Question

Five point charges, each of value $+q$ are placed on five vertices of a regular hexagon of side $L$. What is the magnitude of the force on a point charge of value $-q$ placed at the centre of the hexagon?

Answer 1

Since, all the five charges (+q) are placed at the vertices of the hexagon. So the forces due to opposite charges will cancel out each other and the net force is due to only one +q charge.

So the net force is

$F_{net}=\frac{1}{4\pi {\epsilon }_0}\frac{q\left(q\right)}{L^2}$ (in hexagon equilateral triangle is formed so the r = L)