A parallel plate capacitor of plate separation `2 mm` is connected in an electric circuit having source voltage `400 V`. If the plate area is `60 cm^(2)`, then the value of displacement current for `10^(-6) sec` will be

Options

A. `1.062 amp`
B. `1.062xx10^(-2)amp`
C. `1.062xx10^(-3)amp`
D. `1.062xx10^(-4)amp`

Explanation

To find the value of the displacement current for a parallel plate capacitor, we can follow these steps:

### Step 1: Understand the displacement current formula
The displacement current \( I_d \) can be defined using the formula:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux and \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

### Step 2: Calculate the electric field \( E \)
The electric field \( E \) between the plates of a capacitor can be calculated using:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates and \( d \) is the separation between the plates.

Given:
- \( V = 400 \, \text{V} \)
- \( d = 2 \, \text{mm} = 2 \times 10^{-3} \, \text{m} \)

Substituting the values:
\[ E = \frac{400}{2 \times 10^{-3}} = 200000 \, \text{V/m} \]

### Step 3: Calculate the area \( A \)
Convert the area from \( \text{cm}^2 \) to \( \text{m}^2 \):
\[ A = 60 \, \text{cm}^2 = 60 \times 10^{-4} \, \text{m}^2 = 6 \times 10^{-3} \, \text{m}^2 \]

### Step 4: Calculate the electric flux \( \Phi_E \)
The electric flux \( \Phi_E \) through the capacitor is given by:
\[ \Phi_E = E \cdot A \]
Substituting the values:
\[ \Phi_E = 200000 \times 6 \times 10^{-3} = 1200 \, \text{V m} \]

### Step 5: Calculate the displacement current \( I_d \)
The displacement current can be calculated as:
\[ I_d = \epsilon_0 \frac{\Phi_E}{t} \]
Given \( t = 10^{-6} \, \text{s} \):
\[ I_d = 8.85 \times 10^{-12} \cdot \frac{1200}{10^{-6}} = 8.85 \times 10^{-12} \cdot 1.2 \times 10^{9} \]
Calculating this gives:
\[ I_d = 10.62 \times 10^{-3} \, \text{A} = 10.62 \, \text{mA} \]

### Final Answer
The value of the displacement current for \( 10^{-6} \, \text{s} \) is approximately \( 10.62 \, \text{mA} \).

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