A spherical charged conductor has surface charge density σ.The intensity of electric field and potential on its surface are E and V, now radius is halved keeping surface charge density to be constant. Then the new values will be

Options

A. `E,(V)/(2)`
B. `W, 2 V`
C. `2E, V`
D. `(E)/(2),V`

Answer

To solve the problem, we will analyze the changes in the electric field intensity \( E \) and the electric potential \( V \) when the radius of a spherical charged conductor is halved while keeping the surface charge density \( \sigma \) constant.

### Step-by-Step Solution:

1. **Understanding Surface Charge Density**:
The surface charge density \( \sigma \) is defined as the charge per unit area on the surface of the conductor. For a sphere, the surface area \( A \) is given by:
\[ A = 4\pi r^2 \]
Therefore, the total charge \( Q \) on the sphere can be expressed as:
\[ Q = \sigma \cdot A = \sigma \cdot (4\pi r^2) = 4\pi r^2 \sigma \]

2. **Electric Field Intensity**:
The electric field \( E \) just outside the surface of a charged conductor is given by:
\[ E = \frac{kQ}{r^2} \]
where \( k \) is Coulomb's constant. Substituting the expression for \( Q \):
\[ E = \frac{k(4\pi r^2 \sigma)}{r^2} = 4\pi k \sigma \]
This shows that the electric field intensity \( E \) depends only on the surface charge density \( \sigma \) and is independent of the radius \( r \).

3. **Electric Potential**:
The electric potential \( V \) at the surface of the sphere is given by:
\[ V = \frac{kQ}{r} \]
Substituting the expression for \( Q \):
\[ V = \frac{k(4\pi r^2 \sigma)}{r} = 4\pi k \sigma r \]
Here, the potential \( V \) is directly proportional to the radius \( r \).

4. **Halving the Radius**:
Now, if the radius is halved (i.e., \( r' = \frac{r}{2} \)), we need to find the new charge \( Q' \), electric field \( E' \), and potential \( V' \).

- **New Charge**:
Since the surface charge density \( \sigma \) remains constant, the new charge \( Q' \) is:
\[ Q' = \sigma \cdot (4\pi (r/2)^2) = \sigma \cdot (4\pi \frac{r^2}{4}) = \sigma \cdot \pi r^2 = \frac{Q}{4} \]

- **New Electric Field**:
The new electric field \( E' \) at the new radius \( r' \) is:
\[ E' = \frac{kQ'}{(r/2)^2} = \frac{k(\frac{Q}{4})}{\frac{r^2}{4}} = \frac{kQ}{r^2} = E \]
Thus, the electric field remains the same: \( E' = E \).

- **New Potential**:
The new potential \( V' \) at the new radius \( r' \) is:
\[ V' = \frac{kQ'}{r/2} = \frac{k(\frac{Q}{4})}{\frac{r}{2}} = \frac{kQ}{2r} = \frac{V}{2} \]

5. **Final Results**:
- The new electric field intensity \( E' = E \)
- The new electric potential \( V' = \frac{V}{2} \)

### Summary of Results:
- New Electric Field Intensity: \( E' = E \)
- New Electric Potential: \( V' = \frac{V}{2} \)