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(C) $1$. The charge $+Q$ at the center $O$ contributes a total flux of $\frac{Q}{\varepsilon_{0}}$ through the entire cube. By symmetry, the flux thro

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$\frac{10Q}{6\varepsilon_{0}}$

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(C) $1$. The charge $+Q$ at the center $O$ contributes a total flux of $\frac{Q}{\varepsilon_{0}}$ through the entire cube. By symmetry, the flux through each of the $6$ faces is $\frac{Q}{6\varepsilon_{0}}$.$2$. The charge $+Q$ at point $P$ is located at a distance $a/2$ from the center of the face $ABCD$. This charge creates an electric field. The flux due to charge $+Q$ at $P$ through the face $ABCD$ is $\frac{Q}{2\varepsilon_{0}}$ (since the charge is at a distance $a/2$ from the center of the face, it subtends a solid angle of $2\pi$ steradians at the face, or by considering an identical cube placed adjacent to the first one).$3$. The net flux through face $ABCD$ due to both charges is $\phi_{ABCD} = \phi_{O, ABCD} + \phi_{P, ABCD} = \frac{Q}{6\varepsilon_{0}} - \frac{Q}{2\varepsilon_{0}} = -\frac{Q}{3\varepsilon_{0}}$ (the flux from $P$ enters the cube, hence negative).$4$. Total flux through the cube due to both charges is $\phi_{total} = \frac{Q_{enclosed}}{\varepsilon_{0}} = \frac{Q}{\varepsilon_{0}}$.$5$. The flux through the remaining $5$ faces is $\phi_{5} = \phi_{total} - \phi_{ABCD} = \frac{Q}{\varepsilon_{0}} - (-\frac{Q}{3\varepsilon_{0}}) = \frac{4Q}{3\varepsilon_{0}} = \frac{8Q}{6\varepsilon_{0}}$.

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