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 $\overrightarrow E $ at the centre of curvature O ?

Question

${{3\sqrt 3 Q} \over {8\pi {\varepsilon _0}{R^2}}}(\widehat i)$

${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$

${{3\sqrt 3 Q} \over {16{\pi ^2}{\varepsilon _0}{R^2}}}(\widehat i)$

${{3\sqrt 3 Q} \over {8{\pi ^2}{\varepsilon _0}{R^2}}}( - \widehat i)$

Solution

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