What is the displacement current between the square plate of side `1`cm is of a capacitor, if electric field between the plates is changing at the rate of `3 xx 10^(6) Vm^(-1)s^(-1)`?

Options

A. `2.7 xx 10^(-9)A`
B. `3.2 xx 10^(-5)A`
C. `4.2 xx 10^(-6) A`
D. `4.0 xx 10^(-5) A`

Explanation

To find the displacement current between the square plates of a capacitor, we can use the formula for displacement current \( I_d \):

\[ I_d = \epsilon_0 A \frac{dE}{dt} \]

where:
- \( I_d \) is the displacement current,
- \( \epsilon_0 \) is the permittivity of free space (\( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)),
- \( A \) is the area of the capacitor plates,
- \( \frac{dE}{dt} \) is the rate of change of the electric field.

### Step 1: Calculate the area \( A \) of the capacitor plates
Given that the side of the square plate is \( 1 \, \text{cm} \):
\[ A = \text{side}^2 = (1 \, \text{cm})^2 = (0.01 \, \text{m})^2 = 1 \times 10^{-4} \, \text{m}^2 \]

### Step 2: Use the given rate of change of electric field
The rate of change of electric field is given as:
\[ \frac{dE}{dt} = 3 \times 10^6 \, \text{V/m/s} \]

### Step 3: Substitute values into the displacement current formula
Now, substituting the values into the formula:
\[ I_d = \epsilon_0 A \frac{dE}{dt} \]
\[ I_d = (8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2) \times (1 \times 10^{-4} \, \text{m}^2) \times (3 \times 10^6 \, \text{V/m/s}) \]

### Step 4: Calculate \( I_d \)
Calculating the above expression:
\[ I_d = 8.85 \times 10^{-12} \times 1 \times 10^{-4} \times 3 \times 10^6 \]
\[ I_d = 8.85 \times 3 \times 10^{-10} = 26.55 \times 10^{-10} = 2.655 \times 10^{-9} \, \text{A} \]

### Step 5: Round off the answer
Rounding off gives:
\[ I_d \approx 2.7 \times 10^{-9} \, \text{A} \]

Thus, the displacement current between the square plates of the capacitor is approximately \( 2.7 \, \text{nA} \).

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