A charge q is located at the cneter of a cube .The electric flux through any two consecutive faces will be

Options

A. `(piq)/(6(4piepsilon_(0)))`
B. `(q)/(6(4piepsilon_(0)))`
C. `(2piq)/(6(4piepsilon_(0)))`
D. `(4piq)/(3(4piepsilon_(0)))`

Explanation

To solve the problem of finding the electric flux through any two consecutive faces of a cube with a charge \( q \) at its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a cube with a charge \( q \) placed at its center.
- The cube has 6 faces.

2. **Applying Gauss's Law**:
- According to Gauss's Law, the total electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{q}{\epsilon_0} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

3. **Calculating Total Flux**:
- The total electric flux through the entire cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

4. **Flux Through One Face**:
- Since the cube has 6 faces and the charge is symmetrically placed at the center, the electric flux through each face is equal.
- Therefore, the electric flux through one face \( \Phi_{\text{face}} \) is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

5. **Calculating Flux Through Two Consecutive Faces**:
- The electric flux through two consecutive faces \( \Phi_{\text{two faces}} \) is simply twice the flux through one face:
\[ \Phi_{\text{two faces}} = 2 \times \Phi_{\text{face}} = 2 \times \frac{q}{6\epsilon_0} = \frac{q}{3\epsilon_0} \]

6. **Final Expression**:
- Thus, the electric flux through any two consecutive faces of the cube is:
\[ \Phi_{\text{two faces}} = \frac{q}{3\epsilon_0} \]

### Conclusion:
The electric flux through any two consecutive faces of the cube is \( \frac{q}{3\epsilon_0} \).