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MHT CET 2025 (19 Apr, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from MHT CET 2025 (19 Apr, Shift 2)

Q29MHT CET 2025 · 19 APR · SHIFT 2Medium

Electric Flux & Gauss's Law

A uniformly charged conducting sphere of diameter 3.5 cm has a surface charge density of 20 μC m−220\ \mu\text{C m}^{-2}. The total electric flux leaving the surface of the sphere is nearly [permittivity of free space, ε0=8.85×10−12 SI\varepsilon_0 = 8.85 \times 10^{-12}\text{ SI} unit]

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Answer: (c)

  1. q=σ⋅4πR2q = \sigma\cdot4\pi R^2 with R=1.75R = 1.75 cm =0.0175= 0.0175 m, and ϕ=qε0\phi = \frac{q}{\varepsilon_0}.
  2. q=20×10−6×4π(0.0175)2=7.70×10−8q = 20\times10^{-6}\times4\pi(0.0175)^2 = 7.70\times10^{-8} C
  3. ϕ=7.70×10−88.85×10−12≈8.7×103=87×102\phi = \frac{7.70\times10^{-8}}{8.85\times10^{-12}} \approx 8.7\times10^{3} = 87\times10^{2} (SI unit)
  4. So the answer is (c).
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