An electric field, overrightarrow E =frac 2 hat i +6 hat j +8 hat k √(6) passes through the surface of 4 m 2 area having unit vector hat n = (frac 2 hat i +hat j +hat k √(6) ) . The electric flux for that surface is V m.

Physics · Electrostatics

JEE Main 2024 — 8 April, Shift 1 — Question 52

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

Answer: 12

Numerical answer — enter this value.

Step-by-step solution

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

$\begin{array}{*{35}{r}}{}&~=\left(\frac{2\overset{}{\mathop{i}}\,+6\overset{}{\mathop{j}}\,+8\overset{}{\mathop{k}}\,}{\sqrt{6}}\right)\cdot4\left(\frac{2\overset{}{\mathop{i}}\,+\overset{}{\mathop{j}}\,+\overset{}{\mathop{k}}\,}{\sqrt{6}} \right) \\{} & ~=\frac{4}{6}\times \left( 4+6+8 \right)=12\text{Vm} \\\end{array}$

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Exam: JEE Main 2024
Paper: 8 April, Shift 1
Subject: Physics
Chapter: Electrostatics
Topic: Electric flux and Gauss's Law

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Answer: 12

Step-by-step solution

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