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 xx 10^(-2) A`
B. `2.062 xx 10^(-2) A`
C. `3.062 xx 10^(-2) A`
D. `5.062 xx 10^(-2) A`

Explanation

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

### Step 1: Understand the formula for displacement current
The displacement current \( I_d \) is given by the equation:
\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux.

### Step 2: Determine the electric flux
The electric flux \( \Phi_E \) through the capacitor plates can be expressed as:
\[ \Phi_E = E \cdot A \]
where \( E \) is the electric field and \( A \) is the area of the plates.

### Step 3: Calculate the electric field \( E \)
The electric field \( E \) between the plates of a capacitor is given by:
\[ E = \frac{V}{d} \]
where \( V \) is the voltage across the plates and \( d \) is the separation between the plates.

### Step 4: Substitute the values
Given:
- Voltage \( V = 400 \, V \)
- Plate separation \( d = 2 \, mm = 2 \times 10^{-3} \, m \)
- Plate area \( A = 60 \, cm^2 = 60 \times 10^{-4} \, m^2 \)

Calculating the electric field:
\[ E = \frac{400}{2 \times 10^{-3}} = 200000 \, V/m \]

### Step 5: Calculate the change in electric flux over time
The change in electric flux \( \Delta \Phi_E \) over a time interval \( T \) is:
\[ \Delta \Phi_E = E \cdot A \]
Substituting the values:
\[ \Delta \Phi_E = 200000 \cdot (60 \times 10^{-4}) = 1200 \, V \cdot m^2 \]

### Step 6: Calculate the displacement current
Now, we can find the displacement current using:
\[ I_d = \epsilon_0 \frac{\Delta \Phi_E}{T} \]
where \( T = 10^{-6} \, s \) and \( \epsilon_0 = 8.85 \times 10^{-12} \, F/m \).

Substituting the values:
\[ I_d = 8.85 \times 10^{-12} \cdot \frac{1200}{10^{-6}} = 1.062 \times 10^{-2} \, A \]

### Conclusion
The value of the displacement current is:
\[ I_d \approx 1.062 \times 10^{-2} \, A \]