Three particles have charges + 20 μC 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°. HereF=14πε0×q1q2r2    =9×109×20×10−620×10−6(0⋅5)2   ≈14⋅4NHence, resultant force is given by=F2+F2+2FFcos⁡601/2=3F=14⋅43N