Determine the electric field intensity at a distance of 0.30 m from an isolated positive point charge of 5.0 nC in air. (Take k = 9.0 × 10⁹ N·m²/C²)
Answer and explanation
Correct answer: B. 500 N/C
- Step 1: Electric field formula: E = k q / r². Step 2. Substitute values: E = (9.0 × 10⁹ × 5.0 × 10⁻⁹) / (0.30)² = 45 / 0.09 = 500 N/C. Answer: 500 N/C. Why not the others: 150 N/C divides by 0.30 without squaring; 300 N/C and 450 N/C result from arithmetic errors.
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