A parallel plate capacitor consists of two circular plates each of radius `2 cm`, separated by a distance of `0.1 mm`. Ifvoltage across the plates is varying at the rate of `5xx10^(13) V//s`, then the value of displacement current is :

Options

A. `5.50` A
B. `5.56xx10^(2)` A
C. `5.56xx10^(3)` A
D. `2.28xx10^(4)` A

Explanation

To find the displacement current \( I_D \) in a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the formula for displacement current
The displacement current \( I_D \) is given by the equation:
\[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux and \( \epsilon_0 \) is the permittivity of free space.

### Step 2: Express electric flux in terms of electric field
The electric flux \( \Phi_E \) can be expressed as:
\[ \Phi_E = E \cdot A \]
where \( E \) is the electric field and \( A \) is the area of the plates.

### Step 3: Relate electric field to voltage
The electric field \( E \) between the plates of the capacitor can be expressed as:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates and \( d \) is the separation between the plates.

### Step 4: Differentiate the electric field with respect to time
Substituting \( E \) into the expression for electric flux, we get:
\[ \Phi_E = \frac{V}{d} \cdot A \]
Now, differentiating \( \Phi_E \) with respect to time gives:
\[ \frac{d\Phi_E}{dt} = \frac{A}{d} \frac{dV}{dt} \]

### Step 5: Substitute into the displacement current formula
Now substituting this back into the formula for displacement current, we have:
\[ I_D = \epsilon_0 \frac{A}{d} \frac{dV}{dt} \]

### Step 6: Calculate the area of the plates
The area \( A \) of a circular plate is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius of the plates. Given \( r = 2 \, \text{cm} = 0.02 \, \text{m} \):
\[ A = \pi (0.02)^2 = \pi (0.0004) \approx 1.25664 \times 10^{-3} \, \text{m}^2 \]

### Step 7: Substitute values into the displacement current formula
Now we can substitute the known values:
- \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \)
- \( d = 0.1 \, \text{mm} = 0.1 \times 10^{-3} \, \text{m} = 1 \times 10^{-4} \, \text{m} \)
- \( \frac{dV}{dt} = 5 \times 10^{13} \, \text{V/s} \)

Thus,
\[ I_D = 8.85 \times 10^{-12} \cdot \frac{1.25664 \times 10^{-3}}{1 \times 10^{-4}} \cdot (5 \times 10^{13}) \]

### Step 8: Calculate the displacement current
Calculating the above expression:
\[ I_D = 8.85 \times 10^{-12} \cdot 12.5664 \cdot 5 \times 10^{13} \]
\[ I_D \approx 8.85 \times 10^{-12} \cdot 62.8319 \times 10^{13} \]
\[ I_D \approx 5.56 \times 10^{3} \, \text{A} \]

### Final Answer
Thus, the displacement current \( I_D \) is approximately:
\[ I_D \approx 5.56 \, \text{mA} \]