Answer:
The repulsive electrostatic force between two of the protons is 14.4 N.
Explanation:
Given;
distance between the two protons, r = 4 x 10⁻¹⁵ m
charge of a proton, q = + 1.6 x 10⁻¹⁹ C
The repulsive electrostatic force between two of the protons is calculated by applying Coulomb's law;
[tex]F = \frac{k q_1 q_2 }{r^2} \\\\where;\\k \ is \ Coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2 \\\\F = \frac{9\times 10^9 \ (1.6\times 10^{-19})^2}{(4 \times 10^{-15})^2} \\\\F = 14.4 \ N[/tex]
Therefore, the repulsive electrostatic force between two of the protons is 14.4 N.
