A point charge `q` is placed at a distance `a//2` directly above the center of a square of side `a`. The electric flux through the square is

Options

A. `(q)/(epsilon_(0))`
B. `(q)/(piepsilon_(0))`
C. `(q)/(4epsilon_(0))`
D. `(q)/(6epsilon_(0))`

Explanation

To find the electric flux through the square due to a point charge \( q \) placed at a distance \( \frac{a}{2} \) directly above the center of the square, we can follow these steps:

### Step-by-step Solution:

1. **Understanding the Configuration**:
- We have a square of side \( a \).
- The point charge \( q \) is located at a distance \( \frac{a}{2} \) above the center of the square.

2. **Constructing a Gaussian Surface**:
- To calculate the electric flux through the square, we can use Gauss's law. We will consider a cube (Gaussian surface) that has the square as one of its faces. The cube will have a side length of \( a \).

3. **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 Gaussian surface, which in this case is \( q \).

4. **Calculating Total Flux**:
- Since the charge is at the center of the cube, the total electric flux through the entire surface of the cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

5. **Symmetry Consideration**:
- The cube has 6 faces, and due to symmetry, the electric flux will be equally distributed among all the faces. Therefore, the flux through each face of the cube is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

6. **Identifying the Square Face**:
- Since the square is one of the faces of the cube, the electric flux through the square is:
\[ \Phi_{\text{square}} = \frac{q}{6\epsilon_0} \]

### Final Answer:
The electric flux through the square is:
\[ \Phi_{\text{square}} = \frac{q}{6\epsilon_0} \]