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(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{net}}}{\epsilon_0}$.When a poi

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$\phi = -\frac{1}{2\epsilon_0}$

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(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{net}}}{\epsilon_0}$.When a point charge $q$ is placed at one of the corners of a cube,only $\frac{1}{8}$ of the flux from that charge passes through the cube.Therefore,the flux through the cube due to a charge $q$ at a corner is $\phi = \frac{q}{8\epsilon_0}$.For the given system,the net charge $q_{	ext{net}}$ at the corners is:$q_{	ext{net}} = (+1 - 2 + 3 - 4 - 6 + 7 + 5 - 8) 	ext{ C} = -4 	ext{ C}$.However,since each charge is at a corner,the total flux $\phi$ through the cube is:$\phi = \frac{1}{8\epsilon_0} \sum q_i = \frac{1}{8\epsilon_0} (1 - 2 + 3 - 4 - 6 + 7 + 5 - 8)$$\phi = \frac{-4}{8\epsilon_0} = -\frac{1}{2\epsilon_0}$.

The electric flux passing through the cube for the given arrangement of charges placed at the corners of the cube (as shown in the figure) is:

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