Consider two protons that are separated by 5.9 fm. What is the magnitude of the Coulomb repulsive force between them?
Assume that these protons are on opposite sides of a nucleus and that the strong force on their nearest neighbors is about 2000 N, but nearly zero between nucleons on opposite sides of the nucleus.

Assuming that these protons are on opossite sides of a nucleus, the strong force between them is zero. The magnitude of the repulsive force between them is given by:

Where k is the coulomb constant, e is the proton charge and d its distance of separation. Replacing the given values and solving:

F = 6.604N

Explanation:

To find the electric force between the protons you use the Coulomb's formula:

where you have used that both charges have the same charge.

K: Coulomb's constant = 8.98*10^9 Nm^2/C^2

r: distance between the charges = 5.9fm = 5.9*10^{-15}m

q: 1.6*10^{-19}C

By replacing the values of k, q and r you obtain:

If you compare this force with the nuclear one Fn = 2000N you obtain:

hence, the nuclear force is about 302.8 times stronger than the Coulomb's force

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle