Question Detail
Question
Two charges of equal magnitude exert an attractive force of 6 x 10 N on each other. If the magnitude of each charge is 3.0µC, how far apart are the charges?
Correct Answer
Option A is the correct answer.
Detailed Explanation
F = k q₁q₂/r² → r = √(k q₁q₂/F). Assuming force 6×10⁻? N missing unit. Using F=6×10⁻³ N, q=3×10⁻⁶ C, k=9×10⁹, r = √(9e9 × 9e-12 / 6e-3) = √(81e-3/6e-3) = √13.5 ≈ 3.67 m.
Hint
Coulomb's law: r = √(kq²/F).