# Consider the charges q, q, and –q placed at the vertices of an equilateral triangle, as shown in Figure . What is the force on each charge?

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 Figure .
By the parallelogram law, the total force $F_1$ on the charge q at A is given by $F_1 = F$ where is a unit vector along BC.
The force of attraction or repulsion for each pair of charges has the same magnitude $F= \large\frac{q^2}{4 \pi \in_0 j^2}$
The total force $F_2$ on charge q at B is thus F2 = F , where 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 unit vector along the direction bisecting the $\angle BCA$
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.