The number of lines of force originating from a point charge of 2 × 8.85 × 10⁻⁹ C in a medium of dielectric constant 10 is ( ε₀ = 8.85 × 10⁻¹² SI unit) (a) 100 (b) 200 (c) 400 (d) 2000

Answer: (b). Number of lines of force = flux = (q)/(ε₀ K). N = 2×8.85×10⁻⁹8.85×10⁻¹²×10 = 200 So the answer is (b) 200.

The number of lines of force originating from a point charge of 2 ×… | MHT CET 2026 (15 Apr, Shift 1)

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Electric Flux & Gauss's Law

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

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Answer: (b)

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

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