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Q.

Figure shows a rod AB, which is bent in a $120^{0}$ circular arc of radius R. A charge $(- Q)$ is uniformly distributed over rod AB. What is the electric field $\bar{E}$ at the centre of curvature O?

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a

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

b

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

c

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

d

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

answer is B.

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Detailed Solution

$E = \int d E \cos θ = \int_{\frac{- π}{3}}^{\frac{π}{3}} \frac{K \times (+ Q)}{\frac{2 π}{3} R} \times \frac{R d θ}{R^{2}} \times \cos θ = + \frac{3}{2 π} \frac{K Q}{R^{2}} {[\sin θ]}_{\frac{- π}{3}}^{\frac{π}{3}} ∴ \bar{E} = \frac{3 \sqrt{3}}{8 π^{2} ε_{0} R^{2}} (\hat{i})$

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