B below it; dotted lines from O to A and O to B each make 60 degrees with the x-axis, labelled 60, 60 and 120 degrees. R is marked along the line from O to B.

(1) $\frac{3\sqrt{3}\,Q}{8\pi^2\varepsilon_0R^2}\,(\hat{i})$ Official answer
(2) $\frac{3\sqrt{3}\,Q}{8\pi^2\varepsilon_0R^2}\,(-\hat{i})$
(3) $\frac{3\sqrt{3}\,Q}{8\pi\varepsilon_0R^2}\,(\hat{i})$
(4) $\frac{3\sqrt{3}\,Q}{16\pi^2\varepsilon_0R^2}\,(\hat{i})$

Official answer

Option 1

NTA final key.

Chapter: Electric Charges and Fields
Topic: Electric Field
Idea tested: Charged Arc Field

JEE Main 27 August 2021, Shift 2, Physics Q10

Same idea in other shifts