Electrostatics Part 3

Concept:Gauss's Law: Total electric flux through a closed surface equals the charge enclosed divided by $\epsilon_0$.Explanation:Charge $2q$ is at a vertex of the cube.A vertex is shared by 8 cubes, so only $1/8$ of its flux enters this cube.Flux from $2q$: $\Phi_A = \frac{1}{8} \times \frac{2q}{\epsilon_0} = \frac{q}{4\epsilon_0}$.Charge $q$ is at the centre of face CDEF.A face is shared by 2 cubes, so only $1/2$ of its flux enters this cube.Flux from $q$: $\Phi_{\text{face}} = \frac{1}{2} \times \frac{q}{\epsilon_0} = \frac{q}{2\epsilon_0}$.Total flux: $\Phi_{\text{total}} = \Phi_A + \Phi_{\text{face}} = \frac{q}{4\epsilon_0} + \frac{q}{2\epsilon_0} = \frac{3q}{4\epsilon_0}$.Answer:The electric flux passing through the cube is $\frac{3q}{4\epsilon_0}$, which matches option B.