At the vertices of a regular hexagon, the sides of which are equal to a, the charges + q

At the vertices of a regular hexagon, the sides of which are equal to a, the charges + q, + q, + q, -q, -q, -q were placed in succession. Determine the force acting on the charge + q located in the center of the hexagon

The absolute value of the force f with which each of the charges acts on the central charge q:

f = (kq ^ 2) / a ^ 2;

The two extreme charges from the group of positive charges act at an angle of 120 degrees to each other. If we combine the beginning of the vector f2 with the end of the vector f1, then they, together with the resultant, form an equilateral triangle. Conclusion: the resultant forces f1 and f2 in absolute value is equal to the force of action of one charge f.

f1 + f2 = f;

f3 = f.

The resulting group of three charges:

f1 + f2 + f3 = 2f.

A group of negative charges acts with the same force 2f in the same direction. The resulting force F, with which all charges act:

F = 2f + 2f = 4f.

Answer: F = (4k * q ^ 2) / a ^ 2.