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 _________ \( \text{V m} \).
Correct Answer: 12
Approach Solution - 1
To calculate the electric flux, we use the formula: Φ = E ⋅ A, where Φ is the electric flux, E is the electric field, and A is the area vector, defined as the product of the magnitude of the area and the unit normal vector to the surface. Given the provided vectors:
Electric field, \( \vec{E} = \frac{2\hat{i} + 6\hat{j} + 8\hat{k}}{\sqrt{6}} \)
Unit normal vector, \( \hat{n} = \frac{2\hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \)
We express the area vector as:
A = 4 \( \hat{n} \) = \( \frac{8\hat{i} + 4\hat{j} + 4\hat{k}}{\sqrt{6}} \).
Now, we find the dot product:
E ⋅ A = \( \left(\frac{2}{\sqrt{6}} \hat{i} + \frac{6}{\sqrt{6}} \hat{j} + \frac{8}{\sqrt{6}} \hat{k}\right) \cdot \left(\frac{8}{\sqrt{6}} \hat{i} + \frac{4}{\sqrt{6}} \hat{j} + \frac{4}{\sqrt{6}} \hat{k}\right)\)
= \( \frac{2 \times 8 + 6 \times 4 + 8 \times 4}{6} \)
= \( \frac{16 + 24 + 32}{6} \)
= \( \frac{72}{6} = 12 \).
Hence, the electric flux is 12 \( \text{V m} \). This value falls within the specified range of 12 to 12.
Approach Solution -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: $12 \, \text{V m}$.
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