Physics · Electrostatics

JEE Main 2024 — 9 April, Shift 2 — Question 49

An electric field $\overrightarrow{\mathrm{E}}=(2 x \hat{\mathrm{i}}) \mathrm{NC}^{-1}$ exists in space. A cube of side 2 m is placed in the space as per figure given below.

The electric flux through the cube is $\ldots \ldots . . . . . . . . . . \mathrm{Nm}^{2} / \mathrm{C}$ .

Answer: 16

Numerical answer — enter this value.

Step-by-step solution

$\overrightarrow{\mathrm{E}}=2 x \hat{\mathrm{i}}$

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

$\phi_{\text {in }}=-4 \times 4=-16 \mathrm{Nm}^{2} / \mathrm{c}$

$\phi_{\text {out }}=8 \times 4=32 \mathrm{Nm}^{2} / \mathrm{c}$

$\mathrm{d}_{\text {net }}=\phi_{\text {in }+} \phi_{\text {out }}=-16+32=16 \mathrm{Nm}^{2} / \mathrm{c}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 9 April, Shift 2
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law