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(C) According to Gauss's Law, the net electric flux $\phi_{	ext{net}}$ through the closed surface is $\phi_{	ext{net}} = \oint \vec{E} \cdot d\vec{A

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(C) According to Gauss's Law, the net electric flux $\phi_{	ext{net}}$ through the closed surface is $\phi_{	ext{net}} = \oint \vec{E} \cdot d\vec{A} = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.Given $\vec{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k}$.The flux through the faces perpendicular to the $x$-axis (area $A_x = 2 	imes 3 = 6 \; 	ext{m}^2$): $\phi_x = \phi_{x=1} - \phi_{x=0} = [2(1)^2 	imes 6] - [2(0)^2 	imes 6] = 12 \; 	ext{N}\cdot	ext{m}^2/	ext{C}$.The flux through the faces perpendicular to the $y$-axis (area $A_y = 1 	imes 3 = 3 \; 	ext{m}^2$): $\phi_y = \phi_{y=2} - \phi_{y=0} = [-4(2) 	imes 3] - [-4(0) 	imes 3] = -24 \; 	ext{N}\cdot	ext{m}^2/	ext{C}$.The flux through the faces perpendicular to the $z$-axis (area $A_z = 1 	imes 2 = 2 \; 	ext{m}^2$): $\phi_z = \phi_{z=3} - \phi_{z=0} = [6 	imes 2] - [6 	imes 2] = 0 \; 	ext{N}\cdot	ext{m}^2/	ext{C}$.Total flux $\phi_{	ext{net}} = 12 - 24 + 0 = -12 \; 	ext{N}\cdot	ext{m}^2/	ext{C}$.Since $\phi_{	ext{net}} = \frac{q}{\varepsilon_0}$, we have $q = -12 \varepsilon_0$. The magnitude of the charge is $|q| = 12 \varepsilon_0$.Thus, $n = 12$.

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