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 the given problem, we can follow these steps:

### Step 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 regions where there is no actual current flow. It is given by the equation:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( I_d \) is the displacement current, \( \epsilon_0 \) is the permittivity of free space, and \( \Phi_E \) is the electric flux.

### Step 2: Relate electric flux to charge
From Gauss's law, we know that the electric flux \( \Phi_E \) through a capacitor can be expressed as:
\[ \Phi_E = \frac{Q}{\epsilon_0} \]
where \( Q \) is the charge on the capacitor plates.

### Step 3: Differentiate electric flux with respect to time
To find the displacement current, we need to differentiate the electric flux with respect to time:
\[ \frac{d\Phi_E}{dt} = \frac{1}{\epsilon_0} \frac{dQ}{dt} \]

### Step 4: Substitute into the displacement current formula
Substituting this expression into the displacement current formula gives:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} = \epsilon_0 \cdot \frac{1}{\epsilon_0} \frac{dQ}{dt} = \frac{dQ}{dt} \]

### Step 5: Use the given rate of charge loss
In the problem, it is given that the charge is lost at a rate of:
\[ \frac{dQ}{dt} = -1.8 \times 10^{-8} \, \text{C/s} \]
(Note: The negative sign indicates a loss of charge, but we are interested in the magnitude.)

### Step 6: Calculate the magnitude of displacement current
Thus, the magnitude of the displacement current is:
\[ |I_d| = \left| \frac{dQ}{dt} \right| = 1.8 \times 10^{-8} \, \text{A} \]

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

---