Electric Flux Through Cube Faces with Enclosed Point Charges

To solve this problem, we analyze the charge distribution and apply Gauss's Law along with spatial symmetry arguments.

1. Total Electric Flux Through the Cube (Option C)

According to Gauss's Law, the total electric flux $\Phi_{\text{total}}$ passing through any closed surface is proportional to the total net charge enclosed within that surface: $\Phi_{\text{total}} = \frac{Q_{\text{enclosed}}}{\varepsilon_0}$

The cubical region of side length $a$ centered at the origin encloses three point charges:

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

The net enclosed charge $Q_{\text{enclosed}}$ is: $Q_{\text{enclosed}} = q_1 + q_2 + q_3 = (-q) + (+3q) + (-q) = +q$

Therefore, the total flux through the cubical surface is: $\Phi_{\text{total}} = \frac{q}{\varepsilon_0}$

Thus, Option C is correct.

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2. Comparison of Flux through $x = +a/2$ and $x = -a/2$ (Option A)

All three charges lie on the $y$-axis ($x = 0, z = 0$).

Since the charge distribution is symmetric about the $y$-$z$ plane ($x = 0$), performing a reflection across $x = 0$ ($x \to -x$) leaves the charge distribution completely unchanged.

By this mirror symmetry, the electric field lines exiting through the face at $x = +a/2$ are symmetric to those exiting through the face at $x = -a/2$. Thus, the net electric flux crossing the plane $x = +a/2$ is equal to that crossing the plane $x = -a/2$: $\Phi_{x = +a/2} = \Phi_{x = -a/2}$

Thus, Option A is correct.

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3. Comparison of Flux through $y = +a/2$ and $y = -a/2$ (Option B)

The charges are located at $y = -a/4$, $y = 0$, and $y = +a/4$.

Notice that the charge distribution is symmetric under reflection across the $x$-$z$ plane ($y \to -y$), because:

• The charge at $y = -a/4$ is $-q$.
• The charge at $y = +a/4$ is $-q$.
• The charge at $y = 0$ is $+3q$.

Reflecting the system across $y = 0$ swaps the two $-q$ charges and leaves $+3q$ at the origin, resulting in the exact same charge configuration. Therefore, the net electric flux through the face $y = +a/2$ must be equal to the net electric flux through the face $y = -a/2$: $\Phi_{y = +a/2} = \Phi_{y = -a/2}$

Since the flux through $y = +a/2$ is equal to (not greater than) the flux through $y = -a/2$, Option B is incorrect.

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4. Comparison of Flux through $z = +a/2$ and $x = +a/2$ (Option D)

Since all three charges lie entirely along the $y$-axis, the charge distribution possesses continuous rotational symmetry (cylindrical symmetry) about the $y$-axis.

Rotating the system about the $y$-axis by $90^\circ$ maps the $x$-axis onto the $z$-axis. The square face located at $x = +a/2$ (defined by $x = a/2, -a/2 \le y \le a/2, -a/2 \le z \le a/2$) maps perfectly onto the square face located at $z = +a/2$ (defined by $z = a/2, -a/2 \le x \le a/2, -a/2 \le y \le a/2$).

Due to this axial symmetry about the $y$-axis, the electric flux through the plane $z = +a/2$ is equal to the electric flux through the plane $x = +a/2$: $\Phi_{z = +a/2} = \Phi_{x = +a/2}$

Thus, Option D is correct.

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Conclusion

The correct options are A, C, and D.