# At the vertices of a regular triangle with side a = 10 cm, there are charges Q1 = 10 μC, Q2 = 20 μC, Q3

**At the vertices of a regular triangle with side a = 10 cm, there are charges Q1 = 10 μC, Q2 = 20 μC, Q3 = 30 μC. Determine the force F acting on the charge Q1 from the side of the other two charges.**

To determine the magnitude of the force acting on the charge q1, we use the formula (parallelogram rule): F = √ (F21 ^ 2 + F31 ^ 2 + 2F21 * F31 * cos 60º) = √ ((k * q ^ 2 * q1 / a ^ 2) ^ 2 + ((k * q3 * q1 / a ^ 2) ^ 2 + 2 * (k * q2 * q1 / a ^ 2) * (k * q3 * q1 / a ^ 2) * cos 60º).

Constants and variables: k (proportionality coefficient) = 9 * 10 ^ 9 m / F; q2 is the charge at the second vertex (q2 = 20 μC = 2 * 10 ^ -5 C); q1 is the charge at the first vertex (q1 = 10 μC = 10 ^ -5 C); a – side of the triangle (a = 10 cm = 0.1 m); q3 is the charge at the third vertex (q3 = 30 μC = 3 * 10 ^ -5 C).

Calculation: F = √ ((9 * 10 ^ 9 * 2 * 10 ^ -5 * 10 ^ -5 / 0.1 ^ 2) ^ 2 + ((9 * 10 ^ 9 * 3 * 10 ^ -5 * 10 ^ -5 / 0.1 ^ 2) ^ 2 + 2 * (9 * 10 ^ 9 * 2 * 10 ^ -5 * 10 ^ -5 / 0.1 ^ 2) * (9 * 10 ^ 9 * 3 * 10 ^ -5 * 10 ^ -5 / 0.1 ^ 2) * 0.5) = √ (32400 + 72900 + 48600) = 392.3 N.

Answer: A force of 392.3 N. acts on the first charge.