If the electric flux entering and leaving an enclosed surface respectively is `phi_1` and `phi_2`, the electric charge inside the surface will be

Options

A. `(phi_(1) + phi_(2)) epsi_(0)`
B. `(phi_(2) - phi_(1))epsi_(0)`
C. `(phi_(1) + phi_(2))//epsi_(0)`
D. `(phi_(2) - phi_(1))//epsi_(0)`

Answer

To find the electric charge enclosed within a surface when given the electric flux entering and leaving that surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Variables**:
- Let \( \phi_1 \) be the electric flux entering the surface.
- Let \( \phi_2 \) be the electric flux leaving the surface.

2. **Understand the Sign Convention**:
- According to the convention, the electric flux entering the surface is considered negative, while the electric flux leaving the surface is considered positive.

3. **Express the Total Electric Flux**:
- The total electric flux \( \Phi \) through the surface can be expressed as:
\[ \Phi = \phi_2 - \phi_1 \]
- Here, \( \phi_1 \) is negative, so the equation effectively accounts for the direction of the flux.

4. **Use Gauss's Law**:
- According to Gauss's Law, the total electric flux through a closed surface is related to the charge \( q \) enclosed by that surface:
\[ \Phi = \frac{q}{\epsilon_0} \]
- Where \( \epsilon_0 \) is the permittivity of free space.

5. **Set the Two Expressions for Flux Equal**:
- From the previous steps, we can set the expressions for total electric flux equal to each other:
\[ \phi_2 - \phi_1 = \frac{q}{\epsilon_0} \]

6. **Solve for the Enclosed Charge \( q \)**:
- Rearranging the equation gives:
\[ q = (\phi_2 - \phi_1) \cdot \epsilon_0 \]

### Final Answer:
The electric charge inside the surface is given by:
\[ q = (\phi_2 - \phi_1) \cdot \epsilon_0 \]