Let’s break this problem into steps to compute the total electric flux through the sphere centered at the point charge $q$:

Step 1: Total Electric Flux through a Sphere

Using Gauss's Law: $\Phi = \frac{q_{\text{enclosed}}}{\epsilon_0}$ Here:

• $q_{\text{enclosed}}$: The total charge enclosed within the Gaussian surface (the sphere).
• $\epsilon_0$: The permittivity of free space, $\epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2 / (N \cdot m^2)}$.

Step 2: Enclosed Charge

The sphere is centered at the point charge $q$, and the ring is outside the sphere because its radius $R < a$. The charge on the ring does not contribute to the electric flux through the sphere since no charge lies inside the spherical Gaussian surface.

Thus: $q_{\text{enclosed}} = q$

Step 3: Substitute the Values

Given: $q = 2 \times 10^{-6} \, \mathrm{C}, \quad \epsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2 / (N \cdot m^2)}$ The flux becomes: $\Phi = \frac{q}{\epsilon_0} = \frac{2 \times 10^{-6}}{8.85 \times 10^{-12}}$

Compute: $\Phi = 2.26 \times 10^5 \, \mathrm{N \cdot m^2 / C}$

Final Answer

$\Phi = 2.26 \times 10^5 \, \mathrm{N \cdot m^2 / C}$

Let me know if you'd like further details on the calculations or concepts!

Related Questions:

1. What is the role of the ring's charge density in this problem?
2. How would the flux change if the sphere's radius were increased to $a$?
3. Can we compute the electric field at any point on the sphere?
4. How would the flux change if the ring radius $a$ was smaller than $R$?
5. What is the significance of Gauss's law in simplifying problems with symmetry?

Tip:

Always identify the symmetry in a problem when using Gauss's Law—it greatly simplifies calculations!