Answer:
6.27 x 10^10 N
Explanation:
Since the mass of the proton is 1.602 x 10-19 C, therefore the electric force between two protons separated by a distance:
r = 2.8×10^−15m
is given as following:
F=k*║q_1║*║q_2║/r^2
=k*║q║*║q║/r^2
=k*║q║^2/r^2
=k*q^2/r^2
= 8.99 x 10^9 Nm^2c^2*[(1.67 x 10^-19 c)^2/ (2.8 x 10^-15 m)^2
= 6.27 x 10^10 N
29.4N
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
From the question;
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
Also;
r = 2.8 x 10⁻¹⁵m
Substitute these values into equation (i) s follows;
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
