Consider two concentric spherical surfaces `S_1` with radius a and `S_2` with radius 2a, both centered at the origin. There is a charge +q at the origin and there are no other charges. Compare the flux `phi_1` through `S_1` with the flux `phi_2` through `S_2`.

Options

A. `phi_1 = 4phi_2`
B. `phi_1 = 2phi_2`
C. `phi_1 = phi_2`
D. `phi_1 = phi_2//2`

Explanation

To solve the problem, we will use Gauss's Law, which states that the electric flux (Φ) through a closed surface is proportional to the charge (Q) enclosed by that surface. The formula for Gauss's Law is given by:

\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]

where:
- \( \Phi \) is the electric flux,
- \( Q_{\text{enc}} \) is the charge enclosed within the surface,
- \( \epsilon_0 \) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the Surfaces and Charge**:
- We have two concentric spherical surfaces, \( S_1 \) with radius \( a \) and \( S_2 \) with radius \( 2a \).
- There is a charge \( +q \) located at the origin.

2. **Calculate the Flux through Surface \( S_1 \)**:
- Since \( S_1 \) encloses the charge \( +q \), we can apply Gauss's Law.
- The charge enclosed by \( S_1 \) is \( Q_{\text{enc}} = +q \).
- Therefore, the electric flux through surface \( S_1 \) is:
\[ \Phi_1 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{q}{\epsilon_0} \]

3. **Calculate the Flux through Surface \( S_2 \)**:
- Similarly, surface \( S_2 \) also encloses the same charge \( +q \) located at the origin.
- The charge enclosed by \( S_2 \) is also \( Q_{\text{enc}} = +q \).
- Thus, the electric flux through surface \( S_2 \) is:
\[ \Phi_2 = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{q}{\epsilon_0} \]

4. **Compare the Fluxes**:
- From the calculations, we have:
\[ \Phi_1 = \frac{q}{\epsilon_0} \quad \text{and} \quad \Phi_2 = \frac{q}{\epsilon_0} \]
- Therefore, we can conclude that:
\[ \Phi_1 = \Phi_2 \]

### Final Conclusion:
The electric flux through both surfaces is equal:
\[ \Phi_1 = \Phi_2 \]