Force of attraction between the plates of a parallel plate capacitor is

Options

A. `(q^(2))/(2 epsilon_(0)AK)`
B. `(q^(2))/(epsilon_(0)AK)`
C. `(q)/(2 epsilon_(0)A)`
D. `(q^(2))/(2 epsilon_(0)A^(2)K)`

Explanation

To find the force of attraction between the plates of a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the Electric Field
The electric field (E) between the plates of a parallel plate capacitor can be derived from the charge on the plates. For a single plate with surface charge density σ, the electric field produced by that plate is given by:
\[ E = \frac{\sigma}{2\epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Determine the Total Electric Field
In a parallel plate capacitor, there are two plates, each contributing to the electric field. Therefore, the total electric field (E) between the plates is:
\[ E = \frac{\sigma}{\epsilon_0} \]

### Step 3: Relate Surface Charge Density to Charge
The surface charge density \( \sigma \) is related to the charge \( Q \) on the plates and the area \( A \) of the plates:
\[ \sigma = \frac{Q}{A} \]

### Step 4: Substitute for Electric Field
Substituting \( \sigma \) into the expression for the electric field gives:
\[ E = \frac{Q}{A \epsilon_0} \]

### Step 5: Calculate the Force on One Plate
The force (F) acting on one plate due to the electric field produced by the other plate is given by:
\[ F = Q \cdot E \]
Substituting for E:
\[ F = Q \cdot \frac{Q}{A \epsilon_0} = \frac{Q^2}{A \epsilon_0} \]

### Step 6: Include the Effect of Dielectric
If there is a dielectric material between the plates, the permittivity changes. The new permittivity is given by:
\[ \epsilon = K \epsilon_0 \]
where \( K \) is the dielectric constant. Thus, the force becomes:
\[ F = \frac{Q^2}{A \epsilon} = \frac{Q^2}{A K \epsilon_0} \]

### Final Result
The force of attraction between the plates of a parallel plate capacitor with a dielectric is:
\[ F = \frac{Q^2}{A K \epsilon_0} \]