A spherical charged conductor has surface charge density `sigma`.The intensity of electric field and potential on its surface are `E` and `V` .Now radius of sphere is halved keeping the charge density as constant .The new electric field on the surface and potential at the centre of the sphere are

Options

A. 2E, 2 V
B. E, V/2
C. 4E, 4 V
D. 2E, 4 V

Answer

To solve the problem step by step, we will analyze the situation of a spherical charged conductor and derive the expressions for the electric field and potential when the radius is halved while keeping the surface charge density constant.

### Step 1: Understand the given parameters
We have a spherical charged conductor with:
- Surface charge density: \( \sigma \)
- Electric field on the surface: \( E \)
- Potential on the surface: \( V \)
- Initial radius of the sphere: \( R \)

### Step 2: Relate charge \( Q \) to surface charge density \( \sigma \)
The total charge \( Q \) on the surface of the sphere can be expressed in terms of the surface charge density \( \sigma \):
\[ Q = \sigma \cdot A = \sigma \cdot 4\pi R^2 \]
where \( A \) is the surface area of the sphere.

### Step 3: Express the electric field \( E \) at the surface
The electric field \( E \) at the surface of a charged conductor is given by:
\[ E = \frac{KQ}{R^2} \]
Substituting \( Q \) from Step 2:
\[ E = \frac{K (\sigma \cdot 4\pi R^2)}{R^2} = K \cdot 4\pi \sigma \]
Using Coulomb's constant \( K = \frac{1}{4\pi \epsilon_0} \):
\[ E = \frac{\sigma}{\epsilon_0} \]

### Step 4: Express the potential \( V \) at the surface
The potential \( V \) at the surface of the sphere is given by:
\[ V = \frac{KQ}{R} \]
Substituting \( Q \):
\[ V = \frac{K (\sigma \cdot 4\pi R^2)}{R} = K \cdot 4\pi \sigma R \]
Again substituting \( K \):
\[ V = \frac{\sigma R}{\epsilon_0} \]

### Step 5: Analyze the changes when the radius is halved
Now, we are halving the radius of the sphere, so the new radius \( R' = \frac{R}{2} \). The surface charge density \( \sigma \) remains constant.

### Step 6: Calculate the new electric field \( E' \)
Since the electric field \( E \) is independent of the radius when the charge density is constant, we have:
\[ E' = E = \frac{\sigma}{\epsilon_0} \]

### Step 7: Calculate the new potential \( V' \) at the center
The potential at the center of the sphere is equal to the potential on the surface. Since the radius is halved, we can express the new potential as:
\[ V' = \frac{\sigma R'}{\epsilon_0} = \frac{\sigma \left(\frac{R}{2}\right)}{\epsilon_0} = \frac{V}{2} \]

### Final Results
- The new electric field on the surface remains \( E' = E \).
- The potential at the center of the sphere is \( V' = \frac{V}{2} \).

### Summary
- New Electric Field: \( E' = E \)
- New Potential at Center: \( V' = \frac{V}{2} \)