Question

Five point charges ($+q$ each) are placed at the five vertices of a regular hexagon of side $2a$. What is the magnitude of the net electric field at the centre of the hexagon?

• A. $\frac{1}{4\pi {\epsilon }_0}\frac{q}{a^2}$
• B. $\frac{q}{16\pi {\epsilon }_0{a}^2}$
• C. $\frac{\sqrt{2}q}{4\pi {\epsilon }_0{a}^2}$
• D. $\frac{5q}{16\pi {\epsilon }_0{a}^2}$

Answer 1

Answer: (B)

$\frac{q}{16\pi {\epsilon }_0{a}^2}$

Electric field due $q_1$, $q_2$, $q_4$ and $q_5$ cancel each other and only field due to $q_3$ exists.

As the side of the hexagon is $2a$ we get the distance form $q_3$ to the center as $2a$ only.

Thus we get the electric field as $\frac{kq}{4a^2}$