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

| October 13, 2021 | education

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.



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