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JAMB Physics 1998 Question 36

Question 36

Electrostatics

The force of repulsion between two point positive charges 5 μC and 8 μC separated at a distance of 0.02 m apart is [1/(4πε0) = 9 × 109 N m2 C−2]

  1. A.1.8 × 10−10 N
  2. B.9.0 × 10−8 N
  3. C.9.0 × 102 N
  4. D.4.5 × 103 N
Show the answer and explanation

Answer: C (9.0 × 102 N)

Explanation

F = kq1q2/r2 = 9 × 109 × 5 × 10−6 × 8 × 10−6 ÷ (0.02)2 = 0.36 ÷ 0.0004 = 900 N.