Coulomb’s law gives the force between two charges as F = k q1 q2 / r^2. If q1 = q2 = 3.0 nC and r = 2.0 cm, which is the approximate force magnitude?

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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

Coulomb’s law gives the force between two charges as F = k q1 q2 / r^2. If q1 = q2 = 3.0 nC and r = 2.0 cm, which is the approximate force magnitude?

Explanation:
The force follows F = k q1 q2 / r^2, so convert to SI units and plug in: q1 = q2 = 3.0 nC = 3.0×10^-9 C, r = 2.0 cm = 0.020 m (so r^2 = 4.0×10^-4 m^2). Then F ≈ (8.99×10^9) × (3.0×10^-9) × (3.0×10^-9) / (4.0×10^-4) ≈ (8.99×9)×10^-18+9 / 4×10^-4 ≈ 8.091×10^-8 / 4×10^-4 ≈ 2.0×10^-4 N. So the approximate force is about 2.0×10^-4 newtons.

The force follows F = k q1 q2 / r^2, so convert to SI units and plug in: q1 = q2 = 3.0 nC = 3.0×10^-9 C, r = 2.0 cm = 0.020 m (so r^2 = 4.0×10^-4 m^2). Then F ≈ (8.99×10^9) × (3.0×10^-9) × (3.0×10^-9) / (4.0×10^-4) ≈ (8.99×9)×10^-18+9 / 4×10^-4 ≈ 8.091×10^-8 / 4×10^-4 ≈ 2.0×10^-4 N. So the approximate force is about 2.0×10^-4 newtons.

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