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

Options

A. `(piq)/(6(4pi epsilon_(0)))`
B. `(q)/(6(4pi epsilon_(0)))`
C. `(2pi q)/(6(4pi epsilon_(0))`
D. `(4pi q)/(6(4pi epsilon_(0)))`

Answer

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

### Step-by-Step Solution:

1. **Understand Gauss's Law**:
Gauss's Law states that the electric flux (Φ) through a closed surface is proportional to the charge enclosed (Q_enc) within that surface. Mathematically, it is expressed as:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Charge and Surface**:
In this case, we have a charge `q` located at the center of a cube. The cube has six faces.

3. **Calculate Total Electric Flux through the Cube**:
Since the charge `q` is at the center, the total electric flux through the entire cube is given by:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

4. **Determine the Flux through One Face**:
The cube has six identical faces, and by symmetry, the electric flux will be uniformly distributed across all six faces. Therefore, the electric flux through one face (Φ_face) can be calculated as:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

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

### Final Answer:
The electric flux through any face of the cube is \( \frac{q}{6\epsilon_0} \).