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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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$\frac{q}{8\varepsilon_0}$

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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 a vertex of a cube, the cube occupies only $\frac{1}{8}$ of the total space surrounding the charge because $8$ identical cubes are required to enclose the charge completely at the center of a larger cube.Therefore, the electric flux passing through the cube is $\Phi = \frac{1}{8} 	imes \frac{q}{\varepsilon_0} = \frac{q}{8\varepsilon_0}$.

If charge $q$ is placed on one of the vertex of a cube, then total electric flux passing through the cube is . . . . . . .

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