Radius of a spherical conductor is `2m`, This is kept in dielectric medium of dielectric constant `10^6N//C`. Find
a. capacitance of the conductor
b. maximum charge which can be stored on this conductor.

Explanation

To solve the problem, we need to find two things:

1. The capacitance of the spherical conductor.
2. The maximum charge that can be stored on this conductor.

### Step 1: Calculate the Capacitance of the Spherical Conductor

The formula for the capacitance \( C \) of a spherical conductor in a dielectric medium is given by:

\[ C = 4 \pi \epsilon R \]

Where:
- \( \epsilon \) is the permittivity of the dielectric medium.
- \( R \) is the radius of the spherical conductor.

Given:
- Radius \( R = 2 \, \text{m} \)
- The dielectric constant \( K = 10^6 \, \text{N/C} \)

The permittivity \( \epsilon \) can be calculated as:

\[ \epsilon = K \cdot \epsilon_0 \]

Where \( \epsilon_0 \) (the permittivity of free space) is approximately \( 8.85 \times 10^{-12} \, \text{F/m} \).

Calculating \( \epsilon \):

\[ \epsilon = 10^6 \cdot 8.85 \times 10^{-12} = 8.85 \times 10^{-6} \, \text{F/m} \]

Now substituting \( \epsilon \) and \( R \) into the capacitance formula:

\[ C = 4 \pi (8.85 \times 10^{-6}) (2) \]

Calculating \( C \):

\[ C = 4 \pi (8.85 \times 10^{-6}) (2) \approx 2.22 \times 10^{-5} \, \text{F} = 2.22 \times 10^{-10} \, \text{F} \]

### Step 2: Calculate the Maximum Charge that can be Stored

The maximum charge \( Q_{\text{max}} \) that can be stored on the conductor is given by:

\[ Q_{\text{max}} = C \cdot V_{\text{max}} \]

Where \( V_{\text{max}} \) is the maximum potential difference, which can be derived from the electric field \( E_{\text{max}} \):

\[ E_{\text{max}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q_{\text{max}}}{R^2} \]

Given that \( E_{\text{max}} = 10^6 \, \text{N/C} \), we can rearrange the equation to find \( Q_{\text{max}} \):

\[ 10^6 = \frac{1}{4 \pi (8.85 \times 10^{-12})} \cdot \frac{Q_{\text{max}}}{(2)^2} \]

Solving for \( Q_{\text{max}} \):

\[ Q_{\text{max}} = 10^6 \cdot 4 \pi (8.85 \times 10^{-12}) \cdot 4 \]

Calculating \( Q_{\text{max}} \):

\[ Q_{\text{max}} \approx 4.4 \times 10^{-4} \, \text{C} \]

### Final Answers

a. Capacitance of the conductor: \( C \approx 2.22 \times 10^{-10} \, \text{F} \)

b. Maximum charge that can be stored: \( Q_{\text{max}} \approx 4.4 \times 10^{-4} \, \text{C} \)