A spherical capacitor has the inner sphere of radius `2 cm` and the outerone of `4 cm`. If the inner sphere is earthed and the outer one is charged with a charge of `2muC` and isolated. Calculate ltbr. (a) the potential to which the outer sphere is raised.
(b) the charge retained on the outer surface of the outer sphere.

Explanation

To solve the problem step by step, we will break it down into two parts as required:

### Part (a): Calculate the potential to which the outer sphere is raised.

1. **Identify the given values:**
- Radius of the inner sphere, \( r_1 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Radius of the outer sphere, \( r_2 = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Charge on the outer sphere, \( Q_{\text{outer}} = 2 \, \mu C = 2 \times 10^{-6} \, C \)
- The inner sphere is earthed, so its potential is \( V_{\text{inner}} = 0 \).

2. **Use the formula for potential due to a charged sphere:**
The potential \( V \) at a distance \( r \) from a point charge \( Q \) is given by:
\[ V = \frac{kQ}{r} \]
where \( k = 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \).

3. **Set up the equation for the potential at the inner sphere:**
Since the inner sphere is earthed, its potential is zero:
\[ V_{\text{inner}} = \frac{kQ_{\text{inner}}}{r_1} + \frac{kQ_{\text{outer}}}{r_2} = 0 \]

4. **Let the charge on the inner sphere be \( Q_{\text{inner}} \):**
\[ 0 = \frac{kQ_{\text{inner}}}{0.02} + \frac{k(2 \times 10^{-6})}{0.04} \]

5. **Simplify the equation:**
\[ 0 = \frac{Q_{\text{inner}}}{0.02} + \frac{2 \times 10^{-6}}{0.04} \]
\[ 0 = \frac{Q_{\text{inner}}}{0.02} + 5 \times 10^{-5} \]
\[ Q_{\text{inner}} = -1 \times 10^{-6} \, C = -1 \, \mu C \]

6. **Calculate the potential of the outer sphere:**
The potential at the outer sphere is given by:
\[ V_{\text{outer}} = \frac{kQ_{\text{enclosed}}}{r_2} \]
where \( Q_{\text{enclosed}} = Q_{\text{outer}} + Q_{\text{inner}} = 2 \times 10^{-6} - 1 \times 10^{-6} = 1 \times 10^{-6} \, C \).

7. **Plug in the values:**
\[ V_{\text{outer}} = \frac{(9 \times 10^9)(1 \times 10^{-6})}{0.04} \]
\[ V_{\text{outer}} = \frac{9 \times 10^3}{0.04} = 225000 \, V = 2.25 \times 10^5 \, V \]

### Part (b): Calculate the charge retained on the outer surface of the outer sphere.

1. **Determine the charge on the outer surface:**
The total charge on the outer sphere is \( Q_{\text{outer}} = 2 \, \mu C \). The charge on the inner sphere is \( Q_{\text{inner}} = -1 \, \mu C \).

2. **Use the principle of conservation of charge:**
The charge on the outer surface can be calculated as:
\[ Q_{\text{outer surface}} = Q_{\text{outer}} - Q_{\text{inner}} \]
\[ Q_{\text{outer surface}} = 2 \times 10^{-6} - (-1 \times 10^{-6}) = 2 \times 10^{-6} + 1 \times 10^{-6} = 3 \times 10^{-6} \, C = 1 \, \mu C \]

### Final Answers:
- (a) The potential to which the outer sphere is raised is \( 2.25 \times 10^5 \, V \).
- (b) The charge retained on the outer surface of the outer sphere is \( 1 \, \mu C \).