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 when the electric field is changing, we can follow these steps:

### Step-by-Step Solution:

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

2. **Convert the units**:
- Convert the rate of change of electric field from V/ms to V/s:
\[ \frac{dE}{dt} = 6 \times 10^6 \, \text{V/ms} = 6 \times 10^6 \times 10^3 \, \text{V/s} = 6 \times 10^9 \, \text{V/s} \]

- Convert the area from cm² to m²:
\[ A = 2.0 \, \text{cm}^2 = 2.0 \times 10^{-4} \, \text{m}^2 \]

3. **Use the formula for displacement current**:
The displacement current \( I_d \) can be calculated using the formula:
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]
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**:
\[ I_d = (8.85 \times 10^{-12} \, \text{F/m}) \times (2.0 \times 10^{-4} \, \text{m}^2) \times (6 \times 10^9 \, \text{V/s}) \]

5. **Calculate the displacement current**:
- First, calculate \( 8.85 \times 10^{-12} \times 2.0 \times 10^{-4} \):
\[ = 1.77 \times 10^{-15} \, \text{F m}^2 \]
- Now multiply by \( 6 \times 10^9 \):
\[ I_d = 1.77 \times 10^{-15} \times 6 \times 10^9 = 1.062 \times 10^{-5} \, \text{A} \]

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

### Summary:
The displacement current \( I_d \) between the plates of the capacitor is approximately \( 1.06 \times 10^{-9} \, \text{A} \).

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