By convention, negative flux indicates that there is a net flow of electric field lines into the closed surface. According to Gauss's Law, the net electric flux is proportional to the net charge enclosed ($$\Phi_E = q_{enc}/\epsilon_0$$). If the net flux is negative, the net enclosed charge must also be negative.

The square surface is in the y-z plane at a constant $$x$$ position of 2.0 m. The area vector is $$\vec{A} = L^2 \hat{i}$$. Although the field is non-uniform in space, it is uniform over this specific surface because $$x$$ is constant across it. At $$x = 2.0$$ m, the field is $$\vec{E} = (3(2.0)^2) \hat{i} = 12 \hat{i} \ \text{N/C}$$. The flux is $$\Phi_E = \vec{E} \cdot \vec{A} = (12 \hat{i}) \cdot(L^2 \hat{i}) = 12L^2 \ \text{N} \cdot \text{m}^2 / \text{C}$$.

The rectangular area is in the xz-plane. For any point on this surface, the y-coordinate is zero. The electric field is given by $$\vec{E} = c y^2 \hat{j}$$. Substituting $$y=0$$ into this expression, we find that the electric field is zero everywhere on the surface. Therefore, the electric flux through the surface must be zero.

Initially, the surface normal is parallel to the electric field, so the angle $$\theta$$ between them is 0°. The flux is $$\Phi_0 = EA \cos(0^\circ) = EA$$. After a 90-degree rotation about an axis in its plane, the surface normal becomes perpendicular to the electric field. The new angle is $$\theta' = 90^\circ$$. The new flux is $$\Phi' = EA \cos(90^\circ) = 0$$.

Consider the closed surface formed by the hemisphere and its flat circular base. Since the electric field is uniform, there is no enclosed charge, and the net flux through this closed surface is zero. The flux through the base is $$\Phi_{base} = \vec{E} \cdot \vec{A}{base}$$. The area vector for the base points outward, so $$\vec{A}{base} = -\pi R^2 \hat{k}$$. Thus, $$\Phi_{base} = (E_0 \hat{k}) \cdot(-\pi R^2 \hat{k}) = -\pi R^2 E_0$$. Since $$\Phi_{net} = \Phi_{curved} + \Phi_{base} = 0$$, the flux through the curved surface is $$\Phi_{curved} = -\Phi_{base} = \pi R^2 E_0$$.

Electric flux is given by $$\Phi_E = \vec{E} \cdot \vec{A} = EA \cos\theta$$. Since the field is perpendicular to the plane of the surface, it is parallel to the normal vector $$\vec{A}$$. Thus, the angle $$\theta$$ between $$\vec{E}$$ and $$\vec{A}$$ is 0°, and $$\cos(0°) = 1$$. The area of the circular surface is $$A = \pi r^2$$. Therefore, the flux is $$\Phi_E = E(\pi r^2)(1) = E \pi r^2$$.

Electric flux is given by $$\Phi_E = EA \cos\theta$$. To get the largest negative value, $$\cos\theta$$ must be -1, which occurs when $$\theta = 180°$$. This means the electric field vector and the surface normal vector must be anti-parallel. Since the electric field points upward, the paper's normal vector must point downward. This corresponds to holding the paper horizontally.

This question is a statement of Gauss's Law, which states that the net electric flux through any closed surface is directly proportional to the net electric charge enclosed within that surface ($$\Phi_E = q_{enc} / \epsilon_0$$). Flux does not generally depend directly on the surface area, field magnitude, or volume in this manner.

The surface is in the xy-plane, so the differential area vector is $$d\vec{A} = dx,dy,\hat{k}$$. The flux is the surface integral of $$\vec{E} \cdot d\vec{A}$$. $$\Phi_E = \int \vec{E} \cdot d\vec{A} = \int_0^2 \int_0^2 (5y \hat{k}) \cdot(dx,dy,\hat{k}) = \int_0^2 dx \int_0^2 5y ,dy$$. Evaluating the integrals gives: $$\Phi_E = [x]_0^2 \cdot[5y^2/2]_0^2 = (2) \cdot(5(2^2)/2) = (2)(10) = 20 \text{ N} \cdot \text{m}^2/\text{C}$$

For any closed surface in a uniform electric field, the net electric flux is zero. This is because every field line that enters the surface at one point must also leave it at another. The flux entering through one face is negative, and the flux leaving through the opposite face is positive and equal in magnitude. The flux through the other four faces is zero.