If charge q1 and q2 lies inside and outside respectively of a closed surface S. Let E be the electric field at any point on S and ϕ be the electric flux over S. Select the incorrect option.

Options

A. If `q_1` = 0 and `q_2 ≠0`,then E=0 but `ϕ ≠ 0`
B. If `q_1` changes,both E and `ϕ `will change
C. If `q_2` changes, E will changes but `ϕ` will not change
D. If `q_1≠0` and `q_2 ≠ 0`,then `ϕ ≠ 0`

Answer

To solve the problem, we need to analyze the relationship between electric field (E) and electric flux (ϕ) in the context of charges inside and outside a closed surface (S).

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two charges: \( q_1 \) (inside the closed surface S) and \( q_2 \) (outside the closed surface S).
- The electric field \( E \) at any point on the surface S is influenced by both charges, but the electric flux \( \phi \) through the surface S depends only on the charge enclosed by the surface.

2. **Applying Gauss's Law**:
- According to Gauss's Law, the electric flux \( \phi \) through a closed surface is given by:
\[ \phi = \frac{q_{\text{enc}}}{\epsilon_0} \]
- Here, \( q_{\text{enc}} \) is the charge enclosed by the surface, which in this case is \( q_1 \). Thus:
\[ \phi = \frac{q_1}{\epsilon_0} \]

3. **Electric Field Contribution**:
- The electric field \( E \) at any point on the surface S is due to both charges \( q_1 \) and \( q_2 \). Therefore, if either charge changes, it can affect \( E \):
- If \( q_1 \) changes, both \( E \) and \( \phi \) will change.
- If \( q_2 \) changes, only \( E \) will change, while \( \phi \) remains constant since \( q_1 \) is still the only charge enclosed.

4. **Analyzing the Options**:
- **Option A**: If \( q_1 = 0 \) and \( q_2 \neq 0 \), then \( E \) is not necessarily zero (due to \( q_2 \)), but \( \phi = 0 \). This option is incorrect because it states \( E = 0 \) which is false.
- **Option B**: If \( q_1 \) changes, both \( E \) and \( \phi \) will change. This is correct.
- **Option C**: If \( q_2 \) changes, \( E \) will change, but \( \phi \) will not change. This is correct.
- **Option D**: If \( q_1 \neq 0 \) and \( q_2 \neq 0 \), then \( \phi \neq 0 \). This is correct.

5. **Conclusion**:
- The incorrect option is **Option A**.