A charge of `24 muC ` is given to a hollow metallic sphere of radius 0.2m. Find the potential
(i) at the surface of sphere
(ii) at a distance of 0.1 cm from the center of sphere.

Explanation

To solve the problem, we need to calculate the electric potential at two different points related to a hollow metallic sphere with a given charge.

### Given Data:
- Charge (Q) = 24 µC = \( 24 \times 10^{-6} \) C
- Radius of the sphere (R) = 0.2 m
- Distance from the center for the second part = 0.1 cm = 0.001 m

### Step 1: Calculate the potential at the surface of the sphere

The formula for electric potential (V) due to a point charge or a charged sphere at its surface is given by:

\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{R} \]

Where:
- \( \epsilon_0 \) (the permittivity of free space) = \( 8.85 \times 10^{-12} \, \text{C}^2/\text{N m}^2 \)
- \( Q \) is the charge
- \( R \) is the radius of the sphere

Substituting the values:

\[ V = \frac{1}{4\pi (8.85 \times 10^{-12})} \cdot \frac{24 \times 10^{-6}}{0.2} \]

Calculating \( \frac{1}{4\pi \epsilon_0} \):

\[ \frac{1}{4\pi \epsilon_0} \approx 9 \times 10^9 \, \text{N m}^2/\text{C}^2 \]

Now substituting this into the equation:

\[ V = (9 \times 10^9) \cdot \frac{24 \times 10^{-6}}{0.2} \]

Calculating \( \frac{24 \times 10^{-6}}{0.2} \):

\[ \frac{24 \times 10^{-6}}{0.2} = 120 \times 10^{-6} = 1.2 \times 10^{-4} \]

Now substituting this back into the potential equation:

\[ V = 9 \times 10^9 \cdot 1.2 \times 10^{-4} = 1.08 \times 10^6 \, \text{V} \]

### Step 2: Calculate the potential at a distance of 0.1 cm from the center of the sphere

For a hollow metallic sphere, the electric potential inside the sphere (at any point inside) is equal to the potential at the surface. Therefore, the potential at a distance of 0.1 cm (0.001 m) from the center of the sphere is the same as the potential at the surface.

Thus,

\[ V = 1.08 \times 10^6 \, \text{V} \]

### Final Answers:
(i) The potential at the surface of the sphere is \( 1.08 \times 10^6 \, \text{V} \).

(ii) The potential at a distance of 0.1 cm from the center of the sphere is also \( 1.08 \times 10^6 \, \text{V} \).