JEE Mains · Physics · STD 12 - Electric charges and fields
- JEE Mains 2021
- Hard
- MCQ
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 \({O}\) ?
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(\frac{3 \sqrt{3} Q}{8 \pi^{2} \varepsilon_{0} R^{2}}(\hat{i})\)
Step-by-step Solution
Detailed explanation
\(\varepsilon=\frac{2 {k} \lambda}{{R}} \sin \left(\frac{\theta}{2}\right)(-\hat{{i}})\)
\(\lambda=\left(\frac{-{Q}}{{R} \theta}\right)=\left(\frac{-{Q}}{{R} \cdot \frac{2 \pi}{3}}\right)\)
\(\lambda=\frac{-3 Q}{2 \pi R}\)
\(\varepsilon=\frac{2 k}{R} \cdot \frac{-3 Q}{2 \pi R} \cdot \sin \left(60^{\circ}\right)(-\hat{i})\)
\(\varepsilon=\frac{3 \sqrt{3} Q}{8 \pi^{2} \in_{0} R^{2}}(+\hat{i})\)
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