Physics · Electrostatics

JEE Main 2024 — 29 January, Shift 2 — Question 49

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

Option A:
90
Option B:
150
Option C:
180
Correct
Option D:
60

Answer: C

Step-by-step solution

$\overrightarrow{\mathrm{E}}=6 \hat{i}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}$ $\overrightarrow{\mathrm{A}}=30 \hat{\mathrm{i}}$

$\phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}$

$\phi=(6 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \cdot(30 \hat{\mathrm{i}})$ $\phi=6 \times 30=180$

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Exam: JEE Main 2024
Paper: 29 January, Shift 2
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law

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Answer: C

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