Ratio of Electric Flux Through Concentric Spheres

To determine the ratio of electric flux through the two concentric spheres, we apply Gauss's Law, $\Phi = \frac{Q_{\text{enc}}}{\varepsilon_0}$, where $Q_{\text{enc}}$ is the total charge enclosed within a sphere centered at the origin.

1. For the first sphere of radius $R_1 = 3\text{ cm}$, only the charge $q_1 = 8\text{ }\mu\text{C}$ at $x = 2\text{ cm}$ lies inside the sphere, so $Q_{\text{enc}, 1} = 8\text{ }\mu\text{C}$.
2. For the second sphere of radius $R_2 = 5\text{ cm}$, both charges $q_1 = 8\text{ }\mu\text{C}$ at $x = 2\text{ cm}$ and $q_2 = -2\text{ }\mu\text{C}$ at $x = 4\text{ cm}$ lie inside the sphere, yielding $Q_{\text{enc}, 2} = 8\text{ }\mu\text{C} + (-2\text{ }\mu\text{C}) = 6\text{ }\mu\text{C}$.

Taking the ratio of the enclosed charges gives: $\frac{\Phi_1}{\Phi_2} = \frac{Q_{\text{enc}, 1}}{Q_{\text{enc}, 2}} = \frac{8}{6} = \frac{4}{3}$

Thus, the ratio of electric flux through the two spheres is $4 : 3$, which corresponds to Option C.