The force of interaction of two charges in the air is f = 3H, Q₁ = 0.6 × 10∧ (-5) C and Q₂ = 3 × 10∧ (-5) C.

The force of interaction of two charges in the air is f = 3H, Q₁ = 0.6 × 10∧ (-5) C and Q₂ = 3 × 10∧ (-5) C. Determine the distance between them. How will the distance between charges change if they are placed in transformer oil (2.3)? water? kerosene (2.1)? alcohol?

According to Coulomb’s law, the force of interaction F of two point charges q₁ = 0.6 × 10 ^ (- 5) C and q₂ = 3 × 10 ^ (- 5) C is directly proportional to the modulus of their product | q₁ ∙ q₂ | and is inversely proportional to the square of the distance R² between them: F = k ∙ | q₁ ∙ q₂ | : (ε ∙ R²), where ε is the dielectric constant of the medium, the coefficient k = 9 ∙ 10 ^ 9 N ∙ m² / Kl². Then R² = (k ∙ | q₁ ∙ q₂ |): (ε ∙ F). From the condition of the problem it is known that the interaction force of two charges in the air is F = 3H, for air the dielectric constant of the medium is ε = 1, then: R² = (9 ∙ 10 ^ 9 ∙ | 0.6 × 10 ^ (- 5) ∙ 3 × 10 ^ (- 5) |): (1 ∙ 3) = 0.54; R ≈ 0.73 m.

If the charges are placed in transformer oil (ε = 2.3), then R ≈ 0.48 m.

If into distilled water (ε = 81), then R ≈ 0.081 ppm.

If in kerosene (ε = 2.1), then R ≈ 0.5 m.

If in alcohol (ε = 27), then R ≈ 0.14 ppm.

Answer: the distance between charges R ≈ 0.73 m; in transformer oil R ≈ 0.48 m; in water R ≈ 0.081 m; in kerosene R ≈ 0.5 m; in alcohol R ≈ 0.14 ppm.