The area of each plate of parallel plated condenser is `144 cm^(2)`. The electrical field in the gap between the plates changes at the rate of `10^(12)Vm^(-1)s^(-1)`. The displacement current is

Options

A. `(4)/(pi)A`
B. `(0.4)/(pi)A`
C. `(40)/(pi)A`
D. `(1)/(10pi)A`

Explanation

To find the displacement current for the given parallel plate condenser, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Area of each plate, \( A = 144 \, \text{cm}^2 \)
- Rate of change of electric field, \( \frac{dE}{dt} = 10^{12} \, \text{V/m/s} \)

2. **Convert Area to Square Meters:**
\[ A = 144 \, \text{cm}^2 = 144 \times 10^{-4} \, \text{m}^2 = 0.0144 \, \text{m}^2 \]

3. **Use the Formula for Displacement Current:**
The displacement current \( I_d \) can be calculated using the formula:
\[ I_d = \epsilon_0 \cdot A \cdot \frac{dE}{dt} \]
where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

4. **Substitute the Values into the Formula:**
\[ I_d = (8.85 \times 10^{-12} \, \text{F/m}) \cdot (0.0144 \, \text{m}^2) \cdot (10^{12} \, \text{V/m/s}) \]

5. **Calculate the Displacement Current:**
\[ I_d = 8.85 \times 10^{-12} \cdot 0.0144 \cdot 10^{12} \]
\[ I_d = 8.85 \times 0.0144 \, \text{A} \]
\[ I_d = 0.12744 \, \text{A} \]

6. **Final Result:**
The displacement current \( I_d \) is approximately:
\[ I_d \approx 0.127 \, \text{A} \quad \text{or} \quad \frac{4}{\pi} \, \text{A} \text{ (as per the video)} \]