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 \( I_d \), which is given by:

\[ I_d = \epsilon_0 \frac{d\Phi_E}{dt} \]

where \( \epsilon_0 \) is the permittivity of free space, and \( \Phi_E \) is the electric flux.

### Step 1: Calculate the Electric Flux
The electric flux \( \Phi_E \) between the plates of the capacitor can be expressed as:

\[ \Phi_E = E \cdot A \]

where \( E \) is the electric field and \( A \) is the area of the plates.

### Step 2: Determine the Area of the Plates
Given that the side of the square plate is \( 1 \, \text{cm} \), we first convert this to meters:

\[ \text{Side} = 1 \, \text{cm} = 0.01 \, \text{m} \]

Now, calculate the area \( A \):

\[ A = \text{Side}^2 = (0.01 \, \text{m})^2 = 1 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Differentiate the Electric Field
The rate of change of the electric field \( \frac{dE}{dt} \) is given as:

\[ \frac{dE}{dt} = 3 \times 10^6 \, \text{V/m/s} \]

### Step 4: Substitute Values into the Displacement Current Formula
Now, substituting the values into the displacement current formula:

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

Substituting \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{F/m} \), \( A = 1 \times 10^{-4} \, \text{m}^2 \), and \( \frac{dE}{dt} = 3 \times 10^6 \, \text{V/m/s} \):

\[ I_d = (8.85 \times 10^{-12}) \cdot (1 \times 10^{-4}) \cdot (3 \times 10^6) \]

### Step 5: Calculate the Displacement Current
Now, performing the multiplication:

\[ I_d = 8.85 \times 10^{-12} \times 1 \times 10^{-4} \times 3 \times 10^6 \]

Calculating this gives:

\[ I_d = 8.85 \times 3 \times 10^{-12} \times 10^{-4} \times 10^6 = 26.55 \times 10^{-10} = 2.655 \times 10^{-9} \, \text{A} \]

Rounding it gives:

\[ I_d \approx 2.7 \times 10^{-9} \, \text{A} \]

### Final Answer
The displacement current between the square plates of the capacitor is approximately:

\[ I_d \approx 2.7 \times 10^{-9} \, \text{A} \]