Consider the charge q, q and -q placed at the vertices of an equilateral triangle, as shown in fig. What is the force on each charge?

Explanation

The forces acting on charge q at A due to charges q at B and -q at C are `F_(12)` along BA and `F_(13)` along AC respectively, as shown in Fig. By the parallelogram law, the total force `F_(1)` on the charge q at A is given by
`F_(1)=F hat(r)_(1)` where `hat(r)_(1)` is a unit vector along BC.
The force of attraction or repulsion for each pair of chrages has the same magnitude
`F=q^(2)/(4pi epsi_(0) l^(2))`
The total force `F_(2)` on charge q at B is thus `F_(2)=F hat(r)_(2)`, where `hat(r)_(2)` is a unit vector along AC.
similarly the total force on charge -q at C is `F_(3)=sqrt(3) F hat(n)`, where `hat(n)` is the vector along the direction bisecting the `angleBCA`.
It is interesting to see that the sum of the forces on the three charges is zero, i.e.,
`F_(1)+F_(2)+F_(3)=0`
The result is not at all surprising. It follows straight from the fact that Coulomb's law is consistent with Newton's third law. The proof is left to you as an exercise.