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MHT CET 2026 (20 Apr, Shift 2) · Electric Charge & Coulomb's Law

Electric Charge & Coulomb's Law question from MHT CET 2026 (20 Apr, Shift 2)

Q13MHT CET 2026 · 20 APR · SHIFT 2Medium

Electric Charge & Coulomb's Law

Two point charges placed in air separated by distance RR exert a force FF on each other. Now the space between those charges is filled with dielectric constant KK. At distance R1R_1 between the charges, the same force is exerted as FF. Then

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Answer: (d)

  1. In air F=kq1q2R2F = \dfrac{kq_1q_2}{R^2}; in a medium F′=kq1q2KR12F' = \dfrac{kq_1q_2}{KR_1^2}.
  2. Same force: 1R2=1KR12⇒R12=R2K\dfrac{1}{R^2} = \dfrac{1}{KR_1^2} \Rightarrow R_1^2 = \dfrac{R^2}{K}
  3. So the answer is (d) R1=RKR_1 = \dfrac{R}{\sqrt{K}}.
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