The voltage between the plates of a parallel plate capacitor of capacitance `1 muF` is changing at the rate of 5 V/s. What is the displacement current in the capacitor?

Answer

To solve the problem, we need to find the displacement current \( I_d \) in a parallel plate capacitor given its capacitance and the rate of change of voltage across its plates.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Capacitance \( C = 1 \, \mu F = 1 \times 10^{-6} \, F \)
- Rate of change of voltage \( \frac{dV}{dt} = 5 \, V/s \)

2. **Understand Displacement Current**:
- The displacement current \( I_d \) can be calculated using the formula:
\[ I_d = C \cdot \frac{dV}{dt} \]
- This formula relates the displacement current to the capacitance and the rate of change of voltage.

3. **Substitute the Values**:
- Substitute the values of \( C \) and \( \frac{dV}{dt} \) into the formula:
\[ I_d = (1 \times 10^{-6} \, F) \cdot (5 \, V/s) \]

4. **Calculate the Displacement Current**:
- Perform the multiplication:
\[ I_d = 1 \times 10^{-6} \cdot 5 = 5 \times 10^{-6} \, A \]
- Convert to microamperes:
\[ I_d = 5 \, \mu A \]

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