A point charge Q is kept at the centre of a square face of edge length a. The amount of electric flux associated with the square face in

Options

A. `Q/(6epsilon_0)`
B. `Q/(24epsilon_0)`
C. `Q/(8epsilon_0)`
D. 0

Explanation

To solve the problem of finding the electric flux associated with a square face of edge length \( a \) when a point charge \( Q \) is placed at its center, we can follow these steps:

### Step 1: Understand the Concept of Electric Flux
Electric flux (\( \Phi \)) through a surface is defined as the product of the electric field (\( E \)) and the area (\( A \)) of the surface, integrated over the surface. Mathematically, it is given by:
\[ \Phi = \int \vec{E} \cdot d\vec{A} \]
Where \( d\vec{A} \) is the differential area vector pointing outward from the surface.

### Step 2: Analyze the Configuration
In this case, we have a point charge \( Q \) located at the center of a square face. The square face has an edge length \( a \). The electric field lines emanating from the point charge will radiate outward in all directions.

### Step 3: Determine the Orientation of Electric Field Lines
The electric field lines will be perpendicular to the surface of the square face at the center. However, as we move away from the center towards the edges of the square face, the electric field lines will not remain perpendicular to the surface.

### Step 4: Calculate the Electric Flux
Since the electric field lines are not crossing the square face perpendicularly (they are diverging), the component of the electric field that is perpendicular to the square face is effectively zero. Therefore, the electric flux through the square face can be concluded to be:
\[ \Phi = 0 \]

### Step 5: Conclusion
The total electric flux associated with the square face is zero because no electric field lines cross the surface perpendicularly.

### Summary
Thus, the amount of electric flux associated with the square face is:
\[ \Phi = 0 \]