Question:
The electric field in a region is given by $\overrightarrow{\mathrm{E}}=\frac{2}{5} \mathrm{E}_{0} \hat{\mathrm{i}}+\frac{3}{5} \mathrm{E}_{0} \hat{\mathrm{j}}$ with $\mathrm{E}_{0}=4.0 \times 10^{3} \frac{\mathrm{N}}{\mathrm{C}}$. The flux of this field through a rectangular surface area $0.4 \mathrm{~m}^{2}$ parallel to the $\mathrm{Y}-\mathrm{Z}$ plane is _____$\mathrm{Nm}^{2} \mathrm{C}^{-1}$.
Solution:
$\phi=\mathrm{E}_{\mathrm{x}} \mathrm{A} \Rightarrow \frac{2}{5} \times 4 \times 10^{3} \times 0.4=640$
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