A charge q is placed at the centre of a cube of side `l` what is the electric flux passing through two opposite faces of the cube ?

Options

A. `(q)/(3epsilon_(0))`
B. `(6qL^(2))/(epsilon_(0))`
C. `(q)/(6L^(2)epsilon_(0))`
D. zero

Answer

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

### Step-by-Step Solution:

1. **Understanding Electric Flux**:
The electric flux \( \Phi \) through a closed surface is given by Gauss's Law:
\[ \Phi = \frac{q_{\text{enc}}}{\epsilon_0} \]
where \( q_{\text{enc}} \) is the charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

2. **Total Flux Through the Cube**:
Since the charge \( q \) is placed at the center of the cube, the total electric flux through the entire surface of the cube can be calculated as:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

3. **Flux Distribution Across Faces**:
A cube has 6 faces, and due to symmetry, the electric flux will be uniformly distributed across all faces. Therefore, the flux through each face of the cube is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

4. **Calculating Flux Through Two Opposite Faces**:
Since we need to find the flux through two opposite faces, we can simply add the flux through each of these two faces:
\[ \Phi_{\text{two faces}} = \Phi_{\text{face}} + \Phi_{\text{face}} = \frac{q}{6\epsilon_0} + \frac{q}{6\epsilon_0} = \frac{2q}{6\epsilon_0} \]

5. **Simplifying the Expression**:
Simplifying the expression gives us:
\[ \Phi_{\text{two faces}} = \frac{2q}{6\epsilon_0} = \frac{q}{3\epsilon_0} \]

### Final Answer:
The electric flux passing through two opposite faces of the cube is:
\[ \Phi = \frac{q}{3\epsilon_0} \]