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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\epsilon_0}$.To calculate the flux through the c

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$\frac{q}{2 \epsilon_0}$

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(B) According to Gauss's Law,the total electric flux through a closed surface is $\frac{q_{enclosed}}{\epsilon_0}$.To calculate the flux through the cube when the charge $q$ is placed on one of its faces,we imagine an identical cube placed on top of the first one such that the charge $q$ is now at the center of a closed Gaussian surface formed by the two cubes.The total flux through this combined Gaussian surface is $\phi_{total} = \frac{q}{\epsilon_0}$.Since the charge is placed symmetrically between the two cubes,the flux through each cube is equal.Therefore,the flux linked with one cube is $\phi = \frac{1}{2} \phi_{total} = \frac{q}{2 \epsilon_0}$.

$A$ charge $q$ is placed at the center of one of the faces of a cube. What is the electric flux linked with the cube?

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