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

Answer

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. **Understanding Displacement Current**:
The displacement current \( I_d \) is defined as the rate of change of electric flux, which can be expressed mathematically as:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \epsilon_0 \) is the permittivity of free space and \( \Phi_E \) is the electric flux.

2. **Relating Electric Flux to Charge**:
From Gauss's law, the electric flux \( \Phi_E \) can be expressed in terms of the charge \( Q \) on the capacitor plates:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]

3. **Finding the Rate of Change of Charge**:
The problem states that the capacitor is losing charge at a rate of \( \frac{dQ}{dt} = -1.8 \times 10^{-8} \, \text{C/s} \). The negative sign indicates a loss of charge.

4. **Substituting into the Displacement Current Formula**:
Since the displacement current can also be expressed as:
\[ I_d = \frac{dQ}{dt} \]
We can directly substitute the given rate of charge loss into this equation:
\[ I_d = -1.8 \times 10^{-8} \, \text{C/s} \]

5. **Magnitude of Displacement Current**:
The magnitude of the displacement current is simply the absolute value of \( I_d \):
\[ |I_d| = 1.8 \times 10^{-8} \, \text{C/s} \]

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

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