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Q.Two point charges are placed in the medium of dielectric constant k1k_1. Electrostatic force between them is FF. If the dielectric constant of the medium changes to k2k_2, the electrostatic force between them will be:

i) k2k1F\dfrac{k_2}{k_1}F
ii) 2k1k2F\dfrac{2k_1}{k_2}F
iii) k1k2F\dfrac{k_1}{k_2}F
iv) 2F2F
Uttar Pradesh UpmspUP Board (UPMSP) Intermediate 2022MCQ· 1mImportance★★★★★
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Coulomb force ∝1/k\propto 1/k, so replacing k1k_1 by k2k_2 scales FF by k1/k2k_1/k_2.

The force between two point charges in a medium is F=14πε0kq1q2r2F=\dfrac{1}{4\pi\varepsilon_0 k}\dfrac{q_1q_2}{r^2}. The charges and their separation are unchanged, so F∝1kF\propto \dfrac{1}{k}.

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