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: $12 \, \text{V m}$.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32