Question Detail
Question
Two charges Q1=500 mC, Q2=100 mC at r1=3j m, r2=4i m. Force on Q2?
Correct Answer
Option D is the correct answer.
Detailed Explanation
Distance = √(4²+3²)=5 m. F = k|Q1Q2|/r² = (9e9 × 0.5 × 0.1)/25 = (4.5e8)/25 = 1.8×10⁷ N — huge. Options are 1.9,90,28,18. Likely Q in μC not mC. If Q1=500μC=5e-4, Q2=100μC=1e-4, then F = (9e9×5e-4×1e-4)/25 = (9e9×5e-8)/25 = 450/25=18 N. Yes D.
Hint
F = kQ1Q2/r².