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(B) Let the charge at the center $O$ be $q_1 = q$ and the charge outside be $q_2 = q$. By symmetry, the electric flux through the face $BGFC$ due to t

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(B) Let the charge at the center $O$ be $q_1 = q$ and the charge outside be $q_2 = q$. By symmetry, the electric flux through the face $BGFC$ due to the charge $q_1$ at the center $O$ is $\phi_1 = q / (6\varepsilon_0)$, because the charge is at the center of the cube and the flux is distributed equally among the $6$ faces. Now consider the charge $q_2$ placed at a distance $L$ from $O$. The face $BGFC$ is located at a distance $L/2$ from the center $O$. The charge $q_2$ is placed at a distance $L$ from $O$ along the line passing through the center of the face $BGFC$. However, a simpler way to view this is by symmetry. The electric field lines from the charge $q_2$ enter the face $BGFC$ and exit through the opposite face. Specifically, for the face $BGFC$, the flux due to the internal charge $q_1$ is $q / (6\varepsilon_0)$. For the external charge $q_2$, the net flux through the entire closed cube is zero because the charge is outside. Due to the specific geometry, the flux through face $BGFC$ due to $q_2$ is exactly $-q / (6\varepsilon_0)$. Therefore, the total flux through the face $BGFC$ is $\phi_{total} = \phi_1 + \phi_2 = q / (6\varepsilon_0) - q / (6\varepsilon_0) = 0$.

$A$ charged particle $q$ is placed at the centre $O$ of a cube of side length $L$ $(A, B, C, D, E, F, G, H)$. Another identical charge $q$ is placed at a distance $L$ from $O$, outside the cube, as shown in the figure. The electric flux through the face $BGFC$ is

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