Answer:
F = 3505.2 N
Explanation:
Coulomb law states that the electrostatic force between two charged particles is directly proportional to the product of the two charges and inversely proportional to the square of the distance between the two particles. The electrostatic force is given by:
[tex]F=K\frac{q_{1}q_{2} }{r^{2} }[/tex]
Given:
coulomb constant(K) = 8.98755 × 10⁹ Nm²/C²
Charge of each particle(q₁ = q₂ = 46e)
e is the charge of each electron = 1.602 × 10⁻¹⁹ C
q₁ = q₂ = 46e = 46 × 1.602 × 10⁻¹⁹ = 73.692 × 10⁻¹⁹ C
The distance between the two particles(r) = r₁ + r₂
r₁ and r₂ are the radius of the protons
Therefore r = r₁ and r₂ = (5.9 × 10⁻¹⁵)m + (5.9 × 10⁻¹⁵)m = (11.8 × 10⁻¹⁵)m
Substituting values:
[tex]F=8.98755*10^{9}( \frac{73.692*10^{-19} *73.692*10^{-19} }{(11.8*10^{-15} )^{2} })[/tex]
F = 3505.2 N
The magnitude of the repulsive force pushing these two spheres apart is 3505.2 N
