Consider the charges $q, q$,and $-q$ placed at the vertices of an equilateral triangle of side $l$,as shown in the figure. What is the force on each charge?

(N/A) Let the vertices of the equilateral triangle be $A, B$,and $C$ with charges $q_1 = q, q_2 = q$,and $q_3 = -q$ respectively. The side length is $l$.
The force on charge $q$ at $A$ $(F_1)$ is the vector sum of the repulsive force from $B$ $(F_{12})$ and the attractive force from $C$ $(F_{13})$. Both have magnitude $F = \frac{q^2}{4 \pi \varepsilon_0 l^2}$. The angle between them is $120^\circ$. Using the parallelogram law,the resultant magnitude is $F_1 = \sqrt{F^2 + F^2 + 2F^2 \cos(120^\circ)} = F$. The direction is along the line parallel to $BC$.
Similarly,the force on charge $q$ at $B$ $(F_2)$ is the vector sum of $F_{21}$ and $F_{23}$. By symmetry,its magnitude is $F_2 = F$,directed along the line parallel to $AC$.
The force on charge $-q$ at $C$ $(F_3)$ is the vector sum of $F_{31}$ and $F_{32}$. Both have magnitude $F$ and the angle between them is $60^\circ$. The resultant magnitude is $F_3 = \sqrt{F^2 + F^2 + 2F^2 \cos(60^\circ)} = \sqrt{3}F$. The direction is along the angle bisector of $\angle BCA$.
The sum of the forces is $F_1 + F_2 + F_3 = 0$,which is consistent with Newton's third law.