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(D) According to Gauss's Law,the total electric flux through a closed surface is $\Phi_{total} = \frac{q}{\varepsilon_0}$.In this problem,the cubical

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$q/{5\varepsilon _0}$

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(D) According to Gauss's Law,the total electric flux through a closed surface is $\Phi_{total} = \frac{q}{\varepsilon_0}$.In this problem,the cubical box has an open top,meaning it is not a fully closed surface. To calculate the flux,we can imagine an identical cubical box placed on top of the existing one to enclose the charge $q$ completely.Now,the charge $q$ is at the center of a larger closed cubical volume formed by two such boxes. The total flux through this combined closed surface is $\Phi_{total} = \frac{q}{\varepsilon_0}$.Due to symmetry,the flux through each of the $10$ faces (excluding the two open faces which are now joined) is equal. However,a simpler way is to consider that the total flux through the $5$ closed faces of the original box is half of the total flux through the closed system of two boxes.Therefore,the flux through the $5$ faces of the original box is $\Phi = \frac{1}{2} 	imes \frac{q}{\varepsilon_0} = \frac{q}{2\varepsilon_0}$.Note: If the charge were at the center of the entire cube,the flux through one face would be $q/6\varepsilon_0$. Since the charge is at the center of the base,the flux through the base is $0$ (as the field lines are parallel to the surface),and the flux is distributed among the $4$ side walls.

$A$ charge $q$ is placed at the centre of the base of a cubical box of side $a$ with the top open. The flux of the electric field through the open surface is not considered,but the flux through the remaining five surfaces of the cubical box is:

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