Question

What is the magnitude of the repulsive force of two protons in a nucleus if the charge on each proton is 1.6 X 10 ^-19 C and the protons are 3 X 10 ^-15 m apart?

Answer 1

$$F=k\frac{q_1q_2}{r^2}$$ $$F=9*10^9*\frac{1.6*10^{-19}*1.6*10^{-19}}{(3*10^{-15})^2}=25.6\:\rm N$$