A uniformly charged square plate having side L carries a uniform surface charge density `sigma`. The plate lies in the `y - z` plane with its centre at the origin. A point charge q lies on the x-axis. The flux of the electric field of q through the plate is `phi0`, while the foce on the point charge q due to the plate is `F_(0)`, along the x-axis. Then,

Options

A. `sigma = (F_(0))/(phi_(0)L)`
B. `sigma = (F_(0))/(phi_(0))`
C. `sigma = (F_(0)L)/(phi_(0))`
D. `sigma = (phi_(0))/(F_(0))`

Explanation

To solve the problem, we need to analyze the relationship between the electric flux through the square plate and the force on the point charge due to the plate.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a uniformly charged square plate with side length \( L \) lying in the \( y-z \) plane, centered at the origin.
- The surface charge density of the plate is \( \sigma \).
- A point charge \( q \) is located on the \( x \)-axis.

2. **Electric Flux (\( \Phi_0 \))**:
- The electric flux \( \Phi_0 \) through the plate due to the point charge \( q \) can be expressed using Gauss's law:
\[ \Phi_0 = \frac{q}{\epsilon_0} \]
- Here, \( \epsilon_0 \) is the permittivity of free space.

3. **Force (\( F_0 \)) on the Point Charge**:
- The force \( F_0 \) on the point charge \( q \) due to the electric field \( E \) created by the charged plate can be calculated using:
\[ F_0 = qE \]
- The electric field \( E \) due to a uniformly charged infinite plane sheet is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
- However, since we have a square plate, we need to consider the effective area and the distance from the charge to the plate.

4. **Relating \( F_0 \) and \( \Phi_0 \)**:
- We can write \( F_0 \) in terms of \( \Phi_0 \):
\[ F_0 = q \left( \frac{\sigma}{2\epsilon_0} \right) \]
- Substituting \( \Phi_0 \) into the equation gives:
\[ F_0 = q \left( \frac{\sigma}{2\epsilon_0} \right) \]

5. **Dimensional Analysis**:
- The dimensions of \( \sigma \) (surface charge density) can be expressed as:
\[ \sigma = \frac{Q}{L^2} \quad \text{(where \( Q \) is charge)} \]
- The dimensions of force \( F_0 \) are \( [M L T^{-2}] \) and the dimensions of electric flux \( \Phi_0 \) are \( [E][A] = [M L^3 T^{-3}] \).
- From the above relationships, we can derive that:
\[ \sigma = \frac{F_0}{\Phi_0} \]

6. **Conclusion**:
- The relationship between the surface charge density \( \sigma \), the electric flux \( \Phi_0 \), and the force \( F_0 \) is:
\[ \sigma = \frac{F_0}{\Phi_0} \]

### Final Answer:
The correct statement for the surface charge density \( \sigma \) is:
\[ \sigma = \frac{F_0}{\Phi_0} \]