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JEE Main 20257 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 7 April
Shift: Evening Shift
Subject: Physics
Chapter: Electrostatics

JEE Main 2025 · 7 April · Evening Shift · Physics · Electrostatics

The electric field in a region is given by E =(2 i +4 j +6 k ) ×...

The electric field in a region is given by

\overrightarrow{\mathrm{E}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \mathrm{~N} / \mathrm{C}

. The flux of the field through a rectangular surface parallel to

x-z

plane is

6.0 \mathrm{Nm}^2 \mathrm{C}^{-1}

. The area of the surface is ________

\mathrm{cm}^2

.

Correct answer

0

Step-by-step solution

\begin{aligned} & \phi=\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}=(2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}) \times 10^3 \cdot \mathrm{~A} \hat{\mathrm{j}} \\ & 6=4 \times 10^3 \mathrm{~A} \\ & \mathrm{~A}=1.5 \times 10^{-3} \mathrm{~m}^2 \\ & =15 \mathrm{~cm}^2\end{aligned}

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