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Q.The electrostatic force between two charges distance r apart in vacuum is F. The force between the same charges distance r/2 apart in a medium of dielectric constant 2 is

(i) F/2
(ii) 2F
(iii) F/4
(iv) 4F
Odisha ChseOdisha CHSE +2 Science Board Exam 2020MCQ· 1mImportance★★★★★
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New force = 2F, since halving the distance multiplies the force by 4 but the dielectric medium (K=2) divides it by 2, net factor 2.

Coulomb's law in a medium of dielectric constant KK is:

F=14πε0Kq1q2r2F = \frac{1}{4\pi\varepsilon_0 K}\frac{q_1 q_2}{r^2}

Initially in vacuum (K=1K=1) at separation rr:

F=14πε0q1q2r2F = \frac{1}{4\pi\varepsilon_0}\frac{q_1q_2}{r^2}

Now the separation is r/2r/2 and the medium has K=2K = 2: …

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