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Electrostatics

JEE Main level, single correct, medium

306 of 311

Question 306 of 311

The electric field in a region of space is given by, $\vec{E}={E}_{0}\hat{i}+2{E}_{0}\hat{j}$ where ${E}_{0}=100N{C}^{-1}$ . The flux of this field through a circular surface of radius $0.02m$ parallel to the $Y‐Z$ plane is nearly

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Answer: (C) $0.125N{m}^{2}{C}^{-1}$

\phi =\vec{E}\cdot \vec{A}=\left({E}_{0}\hat{i}+2{E}_{0}\hat{j}\right)\cdot \left(\pi {r}^{2}\hat{i}\right)

={E}_{0}\pi {r}^{2}=0.125N{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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