The electric field in a region is given by $$\vec{E} = \frac{2}{5}E_0\hat{i} + \frac{3}{5}E_0\hat{j}$$ with $$E_0 = 4.0 \times 10^3$$ N C$$^{-1}$$. 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$$^{-1}$$.

The electric field is $$\vec{E} = \frac{2}{5}E_0\,\hat{i} + \frac{3}{5}E_0\,\hat{j}$$ with $$E_0 = 4.0 \times 10^3$$ N/C. The rectangular surface of area 0.4 m$$^2$$ is parallel to the $$Y$$-$$Z$$ plane, so its outward normal is along $$\hat{i}$$.

The flux through the surface is $$\Phi = \vec{E} \cdot \hat{i}\,A = \frac{2}{5}E_0 \times 0.4$$. Substituting, $$\Phi = \frac{2}{5} \times 4.0 \times 10^3 \times 0.4 = 1600 \times 0.4 = 640$$ N m$$^2$$ C$$^{-1}$$.