Two isolated metallic solid spheres of radii `R` and `2R` are charged such that both have same charge density `sigma`.The spheres are then connected by a thin conducting wire.If the new charge density of the bigger sphere is `sigma'`.The ratio `(sigma')/(sigma)` is

Options

A. `5/3`
B. `9/4`
C. `5/6`
D. `4/3`

Answer

To solve the problem, we need to analyze the situation step by step.

### Step 1: Determine the initial charges on the spheres
Given two metallic spheres with radii \( R \) and \( 2R \) and the same surface charge density \( \sigma \):

1. **Charge on the smaller sphere (radius \( R \))**:
\[ Q_1 = \sigma \times \text{Surface Area} = \sigma \times 4\pi R^2 \]

2. **Charge on the larger sphere (radius \( 2R \))**:
\[ Q_2 = \sigma \times \text{Surface Area} = \sigma \times 4\pi (2R)^2 = \sigma \times 4\pi \times 4R^2 = 4\sigma \times 4\pi R^2 = 16\pi R^2 \sigma \]

### Step 2: Calculate the total charge when the spheres are connected
When the two spheres are connected by a thin conducting wire, charge will redistribute between them until they reach the same potential.

3. **Total charge before connection**:
\[ Q_{\text{total}} = Q_1 + Q_2 = 4\pi R^2 \sigma + 16\pi R^2 \sigma = 20\pi R^2 \sigma \]

### Step 3: Determine the new charges after connection
Let \( Q_1' \) and \( Q_2' \) be the new charges on the smaller and larger spheres, respectively. The potential \( V \) on both spheres must be equal after they are connected.

4. **Potential of the smaller sphere**:
\[ V_1 = \frac{Q_1'}{4\pi R^2} \]

5. **Potential of the larger sphere**:
\[ V_2 = \frac{Q_2'}{4\pi (2R)^2} = \frac{Q_2'}{16\pi R^2} \]

Setting the potentials equal:
\[ \frac{Q_1'}{4\pi R^2} = \frac{Q_2'}{16\pi R^2} \]

Cross-multiplying gives:
\[ 16 Q_1' = 4 Q_2' \implies Q_2' = 4 Q_1' \]

### Step 4: Express total charge in terms of \( Q_1' \)
Using the total charge:
\[ Q_{\text{total}} = Q_1' + Q_2' = Q_1' + 4 Q_1' = 5 Q_1' \]
Thus,
\[ 20\pi R^2 \sigma = 5 Q_1' \implies Q_1' = \frac{20\pi R^2 \sigma}{5} = 4\pi R^2 \sigma \]

### Step 5: Find \( Q_2' \)
Using \( Q_2' = 4 Q_1' \):
\[ Q_2' = 4 \times 4\pi R^2 \sigma = 16\pi R^2 \sigma \]

### Step 6: Calculate the new charge density \( \sigma' \) on the larger sphere
The new charge density \( \sigma' \) on the larger sphere is given by:
\[ \sigma' = \frac{Q_2'}{\text{Surface Area of larger sphere}} = \frac{16\pi R^2 \sigma}{16\pi (2R)^2} = \frac{16\pi R^2 \sigma}{64\pi R^2} = \frac{16\sigma}{64} = \frac{\sigma}{4} \]

### Step 7: Find the ratio \( \frac{\sigma'}{\sigma} \)
Finally, we can find the ratio:
\[ \frac{\sigma'}{\sigma} = \frac{\frac{\sigma}{4}}{\sigma} = \frac{1}{4} \]

### Conclusion
The ratio \( \frac{\sigma'}{\sigma} \) is \( \frac{1}{4} \).