A parallel plate capacitor consits of two circular plates of radius 0.5m. If electric field between
the plates changed as `(dE)/(dt) = 10^(10). (V)/(m-c)`, then find displacement current between the plates

Options

A. 7 mA
B. 0.07 mA
C. 0.7 mA
D. 70 mA

Answer

To find the displacement current between the plates of a parallel plate capacitor, 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 conduction current. 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: Calculate 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 between the plates and \(A\) is the area of the plates.

### Step 3: Determine the Area of the Plates
The area \(A\) of a circular plate is given by:

\[ A = \pi r^2 \]

Given that the radius \(r = 0.5 \, m\):

\[ A = \pi (0.5)^2 = \pi \cdot 0.25 \approx 0.7854 \, m^2 \]

### Step 4: Differentiate the Electric Flux
To find the displacement current, we need to differentiate the electric flux with respect to time:

\[ \frac{d\Phi_E}{dt} = A \frac{dE}{dt} \]

### Step 5: Substitute the Values
We know that:

- \(\frac{dE}{dt} = 10^{10} \, V/m/s\)
- \(A \approx 0.7854 \, m^2\)

Now substituting the values:

\[ \frac{d\Phi_E}{dt} = 0.7854 \cdot 10^{10} \]

### Step 6: Calculate the Displacement Current
Now substituting this into the formula for displacement current:

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

Where \(\epsilon_0 \approx 8.85 \times 10^{-12} \, F/m\):

\[ I_d = 8.85 \times 10^{-12} \cdot (0.7854 \cdot 10^{10}) \]

Calculating this gives:

\[ I_d \approx 8.85 \times 10^{-12} \cdot 7.854 \times 10^{9} \approx 6.95 \times 10^{-2} \, A \]

### Step 7: Convert to Milliamperes
To express this in milliamperes:

\[ I_d \approx 69.5 \, mA \approx 0.07 \, A \]

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

\[ I_d \approx 0.07 \, A \text{ or } 70 \, mA \]