Two charges, each q, are separated by a distance r, and exert mutual attractive forces of F on each other. If both charges become 2q and the distance becomes 3r, what are the new mutual forces

F = ⅔ F₀

Explanation:

For this exercise we use Coulomb's law

         F = k q₁q₂ / r²

let's use the subscript "o" for the initial conditions

          F₀ = k q² / r²

now the charge changes q₁ = q₂ = 2q and the new distance is r = 3 r

we substitute

          F = k 4q² / 9 r²

          F = k q² r² 4/9

          F = ⅔ F₀

answer: decibel

explanation: