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MHT CET 2024 (2 May, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from MHT CET 2024 (2 May, Shift 1)

Q51MHT CET 2024 · 2 MAY · SHIFT 1Medium

Electric Flux & Gauss's Law

A uniformly charged conducting sphere of diameter 14 cm has surface charge density of 40 μC m−240\ \mu\text{C m}^{-2}. The total electric flux leaving the surface of the sphere is nearly (Permittivity of free space =8.85×10−12= 8.85 \times 10^{-12} SI unit)

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Answer: (d)

  1. q=σ⋅4πR2q = \sigma\cdot4\pi R^2 with R=7R = 7 cm =0.07= 0.07 m, and ϕ=qε0\phi = \frac{q}{\varepsilon_0}.
  2. q=40×10−6×4π(0.07)2=2.46×10−6q = 40\times10^{-6}\times4\pi(0.07)^2 = 2.46\times10^{-6} C
  3. ϕ=2.46×10−68.85×10−12≈2.8×105=280\phi = \frac{2.46\times10^{-6}}{8.85\times10^{-12}} \approx 2.8\times10^{5} = 280 k (SI unit)
  4. So the answer is (d).
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