A spherical charged conductor has surface density of charge as `sigma`. The electric field intensity on its surface is `E`. If the radius of the surface is doubled, keeping `sigma` uncharged, what will be the electric field intensity on the new sphere ?

Options

A. `E//2`
B. `E//4`
C. `2 E`
D. `E`

Explanation

To solve the problem, we will analyze the relationship between the electric field intensity, surface charge density, and the radius of the spherical charged conductor.

### Step-by-Step Solution:

1. **Understand the Electric Field of a Charged Sphere**:
The electric field \( E \) at 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. **Identify the Given Parameters**:
- The initial surface charge density is \( \sigma \).
- The electric field intensity at the surface is \( E \).
- The radius of the sphere is doubled, but \( \sigma \) remains unchanged.

3. **Analyze the Effect of Doubling the Radius**:
When the radius of the sphere is doubled, the surface area of the sphere increases. However, since the surface charge density \( \sigma \) is defined as charge per unit area, and we are keeping \( \sigma \) constant, the total charge \( Q \) on the sphere must also change to maintain the same surface charge density.

4. **Relate Charge to Surface Area**:
The surface area \( A \) of a sphere is given by:
\[ A = 4\pi r^2 \]
If the radius is doubled (let's denote the new radius as \( r' = 2r \)), the new surface area \( A' \) becomes:
\[ A' = 4\pi (2r)^2 = 16\pi r^2 \]
Since \( \sigma \) remains constant, the total charge \( Q' \) on the new sphere can be expressed as:
\[ Q' = \sigma A' = \sigma (16\pi r^2) \]

5. **Calculate the New Electric Field**:
The new electric field \( E' \) at the surface of the new sphere is given by:
\[ E' = \frac{\sigma}{\epsilon_0} \]
Since \( \sigma \) has not changed, the new electric field intensity remains:
\[ E' = E \]

6. **Conclusion**:
Therefore, the electric field intensity on the new sphere, after doubling the radius while keeping the surface charge density \( \sigma \) unchanged, remains the same:
\[ E' = E \]

### Final Answer:
The electric field intensity on the new sphere is \( E \).