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JEE Mains · Physics · STD 12 - Electric charges and fields
  • JEE Mains 2024
  • Hard
  • Numerical

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 \(\hat{\mathrm{n}}=\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)\). The electric flux for that surface is _______ \(\mathrm{Vm}\)

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Answer & Solution

Correct Answer

(A) \(12\)

Step-by-step Solution

Detailed explanation

\(\phi =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\)

\(=\left(\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}\right) \cdot 4\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)\)

\(=\frac{4}{6} \times(4+6+8)=12 \mathrm{Vm}\)

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