Question

Three equal point charges of charge 5 microcoulomb are placed at the vertices of an equilateral triangle which has sides 10 cm long then what will be the force on each charge

Answer 1

F=k q1.q2/r^2

F=9×10^9(5×10^-6)^2/(10×10^-2)^2

F=225×10^-3/10^-2

F=225×10^-1

F=22.5 N

Answer 2

F=q1.q2/r^2

F=9×10^9 (5×10^-6)^2 (10×10^-2)^2

F=225×10^-1

F=22.5N