Consider three charges $q_{1}, q_{2}, q_{3}$ each equal to $q$ at the vertices of an equilateral triangle of side $l$. What is the force on a charge $Q$ (with the same sign as $q$) placed at the centroid of the triangle,as shown in Figure?

(D) In the given equilateral triangle $ABC$ of side length $l$,let $O$ be the centroid.
The distance of the centroid $O$ from each vertex is $r = \frac{l}{\sqrt{3}}$.
The force on charge $Q$ at $O$ due to each charge $q$ at the vertices is given by Coulomb's law: $F = \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{r^{2}} = \frac{1}{4\pi\varepsilon_{0}} \frac{Qq}{(l/\sqrt{3})^{2}} = \frac{3}{4\pi\varepsilon_{0}} \frac{Qq}{l^{2}}$.
Let the forces due to charges at $A, B,$ and $C$ be $\vec{F}_{1}, \vec{F}_{2},$ and $\vec{F}_{3}$ respectively. These forces are directed away from the vertices along the medians.
By symmetry,the angle between any two force vectors is $120^{\circ}$.
The resultant of $\vec{F}_{2}$ and $\vec{F}_{3}$ is equal in magnitude to $F_{1}$ but directed opposite to it (along $OA$).
Therefore,the net force $\vec{F}_{net} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} = 0$.