The problem asks us to evaluate four statements about the electric flux through the faces of a cube centered at the origin, which encloses three point charges. We will analyze each option using Gauss's Law and symmetry principles.

The given charges are:

• $q_1 = -q$ at $(0, -a/4, 0)$
• $q_2 = +3q$ at $(0, 0, 0)$
• $q_3 = -q$ at $(0, +a/4, 0)$

The cube has side length 'a' and is centered at the origin. Its faces are the planes $x=\pm a/2$, $y=\pm a/2$, and $z=\pm a/2$.

Evaluation of Option C

C: The net electric flux crossing the entire region is ${q \over {{\varepsilon _0}}}$.

1. Gauss's Law: The total electric flux $\Phi_{total}$ through a closed surface is given by Gauss's Law: $\Phi_{total} = \frac{Q_{enc}}{\varepsilon_0}$ where $Q_{enc}$ is the total charge enclosed by the surface.
2. Calculate Enclosed Charge: The cube encloses all three charges. The total enclosed charge is the sum of the individual charges: $Q_{enc} = q_1 + q_2 + q_3 = (-q) + (+3q) + (-q) = q$
3. Calculate Total Flux: Using Gauss's Law, the net electric flux crossing the entire cubical region is: $\Phi_{total} = \frac{q}{\varepsilon_0}$
4. Conclusion: Statement C is correct.

Evaluation of Option A

A: The net electric flux crossing the plane $x=+a/2$ is equal to the net electric flux crossing the plane $x=-a/2$.

1. Symmetry Analysis: Let's consider the symmetry of the charge distribution with respect to the $y-z$ plane (the plane $x=0$). All three charges lie on the y-axis, which is contained within the $y-z$ plane. This charge distribution is symmetric with respect to reflection across the $x=0$ plane.
2. Electric Field Symmetry: Due to this symmetry, the x-component of the electric field must be an odd function of $x$. That is, for any point $(x, y, z)$, the electric field components satisfy $E_x(-x, y, z) = -E_x(x, y, z)$.
3. Flux Calculation:
   • The flux through the face at $x=+a/2$ (let's call it the right face) is given by $\Phi_{right} = \int_{right} \vec{E} \cdot d\vec{A}$. Here, $d\vec{A} = dy dz \, \hat{i}$. So, $\Phi_{right} = \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} E_x(a/2, y, z) \, dy dz$
   • The flux through the face at $x=-a/2$ (left face) is $\Phi_{left} = \int_{left} \vec{E} \cdot d\vec{A}$. Here, $d\vec{A} = dy dz \, (-\hat{i})$. So, $\Phi_{left} = \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} E_x(-a/2, y, z) \, (-1) \, dy dz$
4. Comparing Fluxes: Using the symmetry property $E_x(-a/2, y, z) = -E_x(a/2, y, z)$, we get: $\Phi_{left} = -\int_{-a/2}^{a/2} \int_{-a/2}^{a/2} [-E_x(a/2, y, z)] \, dy dz = \int_{-a/2}^{a/2} \int_{-a/2}^{a/2} E_x(a/2, y, z) \, dy dz = \Phi_{right}$
5. Conclusion: The fluxes are equal. Statement A is correct.

Evaluation of Option B

B: The net electric flux crossing the plane $y=+a/2$ is more than the net electric flux crossing the plane $y=-a/2$.

1. Symmetry Analysis: Let's consider the symmetry of the charge distribution with respect to the $x-z$ plane (the plane $y=0$). The charges are located at $(0, -a/4, 0)$, $(0, 0, 0)$, and $(0, +a/4, 0)$. This distribution is symmetric upon reflection across the $y=0$ plane (i.e., replacing $y$ with $-y$). The two $-q$ charges swap positions, and the $+3q$ charge remains on the plane.
2. Electric Field Symmetry: Due to this symmetry, the y-component of the electric field must be an odd function of $y$: $E_y(x, -y, z) = -E_y(x, y, z)$.
3. Flux Calculation: A similar analysis as for Option A can be done.
   • Flux through the top face ($y=+a/2$): $\Phi_{top} = \int E_y(x, a/2, z) \, dx dz$.
   • Flux through the bottom face ($y=-a/2$): $\Phi_{bottom} = \int E_y(x, -a/2, z) \, (-1) \, dx dz$.
4. Comparing Fluxes: Using $E_y(x, -a/2, z) = -E_y(x, a/2, z)$, we find: $\Phi_{bottom} = -\int [-E_y(x, a/2, z)] \, dx dz = \int E_y(x, a/2, z) \, dx dz = \Phi_{top}$
5. Conclusion: The flux through the top face is equal to the flux through the bottom face. Therefore, statement B is incorrect.

Evaluation of Option D

D: The net electric flux crossing the plane $z = + a/2$ is equal to the net electric flux crossing the plane $x=+a/2$.

1. Symmetry Analysis: Consider the symmetry of the charge distribution with respect to rotations about the y-axis. All charges lie on the y-axis. Therefore, the charge distribution has rotational (cylindrical) symmetry about the y-axis.
2. Geometric Symmetry: The cube itself is symmetric under a $90^\circ$ rotation about the y-axis. This rotation transforms the face at $x=+a/2$ into the face at $z=+a/2$. Similarly, it transforms the faces $z=+a/2 \to x=-a/2 \to z=-a/2 \to x=+a/2$.
3. Comparing Fluxes: Since both the source of the field (the charges) and the geometry of the faces are symmetric under this rotation, the physical situation is identical for the four side faces ($x=\pm a/2, z=\pm a/2$). Therefore, the electric flux through these faces must be related by this symmetry.
4. Specifically, the flux through the face at $x=+a/2$ must be equal to the flux through the face at $z=+a/2$. That is, $\Phi_{x=+a/2} = \Phi_{z=+a/2}$.
5. Conclusion: Statement D is correct.

Summary: Based on the analysis:

• Option A is correct.
• Option B is incorrect.
• Option C is correct.
• Option D is correct.