Coulomb's Law Question

Question

Charges Q1 and Q2 exert repulsive forces of 13N on each other. What is the Repulsive force in Newtons when their separation is decreased so that their final separation is 77% of their initial separation/

Answer 1

The repulsive force is given by Coulomb law to be F = kQ₁Q₂/r², where k = 1/4πε₀ (Coulomb constant)

Now write the Coulomb force for the two cases and solve for the new force i.e

F₁= kQ₁Q₂/r₁² = 13N

F₂ = kQ₁Q₂/(.77r₁)² = kQ₁Q₂/.77²(r₁)² = F₁/.77² = 13/.77²N

Your new force is bigger than the original force as expected!

Answer 2

This question is asking about the inverse square relationship between electrostatic force and separation distance.

Because F_E = k* (Q₁q₂/r²). then F ∝ 1/r² when the charges remain constant. For this example, you can relate the percentage of the separation to the overall force at that distance.

If 13 N = F_i = k* (Q₁q₂/r²), then:

k* (Q₁q₂/(0.77*r)²) = k* (Q₁q₂/0.5929*r²) ⇒ ≅ 1.7 [k* (Q₁q₂/r²)] ≅ 1.7*F_i = 1.7*13 N ≅ 22 N