A parallel plate capacitor is charged to `60 muC`. Due to a radioactive source, the plate losses charge at the rate of `1.8 xx 10^(-8) Cs^(-1)`. The magnitued of displacement current is

Options

A. `1.8 xx 10^(-8) Cs^(-1)`
B. `3.6 xx 10^(-8) Cs^(-1)`
C. `4.1 xx 10^(-11) Cs^(-1)`
D. `5.7 xx 10^(-12) Cs^(-1)`

Explanation

To find the magnitude of the displacement current in a parallel plate capacitor that is losing charge, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Displacement Current**:
Displacement current is a term introduced by James Clerk Maxwell to account for the changing electric field in situations where the electric field is changing with time, such as in a capacitor. The displacement current \( I_d \) can be expressed as:
\[ I_d = \epsilon \frac{d\Phi}{dt} \]
where \( \epsilon \) is the permittivity of the medium and \( \Phi \) is the electric flux.

2. **Identify the Rate of Charge Loss**:
The problem states that the capacitor is losing charge at a rate of \( 1.8 \times 10^{-8} \, \text{C/s} \). This rate of charge loss can be represented as:
\[ \frac{dQ}{dt} = -1.8 \times 10^{-8} \, \text{C/s} \]
(The negative sign indicates a loss of charge.)

3. **Relate Displacement Current to Rate of Change of Charge**:
In a capacitor, the displacement current \( I_d \) can also be defined as:
\[ I_d = \frac{dQ}{dt} \]
Since the displacement current is equal to the rate of change of charge, we can substitute the value we have:
\[ I_d = 1.8 \times 10^{-8} \, \text{A} \]

4. **Conclusion**:
Therefore, the magnitude of the displacement current is:
\[ I_d = 1.8 \times 10^{-8} \, \text{A} \]

### Final Answer:
The magnitude of the displacement current is \( 1.8 \times 10^{-8} \, \text{A} \).