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

Options

A. `(piq)/(6 (pi epsilon_(0)))`
B. `q/(6(4 pi epsilon_(0)))`
C. `(2piq)/(6(4pi epsilon_(0)))`
D. `(4pi q)/(6(4 pi epsilon_(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 1: Understand Gauss's Law
Gauss's Law states that the total electric flux (Φ) through a closed surface is proportional to the charge enclosed (q) within that surface. Mathematically, it is expressed as:
\[ \Phi = \frac{q}{\epsilon_0} \]
where:
- Φ is the total electric flux,
- q is the charge enclosed,
- ε₀ is the permittivity of free space.

### Step 2: Calculate Total Electric Flux through the Cube
Since the charge `q` is at the center of the cube, we can apply Gauss's Law to find the total electric flux through the entire surface of the cube:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

### Step 3: Determine the Flux through One Face
The cube has 6 faces, and due to symmetry, the electric flux is uniformly distributed across all faces. Therefore, the electric flux through one face of the cube can be calculated by dividing the total flux by the number of faces:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{1}{6} \cdot \frac{q}{\epsilon_0} \]

### Step 4: Final Expression for Electric Flux through One Face
Thus, the electric flux through any face of the cube is given by:
\[ \Phi_{\text{face}} = \frac{q}{6\epsilon_0} \]

### Summary
The electric flux through any face of the cube when a charge `q` is located at the center is:
\[ \Phi_{\text{face}} = \frac{q}{6\epsilon_0} \]