An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ 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 ______ $\mathrm{Vm}$.

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Aicharya · Accepted Answer

$\begin{aligned} \phi & =\overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{A}} \ & =\left(\frac{2 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}+8 \hat{\mathrm{k}}}{\sqrt{6}}ight) \cdot 4\left(\frac{2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{6}}ight) \ & =\frac{4}{6} 	imes(4+6+8)=12 \mathrm{Vm}\end{aligned}$

An electric field, $\overrightarrow{\mathrm{E}}=\frac{2 \hat{i}+6 \hat{j}+8 \hat{k}}{\sqrt{6}}$ passes…

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