Two point charges, q1 = +3.0 μC and q2 = -1.5 μC, are placed 0.20 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. 1.0 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 charges: q1 = 3.0 × 10^-6 C, q2 = 1.5 × 10^-6 C. F = (9.0 × 10^9) × (3.0 × 10^-6) × (1.5 × 10^-6) / (0.20)^2 = (9.0 × 10^9 × 4.5 × 10^-12) / 0.04 = (40.5 × 10^-3) / 0.04 = 1.0125 N ≈ 1.0 N.
Example
For q1=3 μC, q2=1.5 μC, r=0.2 m, the force is about 1.0 N.
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