Three particles have charges + 20 $\mathrm{\mu }$C each and they are fixed at the comers of a:r equilateral triangle of side 0.5 m. The force on each of the particles has magnitude

On every particle, there will be two forces of equal magnitude F and inclined at 60°. Here
$$\begin{gathered} \mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2} \\ =\left(9\times {10}^9\right)\times \frac{\left(20\times {10}^{-6}\right)\left(20\times {10}^{-6}\right)}{(0\cdot 5{)}^2} \\ \approx 14\cdot 4\mathrm{N} \end{gathered}$$
Hence, resultant force is given by
$\begin{array}{l}={\left[{\mathrm{F}}^2+{\mathrm{F}}^2+2\mathrm{FFcos}60\right]}^{1/2}=\sqrt{3}\mathrm{F}\\ =14\cdot 4\sqrt{3}\mathrm{N}\end{array}$