Three charges are placed at the vertices of an equilateral triangle of side a as shown in the following figure. The force experienced by the charge placed at the vertex A in a direction normal to BC is

The various forces acting at charge +Q are:

$$\begin{gathered} {\overrightarrow{\mathrm{F}}}_{\mathrm{B}}\left(\text{ force of attraction }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{{\mathrm{Q}}^2}{{\mathrm{a}}^2} \\ {\overrightarrow{\mathrm{F}}}_{\mathrm{C}}\left(\text{ force of repulsion }\right)=\frac{1}{4{\mathrm{\pi \epsilon }}_0}·\frac{{\mathrm{Q}}^2}{{\mathrm{a}}^2} \end{gathered}$$

Resolving the force, we have F_B sin 60° along -y  direction and F_c sin 60° along +y direction. Two equal and opposite forces which cancel each other. Hence, net force is zero.