Two identical metal balls are charged: one with a charge of +3000 e, the other +7000 e, where e is an elementary

Two identical metal balls are charged: one with a charge of +3000 e, the other +7000 e, where e is an elementary electric charge. The balls are brought into contact and then pulled apart. What are the charges of the balls after they are diluted? Has the nature of the interaction of the balls changed?

Elementary electric charge e = 1.602 * 10-19 C.
When the balls come into contact and dilute them, the charge is added up and divided equally between them.
Total charge:
q = q1 + q2 = (+ 3000 * 1.602 * 10-19) + (+ 7000 * 1.602 * 10-19) = 1.602 * 10 ^ 15 Cl.
After contact, the charges:
q1` = q2` = q / 2 = 1.602 * 10 ^ 15 C / 2 = 0.8 * 10 ^ 15 C.
Strength of interaction before contact:
F1 = k * q1 * q2 / L²
Interaction force after contact:
F2 = k * q1`² / L²
Let’s split F2 / F1:
F2 / F1 = (k * q1`² / L²) / (k * q1 * q2 / L²) = q1`² / q1 * q2
Substitute the numbers and get:
F2 / F1 = q1`² / q1 * q2 = (0.8 * 10 ^ 15) / ((+ 3000 * 1.602 * 10-19) * (+ 7000 * 1.602 * 10-19)) = 1.19
This means that the strength of interaction has increased.
Answer: after contact, the charges are equal to each other q1` = q2` = 0.8 * 10 ^ 15 C, and the force of interaction has increased.