Three equal charges (q) are placed at corners of an equilateral triangle. The force on any charge is

The force on charge at A,

${\overrightarrow{\mathrm{F}}}_{\mathrm{AB}}=\frac{\mathrm{K}\left(\mathrm{q}\right)\left(\mathrm{q}\right)}{{\mathrm{a}}^2}=\frac{{\mathrm{Kq}}^2}{{\mathrm{a}}^2}$

In the same way,

${\overrightarrow{\mathrm{F}}}_{\mathrm{AC}}=\frac{\mathrm{K}{q}^2}{{\mathrm{a}}^2}$

So, the angle $\mathrm{b}/\mathrm{w}{\overrightarrow{\mathrm{F}}}_{\mathrm{AB}}\mathrm{\&}{\overrightarrow{\mathrm{F}}}_{\mathrm{AC}}={60}^{\circ }$

When adding vectors using a parallelogram, we get,

$\begin{array}{l}{\overrightarrow{\mathrm{F}}}_{{\mathrm{r}}_1}=\sqrt{{\left({\overrightarrow{\mathrm{F}}}_{\mathrm{AB}}\right)}^2+{\left({\overrightarrow{\mathrm{F}}}_{\mathrm{AC}}\right)}^2+2{\overrightarrow{\mathrm{F}}}_{\mathrm{AB}}{\overrightarrow{\mathrm{F}}}_{\mathrm{AC}}\cos \theta }\\ =\sqrt{\left(3\right){\left(\frac{{\mathrm{Kq}}^2}{{\mathrm{a}}^2}\right)}^2}\\ =\frac{{\mathrm{Kq}}^2}{{\mathrm{a}}^2}\sqrt{3}\mathrm{N}\end{array}$

The identical force will be applied to all three charges, but in separate directions.