A air filled parallel plate capacitor having circular plates of diameter D is given a charge Q.The magnitude of the force acting between plates is

Options

A. `(Q^(2))/( 2epsilon_(0))`
B. `(Q^(2))/( 2pi D^(2) epsilon_(0))`
C. `(Q^(2))/( pi D^(2) epsilon_(0))`
D. `(2Q^(2))/( pi epsilon_(0)D^(2))`

Answer

To find the magnitude of the force acting between the plates of an air-filled parallel plate capacitor with circular plates of diameter \( D \) and charge \( Q \), we can follow these steps:

### Step 1: Understand the Formula for Force
The force \( F \) between two charged plates of a capacitor can be expressed as:
\[ F = \frac{Q^2}{2A \epsilon_0} \]
where \( A \) is the area of one of the plates and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Calculate the Area of the Circular Plates
The area \( A \) of a circular plate is given by the formula:
\[ A = \pi r^2 \]
Since the diameter \( D \) is given, we can express the radius \( r \) as:
\[ r = \frac{D}{2} \]
Substituting this into the area formula gives:
\[ A = \pi \left(\frac{D}{2}\right)^2 = \pi \frac{D^2}{4} \]

### Step 3: Substitute the Area into the Force Formula
Now, we can substitute the area \( A \) back into the force formula:
\[ F = \frac{Q^2}{2 \left(\pi \frac{D^2}{4}\right) \epsilon_0} \]
This simplifies to:
\[ F = \frac{Q^2}{\frac{2\pi D^2}{4} \epsilon_0} = \frac{Q^2 \cdot 4}{2\pi D^2 \epsilon_0} = \frac{2Q^2}{\pi D^2 \epsilon_0} \]

### Step 4: Final Expression for the Force
Thus, the magnitude of the force acting between the plates of the capacitor is:
\[ F = \frac{2Q^2}{\pi \epsilon_0 D^2} \]

### Conclusion
The final answer for the magnitude of the force acting between the plates of the capacitor is:
\[ F = \frac{2Q^2}{\pi \epsilon_0 D^2} \]