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

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;

$F = \frac{k q_1 q_2 }{r^2} \\\\where;\\k \ is \ Coulomb's \ constant = 9\times 10^9 \ Nm^2/C^2 \\\\F = \frac{9\times 10^9 \ (1.6\times 10^{-19})^2}{(4 \times 10^{-15})^2} \\\\F = 14.4 \ N$

Therefore, the  repulsive electrostatic force between two of the protons is 14.4 N.