In a cuboid of dimension `2L times 2L times L`,a charge `q` is placed at the centre of the surface 'S' having area of `4 L^(2)`.The flux through the opposite surface to '`S`' is given by

Options

A. `(q)/(12 varepsilon_(0))`
B. `(q)/(2 varepsilon_(0))`
C. `(q)/(3 varepsilon_(0))`
D. `(q)/(6 varepsilon_(0))`

Answer

To solve the problem, we need to find the electric flux through the surface opposite to the surface \( S \) of a cuboid with dimensions \( 2L \times 2L \times L \) where a charge \( q \) is placed at the center of surface \( S \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- The cuboid has dimensions \( 2L \times 2L \times L \).
- The area of surface \( S \) is given as \( 4L^2 \).

2. **Applying Gauss's Law**:
- According to Gauss's Law, the total electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Here, \( Q_{\text{enc}} \) is the charge enclosed by the surface, and \( \epsilon_0 \) is the permittivity of free space.

3. **Total Flux through the Cuboid**:
- The total flux through the entire surface of the cuboid can be calculated considering that the charge \( q \) is at the center of the cuboid. Since the cuboid has 6 faces, the total flux through the cuboid is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

4. **Calculating Flux through One Face**:
- Since the charge is symmetrically placed, the flux is uniformly distributed across all 6 faces of the cuboid. Therefore, the flux through each face is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

5. **Finding Flux through the Opposite Surface**:
- The problem specifically asks for the flux through the surface opposite to \( S \). By symmetry, the flux through the opposite surface will also be:
\[ \Phi_{\text{opposite}} = \frac{q}{6\epsilon_0} \]

6. **Final Answer**:
- Thus, the flux through the opposite surface to \( S \) is:
\[ \Phi_{\text{opposite}} = \frac{q}{6\epsilon_0} \]