Electric flux $\Phi_E$ is defined as the scalar product of the electric field vector $\mathbf{E}$ and the area vector $\mathbf{A}$: $\Phi_E = \mathbf{E} \cdot \mathbf{A} = EA \cos \alpha$

Where $\alpha$ is the angle between the electric field lines and the normal (perpendicular) to the surface area , .

1. Direction of Area Vector ($\,mathbf{A}$): The square surface lies in the plane of the paper. By definition, the area vector is always perpendicular to the surface. Therefore, $\mathbf{A}$ is perpendicular to the plane of the paper .
2. Direction of Electric Field ($\,mathbf{E}$): The problem states that the electric field acts along the same plane (the plane of the paper).
3. Angle $\alpha$: Since the area vector is perpendicular to the paper and the electric field is parallel to the paper, the angle between them is $90^\circ$ ($\pi/2$).

Substituting this into the flux formula: $\Phi_E = EA \cos(90^\circ) = EA(0) = 0$.

Physically, this means the electric field lines graze the surface parallel to it and do not penetrate through it, resulting in zero flux. The angle $\theta$ given in the problem (between $\mathbf{E}$ and the side of the square) is irrelevant to the flux calculation in this orientation.