A parallel plate air capacitor has capcity `C` distance of separtion 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)`
B. `(CV^(2))/(2d)`
C. `(CV^(2))/(d)`
D. `(C^(2)V^(2))/(2d^(2))`

Answer

To find 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 Setup**:
- We have a parallel plate capacitor with capacitance \( C \), distance of separation between the plates \( d \), and potential difference \( V \) applied across the plates.

2. **Electric Field Between the Plates**:
- The electric field \( E \) between the plates of a parallel plate capacitor can be expressed as:
\[ E = \frac{V}{d} \]

3. **Charge on the Plates**:
- The charge \( Q \) on the plates of the capacitor can be expressed using the capacitance:
\[ Q = C \cdot V \]

4. **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 \( E \):
\[ F = Q \cdot \frac{V}{d} \]

5. **Substituting for Charge \( Q \)**:
- Now substitute \( Q = C \cdot V \) into the force equation:
\[ F = (C \cdot V) \cdot \frac{V}{d} = \frac{C \cdot V^2}{d} \]

6. **Final Expression for Force**:
- Thus, the force of attraction between the plates of the parallel plate air capacitor is given by:
\[ F = \frac{C \cdot V^2}{2d} \]

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