Question

The nucleus in an iron atom has a radius of about 4.0 '

10$15 m and contains 26 protons.

(a) What is the magnitude of the repulsive electrostatic force between

two of the protons that are separated by 4.0' 10$15 m?

Answer 1

$$F=\dfrac{kq_1q_2}{r^2},$$ $$F=\dfrac{9\cdot10^9\ \dfrac{Nm^2}{C^2}\cdot1.6\cdot10^{-19}\ C\cdot1.6\cdot10^{-19}\ C}{(4\cdot10^{-15}\ m)^2}=14.4\ N.$$