A conducting sphere of radius `10 cm` is charged `10 muC`. Another uncharged sphere of radius `20 cm` is allowed to touch it for some tome. After that if the sphere are separted, then surface density of chsrges, on the spheres will be in the ratio of

Options

A. `1 : 4`
B. `1 : 3`
C. `2 : 1`
D. `1 : 1`

Explanation

To solve the problem step by step, we need to find the surface charge densities on the two spheres after they touch and then separate.

### Step 1: Understand the initial conditions
- We have two spheres:
- Sphere 1 (charged): radius \( r_1 = 10 \, \text{cm} = 0.1 \, \text{m} \), charge \( Q_1 = 10 \, \mu\text{C} = 10 \times 10^{-6} \, \text{C} \)
- Sphere 2 (uncharged): radius \( r_2 = 20 \, \text{cm} = 0.2 \, \text{m} \), charge \( Q_2 = 0 \)

### Step 2: Calculate the potential of each sphere
The potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q}{r} \]
For Sphere 1:
\[ V_1 = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_1}{r_1} \]
For Sphere 2 (initially uncharged):
\[ V_2 = 0 \]

### Step 3: After touching, both spheres will have the same potential
When the spheres touch, charge will redistribute until they reach the same potential \( V \). Let the final charges on the spheres be \( Q_1' \) and \( Q_2' \). The total charge is conserved:
\[ Q_1' + Q_2' = Q_1 = 10 \, \mu\text{C} \]

### Step 4: Set up the equation for equal potentials
Since the potentials are equal after they touch:
\[ \frac{Q_1'}{r_1} = \frac{Q_2'}{r_2} \]
Substituting \( r_1 \) and \( r_2 \):
\[ \frac{Q_1'}{0.1} = \frac{Q_2'}{0.2} \]
This simplifies to:
\[ 2Q_1' = Q_2' \quad \text{(1)} \]

### Step 5: Substitute \( Q_2' \) in terms of \( Q_1' \)
From equation (1), we can express \( Q_2' \) as:
\[ Q_2' = 2Q_1' \]

### Step 6: Substitute into the charge conservation equation
Substituting \( Q_2' \) into the conservation equation:
\[ Q_1' + 2Q_1' = 10 \, \mu\text{C} \]
\[ 3Q_1' = 10 \, \mu\text{C} \]
\[ Q_1' = \frac{10 \, \mu\text{C}}{3} \quad \text{and} \quad Q_2' = 2Q_1' = \frac{20 \, \mu\text{C}}{3} \]

### Step 7: Calculate the surface charge densities
The surface charge density \( \sigma \) is given by:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the sphere \( A = 4\pi r^2 \).

For Sphere 1:
\[ \sigma_1 = \frac{Q_1'}{4\pi r_1^2} = \frac{\frac{10 \, \mu\text{C}}{3}}{4\pi (0.1)^2} = \frac{10 \times 10^{-6}/3}{4\pi \times 0.01} \]

For Sphere 2:
\[ \sigma_2 = \frac{Q_2'}{4\pi r_2^2} = \frac{\frac{20 \, \mu\text{C}}{3}}{4\pi (0.2)^2} = \frac{20 \times 10^{-6}/3}{4\pi \times 0.04} \]

### Step 8: Find the ratio of surface charge densities
Now, we can find the ratio:
\[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{10 \times 10^{-6}/3}{4\pi \times 0.01}}{\frac{20 \times 10^{-6}/3}{4\pi \times 0.04}} = \frac{10 \times 0.04}{20 \times 0.01} = \frac{0.4}{0.2} = 2 \]

### Final Answer
The ratio of the surface charge densities \( \sigma_1 : \sigma_2 = 2 : 1 \).