The force between the plates of a parallel plate capacitor of capacitance `C` and distance of separation of the plates `d` with a potential difference `V` between the plates, is.

Options

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

Explanation

To find the force between the plates of a parallel plate capacitor with capacitance \( C \), distance of separation \( d \), and potential difference \( V \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Electric Field**:
The electric field \( E \) between the plates of a parallel plate capacitor is given by the formula:
\[ E = \frac{V}{d} \]
where \( V \) is the potential difference and \( d \) is the distance between the plates.

**Hint**: Recall that the electric field in a capacitor is uniform and depends on the potential difference and the separation between the plates.

2. **Charge on the Plates**:
The charge \( Q \) on the plates of the capacitor can be expressed in terms of capacitance \( C \) and potential difference \( V \):
\[ Q = C \cdot V \]

**Hint**: Remember the relationship between charge, capacitance, and voltage in a capacitor.

3. **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 \( E \) from step 1:
\[ F = Q \cdot \frac{V}{d} \]

**Hint**: The force on a charged plate in an electric field is the product of the charge and the electric field strength.

4. **Substituting Charge**:
Now, substituting \( Q \) from step 2 into the force equation:
\[ F = (C \cdot V) \cdot \frac{V}{d} \]
Simplifying this gives:
\[ F = \frac{C \cdot V^2}{d} \]

**Hint**: Make sure to keep track of units and dimensions when substituting values.

5. **Final Expression**:
The final expression for the force between the plates of the capacitor is:
\[ F = \frac{C \cdot V^2}{2d} \]

**Hint**: The factor of \( \frac{1}{2} \) comes from the consideration of the electric field due to one plate acting on the charge on the other plate.

### Conclusion:
Thus, the force between the plates of a parallel plate capacitor with capacitance \( C \), separation \( d \), and potential difference \( V \) is given by:
\[ F = \frac{C \cdot V^2}{2d} \]