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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_{0}}$.When a charge $q$ is placed at

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$\frac{q}{24 \varepsilon_{0}}$

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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\varepsilon_{0}}$.When a charge $q$ is placed at one corner of a cube,it is shared equally by $8$ such identical cubes to enclose the charge completely.Therefore,the flux through one cube is $\Phi_{cube} = \frac{q}{8 \varepsilon_{0}}$.$A$ cube has $6$ faces. The charge $q$ is located at a corner. The three faces meeting at this corner (in this case,faces $ADHE$,$ABFE$,and $ABCD$ if the charge is at $A$) contain the charge $q$ on their plane. For these faces,the electric field lines are parallel to the surface,meaning the angle between the area vector and the electric field is $90^{\circ}$. Thus,the flux through these $3$ faces is $0$.The remaining $3$ faces share the total flux $\Phi_{cube}$ equally due to symmetry.Therefore,the flux through each of the remaining $3$ faces is $\Phi_{face} = \frac{\Phi_{cube}}{3} = \frac{q / 8 \varepsilon_{0}}{3} = \frac{q}{24 \varepsilon_{0}}$.Thus,the electric flux through the face $ABCD$ is $\frac{q}{24 \varepsilon_{0}}$.

$A$ point charge $q$ is placed at the corner of a cube of side $a$ as shown in the figure. What is the electric flux through the face $ABCD$?

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