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  • Figure shows a rod , which is bent in a circular arc of radius

Figure shows a rod \mathrm{AB}, which is bent in a 120^{\circ} circular arc of radius \mathrm{R}$. A charge $(-\mathrm{Q}) is uniformly distributed over rod A B. What is the electric field \vec{E} at the centre of curvature \mathrm{O} ?
Option: 1 \begin{aligned} &\frac{3 \sqrt{3} \mathrm{Q}}{8 \pi \varepsilon_{0} \mathrm{R}^{2}}(\hat{i}) \\ \end{aligned}
Option: 2 \frac{3 \sqrt{3} \mathrm{Q}}{8 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}(\hat{i}) \\
Option: 3 \frac{3 \sqrt{3} \mathrm{Q}}{8 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}(-\hat{i}) \\
Option: 4 \frac{3 \sqrt{3} \mathrm{Q}}{16 \pi^{2} \varepsilon_{0} \mathrm{R}^{2}}(\hat{i})

Answers (1)

best_answer

Electric field due to circular arc is

\bar{E}= \frac{\lambda }{2\pi \in_{0}r}\sin\left ( \frac{\theta}{2} \right )\left ( +\hat{i} \right )   ( By symmetry )


Inward field for negative charge distribution

\bar{E}= \frac{(Q)/l}{2\pi \in_{0}r}\times \sin\left ( \frac{120^{\circ}}{2} \right )\left ( \hat{i} \right )

l= r\theta= \frac{2\pi}{3}\times R

\bar{E}= \frac{(Q)/\frac{2\pi}{3}R}{2\pi \in_{0}R}\times \left ( \frac{\sqrt{3}}{2} \right )\left ( \hat{i} \right )

\bar{E}= \frac{3\sqrt{3}Q}{8\pi ^{2}\in_{0}R^{2}}\left ( \hat{i} \right )

Posted by

vishal kumar

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