Two spheres of radii 2 cm and 3 cm are charged to the same potential. If `sigma` and `sigma_2` be respectively the values of surface charge density on the conductors, then the ratio `(sigma_1)/(sigma_2)` will be

Options

A. `4/9`
B. `2/3`
C. `3/2`
D. `9/4`

Answer

To solve the problem, we need to find the ratio of the surface charge densities (\( \sigma_1 \) and \( \sigma_2 \)) of two spheres with radii \( R_1 = 2 \, \text{cm} \) and \( R_2 = 3 \, \text{cm} \) that are charged to the same potential \( V \).

### Step-by-Step Solution:

1. **Understand the Relationship between Charge, Radius, and Potential**:
The potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( R \) is the radius of the sphere.

2. **Express Charge in Terms of Potential and Radius**:
Rearranging the formula for potential, we can express the charge \( Q \) as:
\[ Q = \frac{VR}{k} \]

3. **Calculate Surface Charge Density**:
The surface charge density \( \sigma \) is defined as the charge per unit area:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the surface area of the sphere, given by \( A = 4\pi R^2 \). Therefore, we can write:
\[ \sigma = \frac{Q}{4\pi R^2} \]

4. **Substituting for Charge**:
Substituting the expression for \( Q \) into the surface charge density formula, we get:
\[ \sigma = \frac{VR}{k \cdot 4\pi R^2} = \frac{V}{4\pi k R} \]

5. **Finding the Ratio of Surface Charge Densities**:
Now, we can find the ratio of the surface charge densities \( \sigma_1 \) and \( \sigma_2 \):
\[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{V}{4\pi k R_1}}{\frac{V}{4\pi k R_2}} = \frac{R_2}{R_1} \]

6. **Substituting the Values of Radii**:
Given \( R_1 = 2 \, \text{cm} \) and \( R_2 = 3 \, \text{cm} \):
\[ \frac{\sigma_1}{\sigma_2} = \frac{3}{2} \]

### Final Answer:
The ratio of the surface charge densities is:
\[ \frac{\sigma_1}{\sigma_2} = \frac{3}{2} \]