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(C) According to Gauss's Law,the total electric flux through a closed surface enclosing a charge $q$ is $\Phi_{total} = q/\varepsilon_0$.In this probl

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$q/{6\varepsilon _0}$

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(C) According to Gauss's Law,the total electric flux through a closed surface enclosing a charge $q$ is $\Phi_{total} = q/\varepsilon_0$.In this problem,the cubical box has one side open,meaning it is not a fully closed surface. However,the charge $q$ is placed at the center of the cube. By symmetry,the total flux through the entire cube (if it were closed) would be $q/\varepsilon_0$.Since the cube has $6$ identical faces and the charge is at the center,the flux through each face is $\Phi_{face} = \frac{1}{6} \Phi_{total} = \frac{q}{6\varepsilon_0}$.Therefore,the flux through any one of the five closed surfaces is $q/(6\varepsilon_0)$.

$A$ charge $q$ is placed at the center of a cubical box of side $a$ with the top open. The flux of the electric field through one of the five closed surfaces of the cubical box is:

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