Two point charges, q1 = +4.0 μC and q2 = -2.0 μC, are placed 0.30 m apart in vacuum. What is the magnitude of the electric force that q1 exerts on q2? (Use k = 9.0 × 10^9 N·m^2/C^2)
Show the answer and explanation
Correct answer
A. 0.80 N
Principle or equation
Coulomb's law: F = k |q1 q2| / r^2, where k = 9.0 × 10^9 N·m^2/C^2.
Why this answer is correct
Convert microcoulombs to coulombs: q1 = 4.0 × 10^-6 C, q2 = 2.0 × 10^-6 C (magnitude). Then F = (9.0 × 10^9) × (4.0 × 10^-6) × (2.0 × 10^-6) / (0.30)^2 = (9.0 × 10^9) × (8.0 × 10^-12) / 0.09 = (72 × 10^-3) / 0.09 = 0.80 N.
Example
For q1 = 1 μC, q2 = 1 μC, r = 0.1 m, F = 9×10^9 × 10^-12 / 0.01 = 0.9 N.
This published item includes a stored explanation and passed the platform’s publication workflow. It is independent preparation material, not a claim of an official or recalled examination question.