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(C) According to Gauss's Law,the total flux through a closed surface is $\frac{q}{\varepsilon_0}$.To enclose a charge $q$ placed at a vertex of a cube

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$\frac{q}{24 \varepsilon_0}$

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(C) According to Gauss's Law,the total flux through a closed surface is $\frac{q}{\varepsilon_0}$.To enclose a charge $q$ placed at a vertex of a cube,we need $8$ identical cubes to form a larger symmetric cube such that the charge $q$ is at the center of this larger cube.The total flux through the large cube is $\frac{q}{\varepsilon_0}$.Since the charge is at the corner,the flux through the original cube is $\frac{1}{8}$ of the total flux,which is $\frac{q}{8 \varepsilon_0}$.Out of the $6$ faces of the original cube,$3$ faces meet at the vertex where the charge is placed. The electric field lines are parallel to these $3$ faces,so the flux through these $3$ faces is $0$.The remaining $3$ faces share the flux equally.Therefore,the flux through any one of these $3$ faces is $\frac{1}{3} 	imes \frac{q}{8 \varepsilon_0} = \frac{q}{24 \varepsilon_0}$.

If a charge $q$ is placed on one of the vertices of a cube,then the flux passing through any one face of the cube is . . . . . . .

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