Two point charges q1 = -20 nC and q2 = +34 nC are separated by r = 0.50 m. What is the magnitude of the Coulomb force between them?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

Two point charges q1 = -20 nC and q2 = +34 nC are separated by r = 0.50 m. What is the magnitude of the Coulomb force between them?

Explanation:
Coulomb's law gives the force between two point charges as F = k |q1 q2| / r^2. Here q1 = -20 nC and q2 = +34 nC, so in coulombs q1 = -20×10^-9 C and q2 = 34×10^-9 C. The product's magnitude is |q1 q2| = (20×10^-9)(34×10^-9) = 680×10^-18 = 6.80×10^-16 C^2. With k ≈ 8.99×10^9 N·m^2/C^2 and r = 0.50 m (so r^2 = 0.25 m^2), the force is F = (8.99×10^9)(6.80×10^-16) / 0.25 ≈ (8.99×6.80)×10^-7 / 0.25 ≈ 61.1×10^-7 / 0.25 ≈ 24.4×10^-6 N = 2.44×10^-5 N. The charges have opposite signs, so the force is attractive, pulling the charges toward each other.

Coulomb's law gives the force between two point charges as F = k |q1 q2| / r^2. Here q1 = -20 nC and q2 = +34 nC, so in coulombs q1 = -20×10^-9 C and q2 = 34×10^-9 C. The product's magnitude is |q1 q2| = (20×10^-9)(34×10^-9) = 680×10^-18 = 6.80×10^-16 C^2. With k ≈ 8.99×10^9 N·m^2/C^2 and r = 0.50 m (so r^2 = 0.25 m^2), the force is F = (8.99×10^9)(6.80×10^-16) / 0.25 ≈ (8.99×6.80)×10^-7 / 0.25 ≈ 61.1×10^-7 / 0.25 ≈ 24.4×10^-6 N = 2.44×10^-5 N. The charges have opposite signs, so the force is attractive, pulling the charges toward each other.

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