Two identical small metal balls with charges + q and -3q were brought into contact and then moved

Two identical small metal balls with charges + q and -3q were brought into contact and then moved apart at the same distance, much larger than the diameters of the balls. Interaction forces between balls …?

The force of interaction between two charged balls:
F = kq₁q₂ / r², where k is the proportionality coefficient, r is the distance between the centers of the balls.
In the first case, the force of attraction between the balls will be proportional to the product
charges + q and -3q:
F₁ = kq (-3q) / r² = -3kq² / r².
After contact, the sum of the charges was equally divided between the balls:
q₁ = q₂ = (-3q + q) / 2 = -q.
The balls began to push off with force:
F₂ = k (-q) (- q) / r² = kq² / r².
The ratio of forces:
F₁ / F₂ = (-3kq² / r²) / (kq² / r²) = -3.
F₁ = -3F₂.
Answer: After the balls touched, the force of interaction decreased three times and the attraction was replaced by repulsion.