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Q.On inserting a dielectric material between two positive charges in air, the value of repulsive force will (A) increase
(B) decrease
(C) remain same
(D) become zero

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A dielectric weakens the field between charges by a factor K, so the repulsive force falls to F₀/K.

The Coulomb force between two charges in air is F₀ = (1/4πε₀)·q₁q₂/r². Filling the space with a dielectric of relative permittivity K replaces ε₀ with Kε₀:

F = (1/4πKε₀)·q₁q₂/r² = F₀/K

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