If the radius of gaussion surface increases, the electric flux is:

Options

A. doubled
B. remains the same
C. halved
D. decreases

Explanation

To solve the question regarding the effect of increasing the radius of a Gaussian surface on electric flux, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Gauss's Law**:
Gauss's Law states that the electric flux (Φ) through a closed surface is directly proportional to the charge (Q) enclosed within that surface. The mathematical expression of Gauss's Law is given by:
\[ \Phi = \frac{Q_{in}}{\epsilon_0} \]
where \( \Phi \) is the electric flux, \( Q_{in} \) is the total charge enclosed by the Gaussian surface, and \( \epsilon_0 \) is the permittivity of free space.

**Hint**: Remember that Gauss's Law relates the electric flux to the charge enclosed, not the size of the surface.

2. **Consider the Effect of Changing Radius**:
When the radius of the Gaussian surface is increased, we need to consider whether this affects the charge enclosed. If the charges inside the Gaussian surface remain the same, then the enclosed charge \( Q_{in} \) does not change.

**Hint**: Focus on whether the charges inside the Gaussian surface are changing or not.

3. **Analyze the Situation**:
- If the radius is increased, the Gaussian surface can encompass more area, but it does not mean that more charge is enclosed unless additional charges are introduced within that surface.
- The electric flux depends solely on the charge enclosed. If the enclosed charge remains constant, the electric flux will also remain constant.

**Hint**: Think about how the electric field behaves with respect to the charges inside the surface.

4. **Conclusion**:
Since the charge enclosed \( Q_{in} \) remains the same when only the radius of the Gaussian surface is increased, the electric flux \( \Phi \) also remains the same. Therefore, the correct answer to the question is that the electric flux remains the same.

**Final Answer**: The electric flux is **B) remains the same**.