JEE Physics · Easy

Electrostatics: Coulomb's Law MCQ

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Coulomb's LawEasyQuestion 31461

Question

Force between charges of +4 μC and −2 μC placed 0.3 m apart is (k = 9×10⁹):
  1. A
    0.8 N attractive
    Correct
  2. B
    0.8 N repulsive
  3. C
    1.6 N attractive
  4. D
    0.4 N repulsive

Correct answer

0.8 N attractive

Explanation

F = kq₁q₂/r² = 9×10⁹ × 4×10⁻⁶ × 2×10⁻⁶ / (0.09) = 9×10⁹ × 8×10⁻¹² / 0.09 = 72×10⁻³/0.09 = 0.8 N. Unlike charges → attractive.