Two conducting spheres of radii `r_(1)` and `r_(2)` are charged to the same surface charge density . The ratio of electric field near their surface is

Options

A. `r_(1)^(2)//r_(2)^(2)`
B. `r_(2)^(2)//r_(1)^(2)`
C. `r_(1)//r_(2)`
D. `1:1`

Answer

To solve the problem of finding the ratio of the electric fields near the surfaces of two conducting spheres with radii \( r_1 \) and \( r_2 \) that are charged to the same surface charge density \( \sigma \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Concept of Electric Field Near Conductors:**
The electric field \( E \) just outside the surface of a charged conductor is given by the formula:
\[ E = \frac{\sigma}{\epsilon_0} \]
where \( \sigma \) is the surface charge density and \( \epsilon_0 \) is the permittivity of free space.

2. **Apply the Formula to Both Spheres:**
For the first sphere with radius \( r_1 \):
\[ E_1 = \frac{\sigma}{\epsilon_0} \]
For the second sphere with radius \( r_2 \):
\[ E_2 = \frac{\sigma}{\epsilon_0} \]

3. **Calculate the Ratio of Electric Fields:**
Since both spheres have the same surface charge density \( \sigma \), we can set up the ratio of the electric fields:
\[ \frac{E_1}{E_2} = \frac{\frac{\sigma}{\epsilon_0}}{\frac{\sigma}{\epsilon_0}} = 1 \]

4. **Conclusion:**
Therefore, the ratio of the electric fields near the surfaces of the two spheres is:
\[ \frac{E_1}{E_2} = 1 \]

### Final Answer:
The ratio of electric fields near their surfaces is \( 1:1 \).

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