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 $\mathrm{E}_{0}=4.0 \times 10^{3} \frac{\mathrm{N}}{\mathrm{C}} \cdot$ 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:
(640)
$\phi=\mathrm{E}_{\mathrm{x}} \mathrm{A} \Rightarrow \frac{2}{5} \times 4 \times 10^{3} \times 0.4=640$
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