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MHT CET 2025 (22 Apr, Shift 1) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from MHT CET 2025 (22 Apr, Shift 1)

Q20MHT CET 2025 · 22 APR · SHIFT 1Easy

Electric Flux & Gauss's Law

Four charges 2 μC, −3 μC, 4 μC, −4 μC2\ \mu\text{C},\ -3\ \mu\text{C},\ 4\ \mu\text{C},\ -4\ \mu\text{C} and −1 μC-1\ \mu\text{C} are enclosed by the Gaussian surface of radius 2 m. Net outward flux through the Gaussian surface is (in μV-m\mu\text{V-m}) [ ε0=\varepsilon_0 = permittivity of free space]

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Answer: (a)

  1. Net enclosed charge =2−3+4−4−1=−2 μ= 2 - 3 + 4 - 4 - 1 = -2\ \muC.
  2. Gauss's law: ϕ=qencε0=−2ε0\phi = \dfrac{q_{enc}}{\varepsilon_0} = -\dfrac{2}{\varepsilon_0} (in μ\muV m); the radius does not matter.
  3. The magnitude is 2ε0\dfrac{2}{\varepsilon_0}, which matches option (a).
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