A parallel plate air capacitor has capacity `C` distance of separation between plates is `d` and potential difference `V` is applied between the plates force of attraction between the plates of the parallel plate air capacitor is

Options

A. `(C^(2)V^(2))/(2d^(2))`
B. `(C^(2)V^(2))/(2d)`
C. `(CV^(2))/(2d)`
D. `(CV^(2))/(d)`

Explanation

To solve the problem of finding the force of attraction between the plates of a parallel plate air capacitor, we can follow these steps:

### Step-by-Step Solution

1. **Understanding the Capacitor Configuration**:
- We have a parallel plate capacitor with capacitance \( C \), plate separation \( d \), and potential difference \( V \) applied between the plates. The plates carry charges \( +Q \) and \( -Q \).

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 \( \sigma = \frac{Q}{A} \) (with \( A \) being the area of the plates), and \( \epsilon_0 \) is the permittivity of free space.

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

4. **Force on One Plate**:
- The force \( F \) on one plate due to the electric field \( E \) created by the other plate is given by:
\[ F = Q \cdot E \]
- Substituting \( E \) into the force equation:
\[ F = Q \cdot \frac{Q}{\epsilon_0 A} = \frac{Q^2}{\epsilon_0 A} \]

5. **Relating Charge to Capacitance**:
- We know that the charge \( Q \) on the capacitor is related to capacitance \( C \) and potential difference \( V \) by:
\[ Q = C \cdot V \]

6. **Substituting \( Q \) in the Force Equation**:
- Now substituting \( Q = C \cdot V \) into the force equation:
\[ F = \frac{(C \cdot V)^2}{\epsilon_0 A} = \frac{C^2 V^2}{\epsilon_0 A} \]

7. **Relating Area to Capacitance**:
- From the definition of capacitance, we have:
\[ C = \frac{\epsilon_0 A}{d} \]
- Rearranging gives us:
\[ A = \frac{C \cdot d}{\epsilon_0} \]

8. **Substituting Area Back into the Force Equation**:
- Now substituting \( A \) back into the force equation:
\[ F = \frac{C^2 V^2}{\epsilon_0 \cdot \frac{C \cdot d}{\epsilon_0}} = \frac{C^2 V^2}{C \cdot d} = \frac{C V^2}{d} \]

9. **Final Result**:
- Thus, the force of attraction between the plates of the parallel plate air capacitor is:
\[ F = \frac{C V^2}{2d} \]

### Conclusion
The force of attraction between the plates of the parallel plate air capacitor is given by:
\[ F = \frac{C V^2}{2d} \]