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MHT CET 2024 (3 May, Shift 1) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from MHT CET 2024 (3 May, Shift 1)

Q27MHT CET 2024 · 3 MAY · SHIFT 1Easy

Electric Charge & Coulomb's Law

Two equal point charges ' $q$ ' each exert a force ' $F$ ' on each other, when they are placed distance ' $x$ ' apart in air. When the same charges are placed distance ' $y$ ' apart in a medium of dielectric constant ' $k$ ', they exert the same force. The ratio of distance ' $y$ ' to ' $x$ ' is equal to

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Answer: (a)

$\dfrac{kq^2}{x^2} = \dfrac{kq^2}{k_d\, y^2}$ with $k_d = k$ the dielectric constant.
$y^2 = \dfrac{x^2}{k} \Rightarrow \dfrac{y}{x} = \dfrac{1}{\sqrt{k}}$
So the answer is (a) $\dfrac{1}{\sqrt{k}}$ .

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