Two identical conducting charged balls at a distance of 60 cm are repulsed with a force of 20 μN.

Two identical conducting charged balls at a distance of 60 cm are repulsed with a force of 20 μN. After the balls were brought into contact and removed to the same distance, the repulsive force became equal to 160 μN. Considering the charged balls as point charges, find the charges of the balls q1 and q2 that were on the balls before contact.

Let’s write down the formulas that we will use:
F = (k * q1 * q2) / r1 ^ 2;
q1 * q2 = (F * r1 ^ 2) / k;
q1 * q2 = (q1 + q2) / 2;
Let’s combine the formulas:
F = k * ((q1 + q2) / 2) * (q1 + q2) / 2)) / r1 ^ 2;
(q1 + q2) ^ 2 = F * r ^ 2 / k;
Substitute the values:
q1 * q2 = (20 * 10 ^ -6 * 0.36) / 9 * 10 ^ 9 = 0.8 * 10 ^ -15 = 0.8 * 10 ^ -16;
(q1 + q2) ^ 2 = 160 * 10 ^ -6 * 0.36 / 9 * 10 ^ 9 = 6.4 * 10 ^ -15 = 6.4 * 10 ^ -16;
q1 + q2 / 2 = 0.8 * 10 ^ -8;
q1 + q2 = 16 * 10 ^ -8;
Let’s compose a system of equations:
q1 * q2 = 0.8 * 10 ^ -16;
q1 + q2 = 16 * 10 ^ -8;

q1 = 16 * 10 ^ -8 – q2;
(16 * 10 ^ -8 – q2) * q2 = 0.8 * 10 ^ -16;

Let’s make an equation, i.e. equate to 0.
16 * 10 ^ -8 * q2 – q ^ 2 – 0.8 * 10 ^ -16 = 0;
q ^ 2 – 16 * 10 ^ -8 * q2 + 0.8 * 10 ^ -16 = 0;
q2 = 0.52 * 10 ^ -8 or q2 = 15.48 ^ -8;
q1 = 15.48 * 10 ^ -8 or q1 = 0.52 ^ -8;

Answer: q1 = 0.52 ^ -8; q2 = 15.48 ^ -8.