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

Options

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

Explanation

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

### Step-by-Step Solution:

1. **Understand the Setup**:
We have a parallel plate capacitor with two plates. One plate has charge \( +Q \) and the other has charge \( -Q \). The area of each plate is \( A \).

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

3. **Calculate the Force on One Plate**:
The force \( F \) on one plate due to the electric field created by the other plate can be calculated using the formula:
\[ F = Q \cdot E \]
Substituting the expression for \( E \):
\[ F = Q \cdot \left(\frac{Q}{2A \epsilon_0}\right) = \frac{Q^2}{2A \epsilon_0} \]
Here, we use \( \frac{Q}{2A \epsilon_0} \) because the electric field due to one plate at the location of the other plate is half of the total electric field.

4. **Conclusion**:
The magnitude of the electrostatic force between the plates is:
\[ F = \frac{Q^2}{2A \epsilon_0} \]

5. **Independence from Distance**:
Notably, this expression shows that the force is independent of the distance \( d \) between the plates.

### Final Answer:
The electrostatic force between the metal plates of an isolated parallel plate capacitor is given by:
\[ F = \frac{Q^2}{2A \epsilon_0} \]