Gauss's Law

Consider a cube of side $a$ oriented with its left face at $x = a$ and right face at $x = 2a$, subject to a non-uniform electric field $E_x = \alpha x^{1/2}$ and $E_y = E_z = 0$. Since the electric field has only an $x$ component, the flux through the four faces perpendicular to the $y$ and $z$ axes is zero.

At the left face ($x = a$), the magnitude of the electric field is $E_L = \alpha a^{1/2}$. The angle between $\mathbf{E}$ and the outward normal $\Delta \mathbf{S}$ is $180^\circ$, so the flux is $\phi_L = E_L a^2 \cos 180^\circ = -$ .

At the right face ($x = 2a$), the electric field magnitude is $E_R = \alpha (2a)^{1/2}$. Here, the angle between the field and the outward normal is $0^\circ$, so the flux is $\phi_R = E_R a^2 \cos 0^\circ =$ .

The net flux through the cube is the algebraic sum of these components, which factors into $\phi = \phi_R + \phi_L =$ .