Two identical metal balls are charged with positive charges of 5 nC and 20 nC. The centers of the balls are at a distance

Two identical metal balls are charged with positive charges of 5 nC and 20 nC. The centers of the balls are at a distance of 10 cm from each other. The balls are brought into contact. After that, how far should the centers be separated so that the force of interaction remains the same?

To find out the distance to which it is required to separate the centers of the metal balls taken, consider the equality: k * qн1 * qн2 / rн ^ 2 = Fк (constant force of interaction) = k * q ^ 2 / rх ^ 2, from where we express: rх = √ ( k * q ^ 2 * rn2 / (k * qn1 * qn ^ 2)) = √ (((qn1 + qn2) / 2) ^ 2 * rn2 / (qn1 * qn2)).

Data: qн1 – initial charge of the first ball (qн1 = 5 nC = 5 * 10 ^ -9 C); qn2 – the initial charge of the second ball (qn2 = 20 nC = 20 * 10 ^ -9 C); rн – initial distance (rн = 10 cm; in SI system rн = 0.1 m).

Let’s calculate: rх = √ (((qн1 + qн2) / 2) ^ 2 * rн2 / (qн1 * qн2)) = √ (((5 * 10-9 + 20 * 10-9) / 2) ^ 2 * 0.12 / (5 * 10 ^ -9 * 20 * 10 ^ -9)) = 0.125 m (12.5 cm).

Answer: The centers of the taken metal balls must be diluted by 12.5 cm.