If charge `q_1` and `q_2` 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 situation involving the charges \( q_1 \) and \( q_2 \) in relation to a closed surface \( S \). Here’s the step-by-step solution:

### Step 1: Understand the Setup
- We have two charges: \( q_1 \) is located inside a closed surface \( S \), and \( q_2 \) is located outside of this surface.

### Step 2: Recall Gauss's Law
- According to Gauss's Law, the electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{q_{\text{inside}}}{\epsilon_0} \]
where \( q_{\text{inside}} \) is the total charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Determine the Electric Flux \( \Phi \)
- Since \( q_1 \) is the only charge inside the surface \( S \), the electric flux through the surface is:
\[ \Phi = \frac{q_1}{\epsilon_0} \]
- The charge \( q_2 \) being outside the surface does not contribute to the electric flux through \( S \).

### Step 4: Analyze the Electric Field \( E \)
- The electric field \( E \) at any point on the surface \( S \) is influenced by both charges \( q_1 \) and \( q_2 \).
- Therefore, the electric field \( E \) can be expressed as a combination of the fields due to both charges:
\[ E = E_{q_1} + E_{q_2} \]
where \( E_{q_1} \) is the electric field due to \( q_1 \) and \( E_{q_2} \) is the electric field due to \( q_2 \).

### Step 5: Evaluate the Options
1. **If \( q_1 = 0 \)**:
- Then \( \Phi = \frac{0}{\epsilon_0} = 0 \). This means the electric flux is zero, which is correct.
2. **If \( q_1 \) changes**:
- Both \( E \) and \( \Phi \) will change because \( \Phi \) depends on \( q_1 \). This option is correct.
3. **If \( q_2 \) changes**:
- The electric field \( E \) will change due to \( q_2 \), but \( \Phi \) will remain the same since it only depends on \( q_1 \). This option is correct.
4. **If \( q_1 \neq 0 \) and \( q_2 \neq 0 \)**:
- Then \( \Phi \) will not be zero since \( \Phi = \frac{q_1}{\epsilon_0} \). This option is also correct.

### Conclusion
- The only incorrect option is the first one, which states that if \( q_1 = 0 \), then \( E \) will also be zero. This is misleading because while \( \Phi \) becomes zero, \( E \) can still be influenced by \( q_2 \).