The voltage between the plates of a parallel plate capacitor of capacitance `2 mu F` is changing at the rate of `4 V//s`. What is the displacement current in the capacitor?

Options

A. `5 muA`
B. `6 muA`
C. `7 muA`
D. `8 muA`

Explanation

To find the displacement current in a parallel plate capacitor, we can use the formula for displacement current \( I_d \):

\[ I_d = C \frac{dV}{dt} \]

where:
- \( I_d \) is the displacement current,
- \( C \) is the capacitance of the capacitor,
- \( \frac{dV}{dt} \) is the rate of change of voltage across the capacitor.

### Step-by-step Solution:

1. **Identify the given values**:
- Capacitance \( C = 2 \, \mu F = 2 \times 10^{-6} \, F \)
- Rate of change of voltage \( \frac{dV}{dt} = 4 \, V/s \)

2. **Substitute the values into the displacement current formula**:
\[ I_d = C \frac{dV}{dt} = (2 \times 10^{-6} \, F) \times (4 \, V/s) \]

3. **Calculate the displacement current**:
\[ I_d = 2 \times 10^{-6} \times 4 = 8 \times 10^{-6} \, A \]

4. **Convert the result into microamperes**:
\[ I_d = 8 \, \mu A \]

5. **Final Answer**:
The displacement current in the capacitor is \( 8 \, \mu A \).