Electrostatics Part 3

Concept:The net electric field is the vector sum of fields from each charge. Charges at diametrically opposite positions cancel if they are of same sign and magnitude.Explanation:Six point charges are placed $60^{\circ}$ apart along a circle of radius $R$.The magnitude of field due to a single charge $Q$ at the centre is $E_0 = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$.Charges at $90^{\circ}$ (downward) and $270^{\circ}$ (upward) are opposite vectors, so they cancel.Charges at $30^{\circ}$ (away from centre) and $210^{\circ}$ (toward centre) also cancel.Only charges at $150^{\circ}$ and $330^{\circ}$ remain, each giving field $E_0$ in the same direction (toward $150^{\circ}$).Net field magnitude is $2E_0$.The direction vector for $150^{\circ}$ is $\cos150^{\circ}\,\hat{i} + \sin150^{\circ}\,\hat{j} = -\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}$.Thus $\vec{E}_{\text{net}} = 2E_0\left(-\frac{\sqrt{3}}{2}\hat{i} + \frac{1}{2}\hat{j}\right) = -E_0(\sqrt{3}\hat{i} - \hat{j})$.Substitute $E_0$: $\vec{E}_{\text{net}} = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}(\sqrt{3}\hat{i} - \hat{j})$.Answer:$-\frac{Q}{4\pi\epsilon_0 R^2}(\sqrt{3}\hat{i} - \hat{j})$     (Option C).