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(A) According to Gauss's Law,the electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.$1$. The

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$\frac{-2 	ext{ C}}{\varepsilon_0}$

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(A) According to Gauss's Law,the electric flux $\Phi$ through a closed surface is given by $\Phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.$1$. The disk has a radius $a/4$ and center at $(-a/2, 0, 0)$. The cube extends from $x = -a/2$ to $x = a/2$. The disk lies in the $x-y$ plane. The part of the disk inside the cube is the region from $x = -a/2$ to $x = -a/2 + a/4 = -a/4$. This is exactly half of the disk. Thus,the charge enclosed is $6 	ext{ C} / 2 = 3 	ext{ C}$.$2$. The rod has length $a$ and is placed from $x = a/4$ to $x = 5a/4$. The cube extends up to $x = a/2$. The part of the rod inside the cube is from $x = a/4$ to $x = a/2$,which is a length of $a/4$. Since the total length is $a$,the fraction of charge enclosed is $(a/4) / a = 1/4$. Thus,the charge enclosed is $8 	ext{ C} 	imes (1/4) = 2 	ext{ C}$.$3$. The point charge $-7 	ext{ C}$ is at $(a/4, -a/4, 0)$,which is inside the cube $(|x| $4$. The point charge $3 	ext{ C}$ is at $(-3a/4, 3a/4, 0)$,which is outside the cube.Total enclosed charge $q_{	ext{enclosed}} = 3 	ext{ C} + 2 	ext{ C} - 7 	ext{ C} = -2 	ext{ C}$.Therefore,the electric flux $\Phi = \frac{-2 	ext{ C}}{\varepsilon_0}$.

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