The inward and outward electric flux for a closed surface unit of `N-m^(2)//C` are respectively `8xx10^(3)` and `4xx10^(3)`. Then the total charge inside the surface is [where `epsilon_(0)=` permittivity constant]

Options

A. `4 xx 10^(3)C`
B. `-4 xx 10^(3)C`
C. `((-4 xx 10^(3)))/(epsi)C`
D. `-4 xx 10^(3) epsi_(0)C`

Answer

To solve the problem, we will use Gauss's law, which states that the net electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε₀).

### Step-by-Step Solution:

1. **Identify Given Values**:
- Inward electric flux (Φ_in) = -8 × 10³ N·m²/C (considered negative)
- Outward electric flux (Φ_out) = +4 × 10³ N·m²/C (considered positive)

2. **Calculate Net Electric Flux**:
The net electric flux (Φ_net) through the closed surface is the sum of the inward and outward flux:
\[ Φ_{net} = Φ_{out} + Φ_{in} \]
Substituting the values:
\[ Φ_{net} = 4 × 10³ + (-8 × 10³) = 4 × 10³ - 8 × 10³ = -4 × 10³ \, \text{N·m²/C} \]

3. **Apply Gauss's Law**:
According to Gauss's law:
\[ Φ_{net} = \frac{Q_{enclosed}}{ε₀} \]
Rearranging gives:
\[ Q_{enclosed} = Φ_{net} × ε₀ \]

4. **Substitute the Net Flux**:
Now substituting the value of Φ_net:
\[ Q_{enclosed} = (-4 × 10³) × ε₀ \]

5. **Final Expression for Charge**:
Thus, the total charge inside the surface is:
\[ Q_{enclosed} = -4 × 10³ ε₀ \, \text{C} \]

### Conclusion:
The total charge inside the surface is \( -4 × 10³ ε₀ \, \text{C} \).