In a cuboid of dimension 2 L times 2 L times | vedclass

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(D) According to Gauss's Law,the total flux through a closed surface enclosing a charge $q$ is $\phi_{total} = \frac{q}{\varepsilon_0}$.To calculate t

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$\frac{q}{6 \varepsilon_0}$

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(D) According to Gauss's Law,the total flux through a closed surface enclosing a charge $q$ is $\phi_{total} = \frac{q}{\varepsilon_0}$.To calculate the flux through a specific surface,we can use the method of symmetry. Place an identical cuboid of dimension $2 L 	imes 2 L 	imes L$ on top of the given cuboid such that the charge $q$ lies on the common interface of the two cuboids.Now,the charge $q$ is enclosed by a larger cuboid of dimension $2 L 	imes 2 L 	imes 2 L$,which is effectively a cube of side $2 L$.The total flux through this larger closed surface is $\frac{q}{\varepsilon_0}$.By symmetry,the flux through each of the $6$ faces of this larger cube is equal.Therefore,the flux through one face of the larger cube is $\phi_{face} = \frac{1}{6} \left( \frac{q}{\varepsilon_0} ight) = \frac{q}{6 \varepsilon_0}$.Since the surface ' $S$ ' and its opposite surface are part of the faces of this larger cube,the flux through the opposite surface is $\frac{q}{6 \varepsilon_0}$.

In a cuboid of dimension $2 L \times 2 L \times L$,a charge $q$ is placed at the centre of the surface ' $S$ ' having an area of $4 L^2$. The flux through the opposite surface to ' $S$ ' is given by

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