An electric field is given by (6i + 5j + 3k) N/C. The electric flux through a surface area 30i m² lying in YZ-plane (in SI unit) is: (a) 60 (b) 90 (c) 180 (d) 150

Answer: (c). φ = E· A = (6 i + 5 j + 3 k)· 30 i φ = 6×30 = 180 N m ² /C So the answer is (c) 180.

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

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Answer: (c)

$\phi = \vec E\cdot\vec A = (6\hat i + 5\hat j + 3\hat k)\cdot 30\hat i$
$\phi = 6\times30 = 180$ N m $^2$ /C
So the answer is (c) 180.

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

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