There is one parallel plate capacitor with circular plates of radius R. Capacitor is being charged by connecting to a battery . If I is the instantaneous conduction current flowing in the connecting wires then what will be displacement current flowing in the region between the plates enclosed between radii, `R//2 ` to R

Options

A. `3I//4`
B. `I`
C. `I//4`
D. `3I//5`

Explanation

To solve the problem, we need to find the displacement current flowing in the region between the plates of a parallel plate capacitor, specifically between the radii \( R/2 \) and \( R \).

### Step-by-Step Solution:

1. **Understanding the Setup**:
We have a parallel plate capacitor with circular plates of radius \( R \). The capacitor is being charged, which means that there is a conduction current \( I \) flowing in the connecting wires.

2. **Concept of Displacement Current**:
The displacement current \( I_D \) is related to the changing electric field between the plates due to the charging of the capacitor. According to Maxwell's equations, the displacement current can be expressed as:
\[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
where \( \Phi_E \) is the electric flux.

3. **Electric Field Between the Plates**:
The electric field \( E \) between the plates of a capacitor is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density on the plates. The charge \( Q \) on the plates is related to \( \sigma \) by:
\[ Q = \sigma \cdot A \]
where \( A \) is the area of the plates.

4. **Area of the Plates**:
The area \( A \) of the circular plates is:
\[ A = \pi R^2 \]

5. **Finding the Charge Density**:
The surface charge density \( \sigma \) can be expressed as:
\[ \sigma = \frac{Q}{\pi R^2} \]

6. **Calculating the Electric Flux**:
The electric flux \( \Phi_E \) through the area between the radii \( R/2 \) and \( R \) can be calculated as:
\[ \Phi_E = \int E \, dA \]
where \( dA \) is the differential area element. For a circular ring of radius \( r \) and thickness \( dr \), the area element is:
\[ dA = 2\pi r \, dr \]
Thus, the flux becomes:
\[ \Phi_E = \int_{R/2}^{R} E \cdot 2\pi r \, dr \]

7. **Substituting for Electric Field**:
Substituting \( E = \frac{Q}{\epsilon_0 \pi R^2} \) into the flux equation gives:
\[ \Phi_E = \int_{R/2}^{R} \frac{Q}{\epsilon_0 \pi R^2} \cdot 2\pi r \, dr \]
This simplifies to:
\[ \Phi_E = \frac{Q}{\epsilon_0 R^2} \int_{R/2}^{R} r \, dr \]

8. **Calculating the Integral**:
The integral \( \int_{R/2}^{R} r \, dr \) evaluates to:
\[ \left[ \frac{r^2}{2} \right]_{R/2}^{R} = \frac{R^2}{2} - \frac{(R/2)^2}{2} = \frac{R^2}{2} - \frac{R^2}{8} = \frac{4R^2}{8} - \frac{R^2}{8} = \frac{3R^2}{8} \]

9. **Finding the Displacement Current**:
The displacement current \( I_D \) in the region between \( R/2 \) and \( R \) can now be expressed as:
\[ I_D = \epsilon_0 \frac{d\Phi_E}{dt} \]
Since \( \Phi_E \) depends on \( Q \), and \( Q \) is changing with time as the capacitor charges, we can relate \( I_D \) to the conduction current \( I \).

10. **Using Kirchhoff's Current Law**:
By Kirchhoff's current law, the conduction current \( I \) entering the capacitor must equal the displacement current \( I_D \) flowing through the region. Since we are considering only the area between \( R/2 \) and \( R \), we find that:
\[ I_D = \frac{3}{4} I \]

### Final Answer:
Thus, the displacement current flowing in the region between the plates enclosed between the radii \( R/2 \) and \( R \) is:
\[ \boxed{\frac{3I}{4}} \]