JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Electrostatics

Let $E_0=\frac{Q}{4\pi {\epsilon }_0{R}^2}$ be the magnitude of the field each charge produces at the centre; a positive charge gives a field pointing from itself towards the centre, i.e. along $-{\hat{r}}_i$, and a negative charge along $+{\hat{r}}_i$. Here five charges are $+Q$ and the charge at $150^{\circ }$ is $-Q$. Since the six unit vectors to equally spaced points sum to zero, the contribution of the five positive charges is $-E_0{\sum }_{+}{\hat{r}}_i=E_0{\hat{r}}_{150^{\circ }}$, and the lone $-Q$ adds a further $E_0{\hat{r}}_{150^{\circ }}$. The centre field is therefore $\vec{E}=2E_0{\hat{r}}_{150^{\circ }}=2E_0\left(-\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)=-\frac{Q}{4\pi {\epsilon }_0{R}^2}(\sqrt{3}\hat{i}-\hat{j}).$