The force of interaction of two point charges located at a distance of 200 cm from each other is 1 N.

The force of interaction of two point charges located at a distance of 200 cm from each other is 1 N. What is the value of the second charge if the value of the first is 1.2 μC?

Initial data: r (distance between two point charges) = 200 cm (2 m); F (value of the force of interaction) = 1 N; q1 (value of the first charge) = 1.2 μC (1.2 * 10 ^ -6 C).

Reference data: k (proportionality coefficient) = 9 * 10 ^ 9 N * m2 / Cl2.

We calculate the value of the second charge using Coulomb’s law: F = k * q1 * q2 / r ^ 2, whence q2 = F * r ^ 2 / (k * q1).

Calculation: q2 = 1 * 2 ^ 2 / (9 * 10 ^ 9 * 1.2 * 10 ^ -6) = 3.7 * 10 ^ -4 Cl.

Answer: The value of the second charge is 3.7 * 10 ^ -4 C.