Question

An electric field, $\vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}}$ passes through the surface of $4 \, \text{m}^2$ area having unit vector $\hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}$. The electric flux for that surface is _________ $\text{V m}$.

Answer 1

Electric flux is:
$\phi = \vec{E} \cdot \vec{A}.$
The area vector is:
$\vec{A} = A \hat{n} = 4 \cdot \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}.$
Dot product:
$\phi = \left( \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}} \right) \cdot \left( \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{\sqrt{6}} \right).$
$\phi = \frac{4}{6} (2 \cdot 8 + 6 \cdot 4 + 8 \cdot 4).$
$\phi = \frac{4}{6} (16 + 24 + 32) = \frac{4}{6} \cdot 72 = 12 \, \text{V m}.$
Final Answer: $12 \, \text{V m}$.

Answer 2

Electric flux is:
$\phi = \vec{E} \cdot \vec{A}.$
The area vector is:
$\vec{A} = A \hat{n} = 4 \cdot \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}}.$
Dot product:
$\phi = \left( \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}} \right) \cdot \left( \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{\sqrt{6}} \right).$
$\phi = \frac{4}{6} (2 \cdot 8 + 6 \cdot 4 + 8 \cdot 4).$
$\phi = \frac{4}{6} (16 + 24 + 32) = \frac{4}{6} \cdot 72 = 12 \, \text{V m}.$
Final Answer: $12 \, \text{V m}$.