A number of spherical conductors of different radii are charged to same potential. The surface charge density of each conductor is related with its radius as

Options

A. `sigmaprop(1)/(R^(2))`
B. `sigmaprop(1)/(R )`
C. `sigmapropR`
D. none of these

Answer

To solve the problem of finding the relationship between the surface charge density of spherical conductors charged to the same potential and their radii, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We have spherical conductors of different radii (let's denote them as \( r_1, r_2, \ldots \)) that are charged to the same electric potential.

2. **Potential of a Sphere**: The electric potential \( V \) of a charged sphere is given by the formula:
\[ V = \frac{k \cdot Q}{R} \]
where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( R \) is the radius of the sphere.

3. **Setting Up the Equations**: For two spheres with radii \( r_1 \) and \( r_2 \) and charges \( q_1 \) and \( q_2 \), since they are at the same potential:
\[ V_1 = V_2 \]
This gives us:
\[ \frac{k \cdot q_1}{r_1} = \frac{k \cdot q_2}{r_2} \]

4. **Simplifying the Equation**: We can cancel \( k \) from both sides:
\[ \frac{q_1}{r_1} = \frac{q_2}{r_2} \]
Rearranging gives:
\[ \frac{q_1}{q_2} = \frac{r_1}{r_2} \]
This implies that the charge \( Q \) is directly proportional to the radius \( R \):
\[ q \propto r \]

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

6. **Substituting for Charge**: Since we established that \( Q \propto R \), we can write \( Q = kR \) for some constant \( k \). Substituting this into the equation for \( \sigma \):
\[ \sigma = \frac{kR}{4 \pi R^2} \]

7. **Simplifying the Expression**: This simplifies to:
\[ \sigma = \frac{k}{4 \pi R} \]
Thus, we find that:
\[ \sigma \propto \frac{1}{R} \]

### Conclusion:
The surface charge density \( \sigma \) is inversely proportional to the radius \( R \) of the spherical conductor.