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JEE Main 2025 (29 Jan, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2025 (29 Jan, Shift 2)

Q23JEE Main 2025 · 29 JAN · SHIFT 2Easy

Electric Flux & Gauss's Law

A point charge causes an electric flux of −2×104 N m2C−1-2 \times 10^{4}\text{ N m}^2\text{C}^{-1} to pass through a spherical Gaussian surface of 8.0 cm radius, centred on the charge. The value of the point charge is: (Given ϵ0=8.85×10−12 C2N−1m−2\epsilon_0 = 8.85 \times 10^{-12}\text{ C}^2\text{N}^{-1}\text{m}^{-2})

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Answer: (b)

  1. Gauss's law: q=ε0ϕq = \varepsilon_0 \phi (radius does not matter).
  2. q=8.85×10−12×(−2×104)=−1.77×10−7q = 8.85 \times 10^{-12} \times (-2 \times 10^{4}) = -1.77 \times 10^{-7} C
  3. =−17.7×10−8= -17.7 \times 10^{-8} C
  4. So the answer is (b).
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