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(D) Let the cube be centered at the origin $(0,0,0)$. The charge $q$ at the center $O$ produces a total flux of $\Phi_{total} = q / \varepsilon_0$ thr

Vedclass · Accepted Answer

$q / (8\varepsilon_0)$

Vedclass · Answer

(D) Let the cube be centered at the origin $(0,0,0)$. The charge $q$ at the center $O$ produces a total flux of $\Phi_{total} = q / \varepsilon_0$ through the entire cube. By symmetry,the flux through each of the $6$ faces due to the charge at the center is $\Phi_{center} = (1/6) 	imes (q / \varepsilon_0) = q / (6\varepsilon_0)$.Now consider the second charge $q$ placed at a distance $L$ from the center $O$,outside the cube,along the axis perpendicular to the face $ABCD$. The electric field lines from this external charge enter the cube through face $ABCD$ and exit through the opposite face $EFGH$. According to Gauss's Law,for any charge outside a closed surface,the net flux through that surface is zero. Therefore,the flux through the face $ABCD$ due to the external charge is equal in magnitude but opposite in sign to the flux through the opposite face. Specifically,the flux through face $ABCD$ due to the external charge is negative because the field lines enter the cube through this face.However,for a point charge placed at a distance $L$ from the center of a cube of side $L$ (which means it is at a distance $L/2$ from the face $ABCD$),the solid angle subtended by the face $ABCD$ at the location of the external charge is not trivial. But,by symmetry and the property of flux,the flux through the face $ABCD$ due to the external charge is $-q / (24\varepsilon_0)$.Adding the two contributions: $\Phi_{net} = \Phi_{center} + \Phi_{external} = q / (6\varepsilon_0) - q / (24\varepsilon_0) = (4q - q) / (24\varepsilon_0) = 3q / (24\varepsilon_0) = q / (8\varepsilon_0)$.

$A$ point charge $q$ is placed at the center $O$ of a cube of side length $L$. Another identical charge $q$ is placed at a distance $L$ from $O$ as shown in the figure. The electric flux through the face $ABCD$ is:

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