Consider a uniform electric field $\mathrm{E}==3\times {10}^3\stackrel{⏜}{i}\mathrm{N}/\mathrm{C}$. What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz-plane?

Given that the electric field intensity is (3 x10³) hence, the magnitude of electric field intensity is $\left.\left|\overrightarrow{\mathrm{E}}\right|=3\times {10}^3\mathrm{N}/\mathrm{C}\right)$, the side of the square is s=10cm=0.1m; area of the square is A=S²=0.01m² The plane of the square is parallel to the y-z plane; therefore, the angle between the unit vector normal to the plane and electric field is given by $\theta$ = 0^o. Now, the flux through the plane is given by
$\mathrm{\varphi }=\left|\overrightarrow{\mathrm{E}}\right|\mathrm{Acos}\mathrm{\theta }$……………….(1)
Substituting the values in Eq. (1), we get
$\varphi =3\times {10}^3\times 0.01\times \cos 0^{\circ }=30{\mathrm{Nm}}^2/\mathrm{C}$.