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JEE Main 2024 (29 Jan, Shift 2) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2024 (29 Jan, Shift 2)

Q35JEE Main 2024 · 29 JAN · SHIFT 2Easy

Electric Flux & Gauss's Law

An electric field is given by (6i^+5j^+3k^) N/C(6\hat{i} + 5\hat{j} + 3\hat{k})\text{ N/C}. The electric flux through a surface area 30i^ m230\hat{i}\text{ m}^2 lying in YZ-plane (in SI unit) is:

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Answer: (c)

  1. ϕ=E⃗⋅A⃗=(6i^+5j^+3k^)⋅30i^\phi = \vec E\cdot\vec A = (6\hat i + 5\hat j + 3\hat k)\cdot 30\hat i
  2. ϕ=6×30=180\phi = 6\times30 = 180 N m2^2/C
  3. So the answer is (c) 180.
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