A closed surface in vacuum encloses charges `-q and +3q`. The total electric flux emerging out of the surface is:

Options

A. zero
B. `2q//epsi_(0)`
C. `3q//epsi_(0)`
D. `4q//epsi_(0)`

Explanation

To find the total electric flux emerging out of a closed surface that encloses charges of \(-q\) and \(+3q\), we can use Gauss's Law, which states that the electric flux \(\Phi\) through a closed surface is proportional to the total charge \(Q_{\text{enc}}\) enclosed within that surface. The formula for electric flux is given by:

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

where \(\epsilon_0\) is the permittivity of free space.

### Step-by-Step Solution:

1. **Identify the enclosed charges**:
- The charges enclosed by the surface are \(-q\) and \(+3q\).

2. **Calculate the total charge enclosed**:
- The total charge \(Q_{\text{enc}}\) is the sum of the individual charges:
\[ Q_{\text{enc}} = -q + 3q = 2q \]

3. **Apply Gauss's Law**:
- Substitute \(Q_{\text{enc}}\) into the formula for electric flux:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} = \frac{2q}{\epsilon_0} \]

4. **Final result**:
- Therefore, the total electric flux emerging out of the surface is:
\[ \Phi = \frac{2q}{\epsilon_0} \]

### Summary:
The total electric flux emerging out of the closed surface is \(\frac{2q}{\epsilon_0}\).