Using r = sqrt(k q / E) with k = 8.99×10^9 N·m^2/C^2, q = 24 pC, and E = 60 N/C, what is r in meters?

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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

Using r = sqrt(k q / E) with k = 8.99×10^9 N·m^2/C^2, q = 24 pC, and E = 60 N/C, what is r in meters?

Explanation:
The relation E = k q / r^2 for a point charge links the electric field to distance. Solving for r gives r = sqrt(k q / E). Convert the charge: q = 24 pC = 24 × 10^-12 C. Compute k q: 8.99×10^9 × 24×10^-12 ≈ 0.21576. Divide by the field: 0.21576 / 60 ≈ 0.003596. Take the square root: sqrt(0.003596) ≈ 0.05996 m ≈ 0.060 m. The result has the correct units (meters) and matches the given value.

The relation E = k q / r^2 for a point charge links the electric field to distance. Solving for r gives r = sqrt(k q / E).

Convert the charge: q = 24 pC = 24 × 10^-12 C.

Compute k q: 8.99×10^9 × 24×10^-12 ≈ 0.21576.

Divide by the field: 0.21576 / 60 ≈ 0.003596.

Take the square root: sqrt(0.003596) ≈ 0.05996 m ≈ 0.060 m.

The result has the correct units (meters) and matches the given value.

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