Five point charge (`+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 hexazon?

Options

A. `(1)/(4 pi epsi_(0)) (q)/(a^(2))`
B. `(q)/(16 pi epsi_(0)a^(2))`
C. `(sqrt(2) q)/(4 pi epsi_(0) a^(2))`
D. `(5q)/(16 pi epsi_(0) a^(2))`

Answer

B

Explanation

If 6th charge (+q) is placed at the sixth corner then the net electric field at the centre of the hexagon becomes zero `implies` field due to the five charges is equal and opposite of the field due to the sixth change.
`implies E=q/(4 pi e_0(2a)^2) =q/(16 pi e_0 a^2)`