A charge `q` is located at the centre of a cube. The electric flux through any face is

Options

A. `(2piq)/(6(4piepsilon_0))`
B. `(4piq)/(6(4piepsilon_0))`
C. `(piq)/(6(4piepsilon_0))`
D. `(q)/(6(4piepsilon_0))`

Explanation

To find the electric flux through any face of a cube when a charge `q` is located at the center, we can follow these steps:

### Step-by-Step Solution:

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

2. **Identifying the Enclosed Charge**: In this case, the charge \( q \) is located at the center of the cube. Therefore, the total charge enclosed by the cube is:
\[ Q_{\text{enc}} = q \]

3. **Calculating the Total Electric Flux through the Cube**: Using Gauss's Law, we can find the total electric flux through the entire surface of the cube:
\[ \Phi_{\text{cube}} = \frac{q}{\epsilon_0} \]

4. **Determining the Flux through One Face of the Cube**: The cube has 6 faces, and by symmetry, the electric flux is uniformly distributed across all faces. Thus, the electric flux through one face of the cube can be calculated as:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{cube}}}{6} = \frac{q}{6\epsilon_0} \]

5. **Final Result**: Therefore, the electric flux through any face of the cube is:
\[ \Phi_{\text{face}} = \frac{q}{6\epsilon_0} \]