Electrostatics Part 2

Given, The electric field in the region, $E = \frac{2}{5} E_{0} \hat{i} + \frac{3}{5} E_{0} \hat{j}$ Here, $E_{0} = 4 \times 10^{3} \, \text{N/C}$ Area of the rectangular surface, $A = 0.4 \, \text{m}^{2}$ The direction of electric field vector and area vector is same, so the angle between the electric field vector and area vector is 0 . As we know the expression of electric flux, $\phi = E \cdot A \cos \theta$...(i) Here, $E$ is the electric field vector, and $A$ is the surface area of the surfaces. Consider the surface parallel to the $\text{Y-Z}$ plane, so the area vector, $ $A=0.4 i {m}^{2}$ $ Substituting the values in Eq. (i), we get $\phi = E \cdot A \cos 0^{\circ} \Rightarrow \phi = \frac{2}{5} E_{0} (0.4)$ $\Rightarrow \phi = \frac{2}{5} (4 \times 10^{3})(0.4) = 640 \, \text{Nm}^{2} \, \text{C}^{-1}$ Hence, the electric flux of the surface parallel to the $\text{Y-Z}$ plane is $640 \, \text{Nm}^{2} \, \text{C}^{-1}$.