Eight dipoles of charges of magnitude e are placed inside a cube. The total electric flux coming out of the cube will be

Options

A. `(8e)/( epsi_0)`
B. `(16 e)/( epsi_0)`
C. `(e )/( epsi_0)`
D. Zero

Answer

To solve the problem of finding the total electric flux coming out of a cube containing eight dipoles, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Dipoles**:
Each dipole consists of two charges: one positive charge (+e) and one negative charge (-e). The net charge of each dipole is zero because the magnitudes of the charges are equal and opposite.

2. **Total Charge from Dipoles**:
Since there are 8 dipoles in the cube, we can calculate the total charge enclosed by the cube. The charge from each dipole is:
\[ Q_{\text{dipole}} = +e + (-e) = 0 \]
Therefore, for 8 dipoles:
\[ Q_{\text{total}} = 8 \times Q_{\text{dipole}} = 8 \times 0 = 0 \]

3. **Applying Gauss's Law**:
According to Gauss's Law, the electric flux (\(\Phi_E\)) through a closed surface is given by:
\[ \Phi_E = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
where \(Q_{\text{enclosed}}\) is the total charge enclosed by the surface and \(\epsilon_0\) is the permittivity of free space.

4. **Calculating Electric Flux**:
Since we have established that the total charge enclosed by the cube is zero:
\[ Q_{\text{enclosed}} = 0 \]
Substituting this into Gauss's Law gives:
\[ \Phi_E = \frac{0}{\epsilon_0} = 0 \]

5. **Conclusion**:
Therefore, the total electric flux coming out of the cube is:
\[ \Phi_E = 0 \]

### Final Answer:
The total electric flux coming out of the cube is **0**.