Electric Field & Field Lines question from JEE Main 2026 (22 Jan, Shift 1)

Answer: (c)

1. From the figure: $+Q$ at $90^\circ, 30^\circ, -30^\circ, -90^\circ, 210^\circ$ and $-Q$ at $150^\circ$; let $E_0 = \dfrac{Q}{4\pi\epsilon_0 R^2}$.
2. Equal charges at opposite points cancel: $90^\circ$ with $-90^\circ$, and $30^\circ$ with $210^\circ$.
3. $+Q$ at $-30^\circ$: field points away from it, $E_0\left(-\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$
4. $-Q$ at $150^\circ$: field points towards it, $E_0\left(-\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right)$
5. $\vec{E} = E_0(-\sqrt{3}\hat{i} + \hat{j}) = -\dfrac{Q}{4\pi\epsilon_0 R^2}(\sqrt{3}\hat{i} - \hat{j})$
6. So the answer is (c).