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

ankush singh
30 Points
8 years ago
The answer is kq2/L2 towards the direction of empty vertics.  Because force due to other 4 charges cancel out each other.
Heet Mehta
26 Points
6 years ago
There are five vertices so it must be a Pentagon, four charges cancel each other out and there is one charge left on the remaining vertex so apply the regular formula F = kq2/r2
Ashutosh
12 Points
6 years ago
If there were 6 charges at each corner then the net electric field at the centre would be zero since it is a regular hexagon . the electric field due to 5 charges will be equal and opposite to the net electric field due to sixth charge so as to make the net electric field at the centre to be zero . Hence electric field at the centre due to 5 charges is will be equal to electric field at the centre due to sixth charge. F=(kq^2)/L^2
Ajeet Tiwari
3 years ago
Hello students

Among the five forces on five charges,2pairs(four) of forces are cancelled with each other since they are equal in magnitude and opposite in direction.
The resultant force is due to a single force given below.
F= = k q1q2 /r^2

Hope it helps
Thankyou and Regards