An electric charge q is placed at the centre of a cube of side l. The electric flux through one of its faces will be

Options

A. `Q/(epsilon_(0))`
B. `q/(6epsilon_(0))`
C. `q/(epsilon_(0)l^(2))`
D. `q/(4piepsilon_(0)l^(2))`

Answer

To solve the problem of finding the electric flux through one face of a cube with an electric charge \( q \) placed at its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Electric Flux**:
Electric flux (\( \Phi \)) through a closed surface is given by Gauss's law, which states:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
where \( Q_{\text{enc}} \) is the total charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.

2. **Identify the Charge and Surface**:
In this case, we have a charge \( q \) placed 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 of the cube, the total electric flux through the entire surface of the cube can be calculated as:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

4. **Determine the Flux through One Face**:
The cube has six identical faces. By symmetry, the electric flux will be evenly distributed through all six faces. Therefore, the flux through one face of the cube is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

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