The force between two charges in air is 0.40 N. What is the force when they are placed at the same separation in oil of relative permittivity 4.0?
Correct answer C. 0.10 N
Explanation
The force in a medium is F_air/εᵣ = 0.40/4.0 = 0.10 N. Multiplying by εᵣ instead wrongly gives 1.6 N, but a medium always weakens the force.
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