Electrostatics

JEE Main level, single correct, medium

280 of 311

Question 280 of 311

An electric field is given by (6i^+5j^+3k^)NC−1(6\hat{i}+5\hat{j}+3\hat{k})N{C}^{-1}. The electric flux through a surface area 30i^m230\hat{i}{m}^{2} lying in YZ-plane(in SI unit) is:

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Answer: (C) 180180

Given:
E⃗=(6i^+5j^+3k^)NC−1\vec{E}=(6\hat{i}+5\hat{j}+3\hat{k})N{C}^{-1}
A⃗=30i^m2\vec{A}=30\hat{i}{m}^{2}
ϕ=E⃗⋅A⃗\phi =\vec{E}\cdot \vec{A}
⇒ϕ=(6i^+5j^+3k^)⋅(30i^)=180SI unit\Rightarrow \phi =(6\hat{i}+5\hat{j}+3\hat{k})\cdot (30\hat{i})=180\text{SI unit}

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