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(D) The electric field is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$.Given $E_{0} = 4.0 	imes 10^{3} \, N/

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$640$

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(D) The electric field is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$.Given $E_{0} = 4.0 	imes 10^{3} \, N/C$.The surface area is $A = 0.4 \, m^{2}$ and it is parallel to the $Y-Z$ plane.The area vector $\overrightarrow{A}$ for a surface parallel to the $Y-Z$ plane is directed along the $X$-axis,so $\overrightarrow{A} = 0.4 \hat{i} \, m^{2}$.The electric flux $\phi$ is given by the dot product $\phi = \overrightarrow{E} \cdot \overrightarrow{A}$.$\phi = (\frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}) \cdot (0.4 \hat{i})$.Since $\hat{i} \cdot \hat{i} = 1$ and $\hat{j} \cdot \hat{i} = 0$,we have $\phi = \frac{2}{5} E_{0} 	imes 0.4$.Substituting the value of $E_{0} = 4.0 	imes 10^{3} \, N/C$:$\phi = \frac{2}{5} 	imes (4.0 	imes 10^{3}) 	imes 0.4$.$\phi = 0.4 	imes 4000 	imes 0.4 = 1600 	imes 0.4 = 640 \, N m^{2} C^{-1}$.

The electric field in a region is given by $\overrightarrow{E} = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ with $E_{0} = 4.0 \times 10^{3} \, N/C$. The flux of this field through a rectangular surface area $0.4 \, m^{2}$ parallel to the $Y-Z$ plane is ....... $N m^{2} C^{-1}$.

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