Question

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 $O$ ?

• A. $\frac{3\sqrt{3}Q}{16{\pi }^2{\epsilon }_0{R}^2}\left(\hat{i}\right)$
• B. $\frac{3\sqrt{3}Q}{8{\pi }^2{\epsilon }_0{R}^2}\left(\hat{i}\right)$
• C. $\frac{3\sqrt{3}Q}{8\pi {\epsilon }_0{R}^2}\left(\hat{i}\right)$
• D. $\frac{3\sqrt{3}Q}{8{\pi }^2{\epsilon }_0{R}^2}(-\hat{i})$

Answer 1

The correct option is B $\frac{3\sqrt{3}Q}{8{\pi }^2{\epsilon }_0{R}^2}\left(\hat{i}\right)$

Magnitude of electric field generated at centre of the arc subtending an angle $\theta$ at its centre is
$E=\frac{2k|\lambda |}{R}\sin \left(\frac{\theta }{2}\right)$
As charge is negative in nature, direction of electric field is towards the arc i.e. towards $\hat{i}$

Given that $\theta ={120}^{\circ }$
$|\lambda |=\frac{Q}{R\theta }$
On substitution,
$\overrightarrow{E}=\frac{3\sqrt{3}Q}{8R^2{\pi }^2{ϵ}_0}\hat{i}$