Consider a uniform electric field $E = 3 \times 10^{3} \hat{i} \; N/C$.
$(a)$ What is the flux of this field through a square of $10 \; cm$ on a side whose plane is parallel to the $yz$ plane?
$(b)$ What is the flux through the same square if the normal to its plane makes a $60^{\circ}$ angle with the $x$-axis?

(A) Electric field intensity,$E = 3 \times 10^{3} \hat{i} \; N/C$.
Magnitude of electric field intensity,$|E| = 3 \times 10^{3} \; N/C$.
Side of the square,$s = 10 \; cm = 0.1 \; m$.
Area of the square,$A = s^{2} = 0.01 \; m^{2}$.
The plane of the square is parallel to the $yz$ plane. Hence,the angle between the unit vector normal to the plane and the electric field is $\theta = 0^{\circ}$.
Flux $(\phi)$ through the plane is given by the relation,$\phi = |E| A \cos \theta = 3 \times 10^{3} \times 0.01 \times \cos 0^{\circ} = 30 \; N \cdot m^{2}/C$.
$(b)$ The normal to the plane makes an angle of $60^{\circ}$ with the $x$-axis. Hence,$\theta = 60^{\circ}$.
Flux,$\phi = |E| A \cos \theta = 3 \times 10^{3} \times 0.01 \times \cos 60^{\circ}$.
$\phi = 30 \times \frac{1}{2} = 15 \; N \cdot m^{2}/C$.