A flat, square surface with sides of length L is described by the equations
`x = L, 0le y leL,0lezleL`
The electric flux through the square due to a positive point charge q located at the origin `(x = 0 , y = 0, z = 0)` is

Options

A. `q/(4epsilon_0)`
B. `q/(6epsilon_0)`
C. `q/(24epsilon_0)`
D. `q/(48epsilon_0)`

Explanation

To find the electric flux through a flat square surface due to a positive point charge located at the origin, we can follow these steps:

### Step 1: Understand the Geometry
The square surface is defined by the equation \( x = L \), where \( y \) varies from \( 0 \) to \( L \) and \( z \) varies from \( 0 \) to \( L \). This means the square surface is located in the plane where \( x = L \) and extends from \( (L, 0, 0) \) to \( (L, L, L) \).

### Step 2: Apply Gauss's Law
Gauss's Law states that the electric flux \( \Phi_E \) through a closed surface is equal to the charge enclosed \( Q \) divided by the permittivity of free space \( \epsilon_0 \):
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
In this case, we will consider a Gaussian surface that encloses the charge \( q \) at the origin.

### Step 3: Define the Gaussian Surface
To apply Gauss's Law effectively, we can imagine a cube centered at the origin with a charge \( q \) at its center. The cube has a side length that allows it to enclose the charge completely.

### Step 4: Calculate Total Flux Through the Cube
Since the charge is at the center of the cube, the total electric flux through the entire surface of the cube is:
\[ \Phi_{\text{total}} = \frac{q}{\epsilon_0} \]

### Step 5: Determine the Flux Through the Square Surface
The cube has 6 faces, and due to symmetry, the electric flux is uniformly distributed across all faces. Therefore, the flux through one face (the square surface at \( x = L \)) is:
\[ \Phi_{\text{face}} = \frac{\Phi_{\text{total}}}{6} = \frac{q}{6\epsilon_0} \]

### Conclusion
The electric flux through the square surface due to the point charge \( q \) located at the origin is:
\[ \Phi_E = \frac{q}{6\epsilon_0} \]