Figure shows a rod {AB}, which is bent in a 1 | vedclass

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Question

$\varepsilon=\frac{2 {k} \lambda}{{R}} \sin \left(\frac{	heta}{2}ight)(-\hat{{i}})$

$\lambda=\left(\frac{-{Q}}{{R} 	heta}ight)=\left(\frac{-{

Vedclass · Accepted Answer

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

Vedclass · Answer

$\varepsilon=\frac{2 {k} \lambda}{{R}} \sin \left(\frac{	heta}{2}ight)(-\hat{{i}})$

$\lambda=\left(\frac{-{Q}}{{R} 	heta}ight)=\left(\frac{-{Q}}{{R} \cdot \frac{2 \pi}{3}}ight)$

$\lambda=\frac{-3 Q}{2 \pi R}$

$\varepsilon=\frac{2 k}{R} \cdot \frac{-3 Q}{2 \pi R} \cdot \sin \left(60^{\circ}ight)(-\hat{i})$

$\varepsilon=\frac{3 \sqrt{3} Q}{8 \pi^{2} \in_{0} R^{2}}(+\hat{i})$

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 ${O}$ ?

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