Electric Field & Field Lines question from JEE Main 2021 (27 Aug, Shift 2)

Answer: (b)

1. Arc of angle $2\theta$: $E = \dfrac{2k\lambda\sin\theta}{R}$; here $2\theta = 120^\circ$, $\theta = 60^\circ$.
2. $\lambda = \dfrac{Q}{R(2\pi/3)} = \dfrac{3Q}{2\pi R}$
3. $E = 2\cdot\dfrac{1}{4\pi\varepsilon_0}\cdot\dfrac{3Q}{2\pi R}\cdot\dfrac{\sqrt3}{2}\cdot\dfrac{1}{R} = \dfrac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}$
4. The charge is negative, so the field at O points towards the arc, i.e. along $+\hat i$.
5. So the answer is (b) $\dfrac{3\sqrt3 Q}{8\pi^2\varepsilon_0 R^2}\hat i$.