Consider three charges `q_(1), q_(2), q_(3)` each equal to q at the vertices of an equilateral triangle of side l. What is the force on a charge Q (with the same sign as q) placed at the centroid of the triangle, as shown in Fig. ?

Explanation

In the given equilateral triangle ABC of sides of length `l`, if we draw a perpendicular AD to the side BC,
`AD =AC cos 30^(@)=(sqrt(3)/2)l` and the distance AO of the centroid O from A is
`(2//3) AD=(1/sqrt(3))l`.
By symmetry `AO=BO=CO`.
Thus,
Force `F_(1)` on Q due to charge q at `A=3/(4pi epsi_(0)) (Qq)/l^(2)` along AO
Force `F_(2)` on Q due to charge q at `B =3/(4pi epsi_(0)) (Qq)/l^(2)` along BO
Force `F_(3)` on Q due to charge q at `C=3/(4pi epsi_(0)) (Qq)/l^(2)` along CO
The resultant of forces `F_(2)` and `F_(3)` is `3/(4pi epsi_(0)) (Q q)/l^(2)` along OA, by the parallelogram law. therefore, the total force on `Q=3/(4pi epsi_(0)) (Q q)/l^(2) (hat(r)-hat(r))=0`, where `hat(r)` is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero. suppose that the resultant force was non-zero but in some direction. Consider what would happen if the system was rotated through `60^(@)` about O.