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JAMB UTME - Physics - 1998

Question 1 Report

The force of repulsion between two point positive charges \(5\,\mu\text{C}\) and \(8\,\mu\text{C}\) separated at a distance of \(0.02\) map art is. \(\left[\dfrac{1}{4\pi\varepsilon_o}=9 \times 10^9\text{ Nm}^2\text{C}^{-2}\right]\)

Answer Details
F = 1 4 π ε o  x  q 1 q 2 r 2

F = 5 × 10 × 8 × 10 0.002 × 0.002  x 9 x 10

F = 900N