Question

What will be the electric field experienced by a test charge if q = 7.03 x 10^-6 C from a source charge q = 12 x 10^-6 C in a vacuum when the test charge is placed 0.25 m away from the other charge? 

Answer 1

The Coulomb's law says

$$F=k\frac{q_1q_2}{r^2}\\ =9*10^9*\frac{7.03*10^{-6}*12*10^{-6}}{0.25^2}=12\:\rm N$$