चित्र में दर्शाए अनुसार किसी समबाहु त्रिभुज के शीर्षो पर स्थित आवेशों $q, q,$ तथा $-q$ पर विचार कीजिए। प्रत्येक आवेश पर कितना बल लग रहा है?

The forces acting on charge $q$ at $A$ due to charges $q$ at $B$ and $-q$ at $C$ are $F_{12}$ along $B A$ 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

$F _{1}=F \hat{ r }_{1}$

where $\hat{ r }_{1}$ is a unit vector along $BC$.

The force of attraction or repulsion for each pair of charges has the same magnitude

$F=\frac{q^{2}}{4 \pi \varepsilon_{0} l^{2}}$

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 $F _{3}=\sqrt{3} F$ in , 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.