JEE Mains · Physics · STD 12 - 1. Electric charges and fields
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 _______ \(\mathrm{Vm}\)
- A \(12\)
- B \(13\)
- C \(15\)
- D \(16\)
Answer & Solution
Correct Answer
(A) \(12\)
Step-by-step Solution
Detailed explanation
\(\phi =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}}\) \(=\left(\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}\right) \cdot 4\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}\right)\)…
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