Electrostatics Part 3

Concept:The net electric field at the centre is the vector sum of the fields due to all six charges.A positive charge produces a field directed away from it, while a negative charge produces a field directed toward it.Explanation:From the figure, the charges are placed at $30^\circ,90^\circ,150^\circ,210^\circ,270^\circ,330^\circ$.The charges at $30^\circ,90^\circ,210^\circ,270^\circ,330^\circ$ are $+Q$, and the charge at $150^\circ$ is $-Q$.The fields due to the $+Q$ charges at $90^\circ$ and $270^\circ$ are equal and opposite, so they cancel.The fields due to the $+Q$ charges at $30^\circ$ and $210^\circ$ are also equal and opposite, so they cancel.Thus only the $+Q$ charge at $330^\circ$ and the $-Q$ charge at $150^\circ$ contribute to the net field.The field from the $+Q$ charge at $330^\circ$ points toward the centre, i.e. in the direction $150^\circ$.The field from the $-Q$ charge at $150^\circ$ points toward that charge, also in the direction $150^\circ$.Each of these fields has magnitude $E_1=\frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$.Hence the net field has magnitude $2E_1$ and is directed along $150^\circ$.The unit vector along $150^\circ$ is $\cos150^\circ\,\hat{i}+\sin150^\circ\,\hat{j}=-\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}$.Therefore, $\vec{E}_{\text{net}}=2E_1\left(-\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}\right)=-\frac{Q}{4\pi\epsilon_0 R^2}\left(\sqrt{3}\hat{i}-\hat{j}\right)$.Using the option notation where the permittivity is written as $E_0$, this is $-\frac{Q}{4\pi E_0 R^2}\left(\sqrt{3}\hat{i}-\hat{j}\right)$.Answer:$-\frac{Q}{4\pi E_0 R^2}\left(\sqrt{3}\hat{i}-\hat{j}\right)$   (Option C / option C).