Determine the values of equal charges, which, at a distance of r = 0.12 m from each other, are repelled

Determine the values of equal charges, which, at a distance of r = 0.12 m from each other, are repelled with a force of F = 20 μN.

Initial data: r (distance between equal charges) = 0.12 m; F (repulsive force of charges) = 20 μN = 20 * 10 ^ -6 N; q1 = q2 = q (charge amount).

Constants: k (proportionality coefficient (air)) = 9 * 10 ^ 9 N * m ^ 2 / Kl ^ 2.

We express the value of equal charges from the Coulomb law: F = k * q1 * q2 / r ^ 2 = k * q ^ 2 / r ^ 2 and q = √ (F * r ^ 2 / k).

Let’s calculate: q = √ (20 * 10 ^ -6 * 0.12 ^ 2 / (9 * 10 ^ 9)) = 5.66 * 10 ^ -9 Cl.

Answer: The value of each of the equal charges is 5.66 * 10 ^ -9 C.