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MHT CET 2026 (15 Apr, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from MHT CET 2026 (15 Apr, Shift 1)

Q2MHT CET 2026 · 15 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

The number of lines of force originating from a point charge of 2×8.85×10−92 \times 8.85 \times 10^{-9} C in a medium of dielectric constant 1010 is (ϵ0=8.85×10−12\epsilon_0 = 8.85 \times 10^{-12} SI unit)

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Answer: (b)

  1. Number of lines of force = flux =qε0K= \dfrac{q}{\varepsilon_0 K}.
  2. N=2×8.85×10−98.85×10−12×10=200N = \dfrac{2\times8.85\times10^{-9}}{8.85\times10^{-12}\times10} = 200
  3. So the answer is (b) 200.
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