The electric field components in the figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \; N/C \cdot m^{1/2}.$ Calculate
$(a)$ the flux through the cube,and
$(b)$ the charge within the cube. Assume that $a=0.1 \; m$.

(N/A) Since the electric field has only an $x$ component,for faces perpendicular to the $x$ direction,the angle between $E$ and $\Delta S$ is $\pm \pi/2$. Therefore,the flux $\phi = E \cdot \Delta S$ is zero for each face of the cube except the two shaded ones.
The magnitude of the electric field at the left face is $E_{L} = \alpha x^{1/2} = \alpha a^{1/2}$ (at $x=a$).
The magnitude of the electric field at the right face is $E_{R} = \alpha x^{1/2} = \alpha (2a)^{1/2}$ (at $x=2a$).
The corresponding fluxes are:
$\phi_{L} = E_{L} \cdot \Delta S = E_{L} \Delta S \cos(180^{\circ}) = -E_{L} a^{2} = -\alpha a^{1/2} a^{2} = -\alpha a^{5/2}$.
$\phi_{R} = E_{R} \cdot \Delta S = E_{R} \Delta S \cos(0^{\circ}) = E_{R} a^{2} = \alpha (2a)^{1/2} a^{2} = \alpha \sqrt{2} a^{5/2}$.
Net flux through the cube $\phi = \phi_{R} + \phi_{L} = \alpha a^{5/2} (\sqrt{2} - 1)$.
Substituting the values: $\phi = 800 \times (0.1)^{5/2} \times (1.414 - 1) = 800 \times 0.003162 \times 0.414 \approx 1.05 \; N \cdot m^{2} \cdot C^{-1}$.
$(b)$ Using Gauss's law,the total charge $q$ inside the cube is $q = \phi \varepsilon_{0}$.
$q = 1.05 \times 8.854 \times 10^{-12} \; C \approx 9.27 \times 10^{-12} \; C$.