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

Vedclass · Accepted Answer

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

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(A) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{	ext{enclosed}}}{\varepsilon_0}$.$1$. Disk: The disk has radius $a/4$ and center at $(-a/2, 0, 0)$. The cube extends from $x = -a/2$ to $x = a/2$. Since the disk is in the $x-y$ plane,only the part of the disk with $x > -a/2$ is inside the cube. The disk spans from $x = -a/2 - a/4 = -3a/4$ to $x = -a/2 + a/4 = -a/4$. The portion inside the cube is from $x = -a/2$ to $x = -a/4$. By symmetry,exactly half of the disk is inside the cube. Thus,$Q_{	ext{disk, enclosed}} = 6 	ext{ C} / 2 = 3 	ext{ C}$.$2$. Rod: The rod spans from $x = a/4$ to $x = 5a/4$. The cube boundary is at $x = a/2$. The portion of the rod inside the cube is from $x = a/4$ to $x = a/2$. The length of the rod inside the cube is $a/2 - a/4 = a/4$. Since the total length is $a$,the fraction of charge inside is $(a/4) / a = 1/4$. Thus,$Q_{	ext{rod, enclosed}} = 8 	ext{ C} 	imes (1/4) = 2 	ext{ C}$.$3$. Point charges: The point charge $-7 	ext{ C}$ is at $(a/4, -a/4, 0)$,which is inside the cube (since $|a/4|  a/2$).$4$. Total enclosed charge: $Q_{	ext{enclosed}} = 3 	ext{ C} + 2 	ext{ C} - 7 	ext{ C} = -2 	ext{ C}$.Therefore,the electric flux is $\phi = \frac{-2 	ext{ C}}{\varepsilon_0}$.

List-$I$	List-$II$
$(a)$ Net flux through $S_1$	$(i)$ $\frac{-2Q}{\epsilon_0}$
$(b)$ Net flux through $S_2$	(ii) $\frac{-Q}{\epsilon_0}$
$(c)$ Net flux through $S_3$	(iii) Zero

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