Consider the charges q,q and -q placed at the vertices of an equilateral triangle of each side l. 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 F12 along BA and F13 along AC respectively, as shown in Fig. 1.10. By the parallelogram law, the total force F1 on the charge q at A is given by
F1 = F 1 ˆ r where 1 ˆ r is a unit vector along BC. The force of attraction or repulsion for each pair of charges has the
same magnitude `F=(q^(2))/(4piepsilon_(0)l^(2))`
Similarly the total force on charge `-q at C is F_(3)=sqrt3 F hatn` where `hatn` is the unit vector along the direction bisecting the ∠BCA. 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