The electrostatic force between the metal plate of an isolated parallel plate capacitro `C` having charge `Q` and area `A`, is

Options

A. inveresly proportional to the distance between the plates.
B. independent of the distance between the plates.
C. proportional to the square root of the distance between the plates.
D. linearly proportional to the distance between the plates

Explanation

To find the electrostatic force between the metal plates of an isolated parallel plate capacitor with capacitance \( C \), charge \( Q \), and area \( A \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a parallel plate capacitor with two plates, each carrying a charge \( Q \). The area of each plate is \( A \).

2. **Electric Field Between the Plates**:
- The electric field \( E \) between the plates of a parallel plate capacitor can be expressed as:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density given by \( \sigma = \frac{Q}{A} \), and \( \epsilon_0 \) is the permittivity of free space.

3. **Substituting for Surface Charge Density**:
- Substituting \( \sigma \) into the electric field equation:
\[ E = \frac{Q}{A \epsilon_0} \]

4. **Force on One Plate**:
- The force \( F \) on one plate due to the electric field produced by the other plate is given by:
\[ F = Q \cdot E \]
- Substituting the expression for \( E \):
\[ F = Q \cdot \left(\frac{Q}{A \epsilon_0}\right) \]

5. **Final Expression for Force**:
- Simplifying the equation gives:
\[ F = \frac{Q^2}{A \epsilon_0} \]

6. **Conclusion**:
- The electrostatic force between the metal plates of the capacitor is:
\[ F = \frac{Q^2}{2A \epsilon_0} \]
- Note: The factor of \( \frac{1}{2} \) arises because the electric field is produced by one plate acting on the charge of the other plate.