How many times will the interaction force of two identical point charges decrease if each of them is reduced

How many times will the interaction force of two identical point charges decrease if each of them is reduced by 2 times and transferred from vacuum to a medium with a dielectric constant equal to 2.5. The distance between charges does not change.

Let two identical point charges, q₀, located at a distance R in a vacuum with a dielectric constant ε = 1, interact with a certain force F determined by the Coulomb law:

F = (k ∙ | q₁ | ∙ | q₂ |) / (ε ∙ R ^ 2), where the proportionality coefficient k = 9 ∙ 10 ^ 9 (N ∙ m ^ 2) / Kl ^ 2, q₁ = q₂ = q₀.

We get: F₁ = (k ∙ q₀ ^ 2) / R ^ 2.

After each charge was halved, transferred from a vacuum to a medium with a dielectric constant ε = 2.5, keeping the distance between the charges R, the interaction force became:

F₂ = (k ∙ (q₀ / 2) ^ 2) / (ε ∙ R ^ 2);

F₂ = (k ∙ q₀ ^ 2) / (4 ∙ 2.5 ∙ R ^ 2).

To determine how many times the interaction force of these charges has decreased, we find the ratio:

F₁: F₂ = ((k ∙ q₀ ^ 2) / R ^ 2): ((k ∙ q₀ ^ 2) / (10 ∙ R ^ 2));

F₁: F₂ = 10.

Answer: the force of interaction of charges has decreased 10 times.