The electric flux through a Gaussian surface that encloses three charges given by `q_(1) = -14nC, q_(2) = 78.85 nC, q_(3)= -56nC`

Options

A. `10^(3) Nm^(2) C^(-1)`
B. `10^(3) CN^(-1) m^(-2)`
C. `6.32 xx 10^(3) Nm^(2) C^(-1)`
D. `6.32 xx 10^(3) CN^(-1) m^(-2)`

Answer

To find the electric flux through a Gaussian surface that encloses three charges, we can use Gauss's law, which states that the electric flux (Φ) through a closed surface is equal to the total charge (Q_enc) enclosed by that surface divided by the permittivity of free space (ε₀).

### Step-by-Step Solution:

1. **Identify the Charges**:
- We have three charges:
- \( q_1 = -14 \, \text{nC} = -14 \times 10^{-9} \, \text{C} \)
- \( q_2 = 78.85 \, \text{nC} = 78.85 \times 10^{-9} \, \text{C} \)
- \( q_3 = -56 \, \text{nC} = -56 \times 10^{-9} \, \text{C} \)

2. **Calculate the Total Charge Enclosed (Q_enc)**:
- The total charge enclosed by the Gaussian surface is the sum of the individual charges:
\[ Q_{\text{enc}} = q_1 + q_2 + q_3 \]
\[ Q_{\text{enc}} = (-14 \times 10^{-9}) + (78.85 \times 10^{-9}) + (-56 \times 10^{-9}) \]
\[ Q_{\text{enc}} = (-14 + 78.85 - 56) \times 10^{-9} \]
\[ Q_{\text{enc}} = (8.85) \times 10^{-9} \, \text{C} \]

3. **Use Gauss's Law to Find Electric Flux (Φ)**:
- According to Gauss's law, the electric flux is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \).

4. **Substituting Values**:
\[ \Phi = \frac{8.85 \times 10^{-9}}{8.85 \times 10^{-12}} \]

5. **Simplifying the Expression**:
- The \( 8.85 \) in the numerator and denominator cancels out:
\[ \Phi = \frac{1 \times 10^{-9}}{1 \times 10^{-12}} = 10^{3} \, \text{N m}^2/\text{C} \]

6. **Final Result**:
- Thus, the electric flux through the Gaussian surface is:
\[ \Phi = 1000 \, \text{N m}^2/\text{C} \, \text{or} \, 10^{3} \, \text{N m}^2/\text{C} \]