Two point charges, q1 = +6.0 μC and q2 = -3.0 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
B. 4.05 N
Principle or equation
Coulomb's law: F = k|q1 q2|/r^2.
Why this answer is correct
F = (9.0 × 10^9)(6.0 × 10^-6)(3.0 × 10^-6)/(0.20)^2 = (9.0 × 10^9)(18 × 10^-12)/0.04 = 162 × 10^-3 / 0.04 = 4.05 N.
Example
For q1 = 2 μC, q2 = 3 μC, r = 0.1 m, F = 5.4 N.
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