# 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.