A spherical Gaussian surface encloses a charge of `8.85xx10^(-8) C` (i) Calculate the electric flux passing through the surface (ii) If the radius of Gaussian surface is doubled, how would the flux change ?

Explanation

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

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

where:
- \( \Phi \) is the electric flux,
- \( Q_{\text{enclosed}} \) is the charge enclosed by the Gaussian surface,
- \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

### Step 1: Calculate the Electric Flux

1. **Identify the enclosed charge**:
The charge enclosed by the spherical Gaussian surface is given as \( Q_{\text{enclosed}} = 8.85 \times 10^{-8} \, \text{C} \).

2. **Use Gauss's Law to calculate the flux**:
Substitute the values into the formula:
\[ \Phi = \frac{8.85 \times 10^{-8}}{8.85 \times 10^{-12}} \]

3. **Calculate the electric flux**:
\[ \Phi = \frac{8.85 \times 10^{-8}}{8.85 \times 10^{-12}} = 10^4 \, \text{N m}^2/\text{C} \]

### Step 2: Analyze the Effect of Doubling the Radius

1. **Consider the effect of changing the radius**:
According to Gauss's Law, the electric flux through a closed surface depends only on the charge enclosed within that surface, not on the size or shape of the surface.

2. **Conclusion about the flux when the radius is doubled**:
Since the enclosed charge remains the same (\( Q_{\text{enclosed}} = 8.85 \times 10^{-8} \, \text{C} \)), the electric flux will also remain the same regardless of the radius of the Gaussian surface.

### Final Answers:
(i) The electric flux passing through the surface is \( 10^4 \, \text{N m}^2/\text{C} \).

(ii) If the radius of the Gaussian surface is doubled, the electric flux will remain the same.