A charge is placed at the centre of cube of side a then flux linked with one of its given faces will be

Options

A. `(Q)/(epsi_(0))`
B. `(Q)/(6epsi_(0))`
C. `(Q)/(epsi_(0)a^(2))`
D. `(Q)/(4piepsi_(0)a^(2))`

Explanation

To solve the problem of finding the electric flux linked with one face of a cube when a charge is placed at the center, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**: We have a cube of side length \( a \) with a charge \( Q \) placed at its center. The cube has 6 faces.

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

3. **Calculate Total Flux through the Cube**: Since the charge \( Q \) is at the center of the cube, the total flux through the entire surface of the cube is:
\[ \Phi_{\text{total}} = \frac{Q}{\epsilon_0} \]

4. **Distribute the Flux Among the Faces**: The cube has 6 identical faces. By symmetry, the flux will be evenly distributed across all faces. Therefore, the flux through one face \( \Phi_{\text{face}} \) is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{Q}{\epsilon_0 \cdot 6} \]

5. **Final Expression for Flux through One Face**: Thus, the electric flux linked with one of the faces of the cube is:
\[ \Phi_{\text{face}} = \frac{Q}{6\epsilon_0} \]

### Conclusion:
The flux linked with one of the faces of the cube is \( \frac{Q}{6\epsilon_0} \).