Figure shows a rod AB, which is bent in a 120^ocircular 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?

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$\frac{3 \sqrt{3} Q}{8 π ε_{0} R^{2}} (\hat{i})$

$\frac{3 \sqrt{3} Q}{16 π^{2} ε_{0} R^{2}} (\hat{i})$

$\frac{3 \sqrt{3} Q}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

$\frac{3 \sqrt{3} Q}{8 π^{2} ε_{0} R^{2}} (- \hat{i})$

answer is B.

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Detailed Solution

Charge per unit length of the rod $λ = (\frac{- Q}{Rθ}) = (\frac{- Q}{R \frac{2 π}{3}}) = \frac{- 3 Q}{2 πR}$
$E = \int dEcos θ \Rightarrow E = \int_{- \frac{π}{3}}^{\frac{π}{3}} \frac{k \times (Q)}{\frac{2 π}{3} R} \times \frac{Rdθ}{R^{2}} \cos θ$
$\Rightarrow E = \frac{3}{2 π} \frac{kQ}{R^{2}} [\sin θ]_{- \frac{π}{3}}^{\frac{π}{3}} = \frac{3}{2 π} \frac{kQ}{R} \times \frac{2 \sqrt{3}}{2}$
$\Rightarrow E = \frac{3 \sqrt{3} Q}{8 π^{2} \in_{0} R^{2}} (+ \hat{i})$