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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?

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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.,
The result is not at all surprising. It follows straight from the fact that Coulomb’s law is consistent with Newton’s third law.
answered May 29, 2014 by meena.p
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