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Electrostatics

JEE Main level, numerical answer, medium

293 of 311

Question 293 of 311

The electric field in a region is given by $\vec{E}=\dfrac{2}{5}{E}_{0}\hat{i}+\dfrac{3}{5}{E}_{0}\hat{j}$ with ${E}_{0}=4.0\times {10}^{3}N{C}^{-1}.$ The flux of this field through a rectangular surface, area $0.4{m}^{2}$ parallel to the $Y-Z$ plane is _______ $N{m}^{2}{C}^{-1}$ .

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Answer: 640

\phi ={E}_{x}A\Rightarrow \dfrac{2}{5}\times 4\times {10}^{3}\times 0.4=640N{m}^{2}{C}^{-1}

JEE Main question paper PDF

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JEE Main 2025 Apr 2 Shift 1 question paper

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JEE Main Physics practice question on Electrostatics (Electric Flux and Gauss Law), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.