Q 12-01-124JEE MainJEE Main 2021 (27 Aug, Shift 2)Medium
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 $\vec{E}$ at the centre of curvature $O$?
Answer: (A) $\frac{3\sqrt{3}Q}{8\pi^2\varepsilon_0R^2}\hat{i}$
Linear charge density: $\lambda = \dfrac{Q}{R\cdot\frac{2\pi}{3}} = \dfrac{3Q}{2\pi R}$ (magnitude).
For an arc subtending $2\theta_0$ at the centre, the vertical components cancel and
$$E = \frac{2k\lambda\sin\theta_0}{R} = \frac{2}{4\pi\varepsilon_0}\cdot\frac{3Q}{2\pi R}\cdot\frac{\sin60^\circ}{R} = \frac{3\sqrt{3}Q}{8\pi^2\varepsilon_0R^2}$$
The charge is negative, so the field at $O$ points towards the arc, i.e. along $+\hat{i}$.
Solution by Sreeraj P, M.Sc Physics