If electric field is changing at rate of `6xx10^(6)` V/ms between the plates of a capacitor. Having plate areas `2.0cm^(2)`, then the displacement current is

Options

A. `1.06xx10^(-9)`A
B. `2.7xx10^(-9)A`
C. `5.4xx10^(-7)A`
D. `2xx10^(-6)A`

Answer

To find the displacement current between the plates of a capacitor, we can use the formula derived from Maxwell's equations:

\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

Where:
- \( I_d \) is the displacement current,
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{F/m} \)),
- \( A \) is the area of the plates,
- \( \frac{dE}{dt} \) is the rate of change of the electric field.

### Step-by-step Solution:

1. **Identify the given values:**
- Rate of change of electric field, \( \frac{dE}{dt} = 6 \times 10^6 \, \text{V/m/s} \)
- Area of the plates, \( A = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2 \)

2. **Substitute the values into the formula:**
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]
\[ I_d = (8.85 \times 10^{-12} \, \text{F/m}) \times (2.0 \times 10^{-4} \, \text{m}^2) \times (6 \times 10^6 \, \text{V/m/s}) \]

3. **Calculate the product:**
- First, calculate \( \epsilon_0 A \):
\[ \epsilon_0 A = 8.85 \times 10^{-12} \times 2.0 \times 10^{-4} = 1.77 \times 10^{-15} \, \text{F} \]
- Now multiply by \( \frac{dE}{dt} \):
\[ I_d = 1.77 \times 10^{-15} \times 6 \times 10^6 \]

4. **Perform the final calculation:**
\[ I_d = 1.77 \times 6 \times 10^{-15+6} = 1.062 \times 10^{-9} \, \text{A} \]

5. **Final result:**
\[ I_d \approx 1.06 \times 10^{-9} \, \text{A} \]

### Conclusion:
The displacement current \( I_d \) is approximately \( 1.06 \times 10^{-9} \, \text{A} \).