<h2>
Answer:</h2>
29.4N
<h2>
Explanation:</h2>
By Coulomb's law, the electrical force of attraction/repulsion, F, between two charges of magnitudes Q₁ and Q₂, is directly proportional to the product of the magnitudes of these charges and inversely proportional to the square of the distance, r, between these charges. i.e;
F ∝ Q₁ x Q₂ / r²
F = k x Q₁ x Q₂ / r² -------------------(i)
Where;
k = proportionality constant = 8.99 x 10⁹Nm²/C²
<em>From the question;</em>
Two protons are given;
Let the charge of first proton be Q₁
Let the charge of second proton be Q₂
The charge of a proton = 1.602 x 10⁻¹⁹C
=> Q₁ = Q₂ = 1.602 x 10⁻¹⁹C
<em>Also;</em>
r = 2.8 x 10⁻¹⁵m
<em>Substitute these values into equation (i) s follows;</em>
F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x 1.602 x 10⁻¹⁹ / (2.8 x 10⁻¹⁵)²
F = 8.99 x 10⁹ x 1.602 x 10⁻¹⁹ x 1.602 x 10⁻¹⁹ / (7.84 x 10⁻³⁰)
F = 23.07 x 10⁻²⁹ / (7.84 x 10⁻³⁰)
F = 29.4N
Therefore, the electrical force that the two protons exert on each other is 29.4N
