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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 ' qq ' each exert a force ' FF ' on each other, when they are placed distance ' xx ' apart in air. When the same charges are placed distance ' yy ' apart in a medium of dielectric constant ' kk ', they exert the same force. The ratio of distance ' yy ' to ' xx ' is equal to

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Answer: (a)

  1. kq2x2=kq2kd y2\dfrac{kq^2}{x^2} = \dfrac{kq^2}{k_d\, y^2} with kd=kk_d = k the dielectric constant.
  2. y2=x2k⇒yx=1ky^2 = \dfrac{x^2}{k} \Rightarrow \dfrac{y}{x} = \dfrac{1}{\sqrt{k}}
  3. So the answer is (a) 1k\dfrac{1}{\sqrt{k}}.
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