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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.

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