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

Options

A. `(4piq)/(6(4piepsilon_(0)))`
B. `(piq)/(6(4piepsilon_(0)))`
C. `(q)/(6(4piepsilon_(0)))`
D. `(2piq)/(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 1: Understand Gauss's Law
Gauss's Law states that the total electric flux \( \Phi \) through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi = \frac{Q}{\epsilon_0} \]

### Step 2: Identify the Charge and the Surface
In this case, we have a charge \( q \) located at the center of a cube. The cube is a closed surface surrounding the charge.

### Step 3: Calculate the Total Electric Flux
Using Gauss's Law, the total electric flux through the entire surface of the cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

### Step 4: Determine the Flux Through One Face of the Cube
Since the cube has 6 faces and the charge is symmetrically located at the center, the electric flux is uniformly distributed across all faces. Therefore, the electric flux through one face \( \Phi_{\text{face}} \) can be calculated by dividing the total flux by the number of faces:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

### Conclusion
The electric flux through any face of the cube is:
\[ \Phi_{\text{face}} = \frac{q}{6\epsilon_0} \]