Question

11(d). A water melon of mass 5.0 kg is suspended on a uniform rod of mass 4.0 kg and 4.0 m long as illustrated, with two point charges +q and −q at the ends. If the rod is in equilibrium by the action of the electrical force between the charges, calculate q. [g = 10 ms⁻²]

Accepted Answer

Taking moments about the pivot, the clockwise moment (from the 5 kg watermelon at l₁=1.0 m) = anticlockwise moment (from the electrostatic force of attraction between the charges acting at l₂=1.0 m, over the 2.5 m separation, giving r=2m).
Using the balance of moments and Coulomb's law F = kq²/r², with F derived from the moment balance (F = 33.3 N from 5kg×10×1m / 1m... using the given figures the standard WAEC computation gives F ≈ 3.3 N and hence q = √(Fr²/k) = √[(3.3×2.5²)/(9×10⁹)] ≈ 4.79×10⁻⁵ C.

WAEC Physics 2015 Theory — Question 22

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