Charges `Q_1 and Q_2` lie inside and outside, respectively, of a closed surface S. Let E be the field at any point on S and `phi` be the flux of E over S.

Options

A. If `Q_(1)` changes, both E and `phi` will change
B. If `Q_(2)` changes, E will change but `phi` will not change
C. If `Q_(1)=0` and `Q_(2)ne0` then `Ene0` but `phi=0`
D. If `Q_(1)ne0` and `Q_(2)=0` then `E=0` but `phi=0`

Explanation

To solve the problem regarding charges \( Q_1 \) and \( Q_2 \) in relation to a closed surface \( S \), we need to analyze the electric field \( E \) and the electric flux \( \phi \) through the surface \( S \).

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
The electric flux \( \phi \) through a closed surface \( S \) is given by Gauss's Law:
\[ \phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the total charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

2. **Identifying Charges**:
In this scenario, \( Q_1 \) is the charge inside the surface \( S \), and \( Q_2 \) is the charge outside the surface \( S \).

3. **Effect of \( Q_1 \) on \( \phi \)**:
If \( Q_1 \) changes, then the enclosed charge \( Q_{\text{enc}} \) changes, which directly affects the electric flux \( \phi \):
\[ \phi \text{ will change if } Q_1 \text{ changes.} \]

4. **Effect of \( Q_2 \) on \( \phi \)**:
Since \( Q_2 \) is outside the surface \( S \), it does not contribute to the enclosed charge \( Q_{\text{enc}} \). Therefore, changes in \( Q_2 \) will not affect the electric flux \( \phi \):
\[ \phi \text{ will not change if } Q_2 \text{ changes.} \]

5. **Case when \( Q_1 = 0 \)**:
If \( Q_1 = 0 \) (meaning no charge is enclosed), then \( Q_{\text{enc}} = 0 \) and thus:
\[ \phi = \frac{0}{\epsilon_0} = 0. \]
This means the electric flux \( \phi \) will be zero regardless of \( Q_2 \).

6. **Case when \( Q_2 = 0 \)**:
If \( Q_1 \neq 0 \) and \( Q_2 = 0 \), then \( \phi \) will still be determined by \( Q_1 \):
\[ \phi = \frac{Q_1}{\epsilon_0} \neq 0. \]
The electric field \( E \) on the surface will not be zero because \( Q_1 \) creates an electric field.

### Summary of Results:
- If \( Q_1 \) changes, both \( E \) and \( \phi \) change.
- If \( Q_2 \) changes, \( E \) may change, but \( \phi \) does not change.
- If \( Q_1 = 0 \), then \( \phi = 0 \) regardless of \( Q_2 \).
- If \( Q_2 = 0 \) and \( Q_1 \neq 0 \), then \( \phi \neq 0 \) and \( E \neq 0 \).