# Consider three charges q1, q2, q3 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

In the given equilateral triangle ABC of sides of length l, if we draw a perpendicular AD to the side $BC,AD = AC \cos 30 ^{\circ} = ( \sqrt 3 /2) l$ and the distance AO of the centroid O from A is $(2/3)AD =(1/ \sqrt 3) l$.
By symmetry $AO=BO=CO$
Thus,
Force $F_1$ on Q due to charge q at $A =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $AO$
Force $F_2$ on Q due to charge q at $B =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $BO$
Force $F_3$ on Q due to charge q at $C =\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$ along $CO$
The resultant of forces $F_2$ and $F_3$ is $\large\frac{3}{4 \pi \in_0} \frac{Qq}{l^2}$$( \hat {r} - \hat {r})=0$ where $\hat {r}4$ is the unit vector along OA.
It is clear also by symmetry that the three forces will sum to zero. Suppose that the resultant force was non-zero but in some direction.
Consider what would happen if the system was rotated through $60^{\circ}$ about O.