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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 8 April · Morning Shift · Physics · Electrostatics

An electric field, E = 2 i +6 j +8 k 6 passes through the surface...

An electric field, E→=2i^+6j^+8k^6\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}} passes through the surface of 4 m24 \mathrm{~m}^2 area having unit vector n^=(2i^+j^+k^6)\hat{n}=\left(\frac{2 \hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}\right). The electric flux for that surface is ______ Vm\mathrm{Vm}.

Correct answer

0

Step-by-step solution

ϕ=E→⋅A→=(2i^+6j^+8k^6)⋅4(2i^+j^+k^6)=46×(4+6+8)=12Vm\begin{aligned} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \\ & =\left(\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}\right) \cdot 4\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right) \\ & =\frac{4}{6} \times(4+6+8)=12 \mathrm{Vm}\end{aligned}
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