Question

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.

Answer 1

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:

$F=k\frac{e^2}{d^2}$

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

$F=8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(5.9*10^{-15}m)^2}\\F=6.61N$

Answer 2

Answer:

F = 6.604N

Explanation:

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

$F=k\frac{q_1q_2}{r^2}=k\frac{q^2}{r^2}$

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:

$F=(8.98*10^9Nm^2/A^2)\frac{(1.6*10^{-19}C)^2}{(5.9*10^{-15}m)^2}=6.604N$

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

$\frac{F_n}{F}=\frac{2000N}{6.604N}=302.8$

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