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

Options

A. (a) `q//epsi_(0)`
B. (b) `q//2epsi_(0)`
C. (c) `q/(4 epsi_(0))`
D. (d) `q//epsi_(0)`

Answer

To solve the problem of finding the electric flux through a square when a point charge \( q \) is placed above it, we can follow these steps:

### Step 1: Understand the Geometry
We have a square of side \( a \) and a point charge \( q \) located at a distance \( \frac{a}{2} \) directly above the center of the square.

### Step 2: Visualize the Cube
The square can be thought of as one face of a cube. The cube has six faces, and the charge \( q \) is located at the center of the cube, above the square face.

### Step 3: Use Gauss's Law
According to Gauss's law, the total electric flux \( \Phi \) through a closed surface is given by:
\[ \Phi = \frac{q_{\text{enc}}}{\epsilon_0} \]
where \( q_{\text{enc}} \) is the charge enclosed by the surface and \( \epsilon_0 \) is the permittivity of free space.

### Step 4: Calculate Total 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 is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

### Step 5: Calculate Flux Through One Face of the Cube
Since the cube has six faces and the charge is symmetrically placed, the flux will be evenly distributed among the six faces. Therefore, the flux through one face (which is our square) is:
\[ \Phi_{\text{square}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

### Conclusion
Thus, the electric flux through the square is:
\[ \Phi = \frac{q}{6\epsilon_0} \]