A charge Q is uniformly distributed over one large insulating surface and field intensity at a point X near its centre E. If it were a metal plate in place of an insulating surface then the electric field intensity at point X will become

Options

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

Explanation

To solve the problem, we need to analyze the electric field intensity produced by a uniformly distributed charge on both an insulating surface and a conducting (metal) surface.

### Step-by-Step Solution:

1. **Understanding the Insulating Surface**:
- When a charge \( Q \) is uniformly distributed over a large insulating surface, the surface charge density \( \sigma \) can be defined as:
\[ \sigma = \frac{Q}{A} \]
where \( A \) is the area of the surface.

2. **Electric Field from Insulating Surface**:
- The electric field intensity \( E \) at a point near the center of a uniformly charged insulating surface is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
- Substituting the expression for \( \sigma \):
\[ E = \frac{Q}{2 \epsilon_0 A} \]

3. **Understanding the Conducting Surface**:
- If the charge \( Q \) were distributed over a large conducting plate instead of an insulating surface, the charge would redistribute itself. For a large conducting plate, the electric field is uniform and directed away from the surface.
- The surface charge density \( \sigma \) would still be:
\[ \sigma = \frac{Q}{A} \]
- However, the electric field intensity \( E \) due to a conducting plate is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]

4. **Calculating Electric Field for Conducting Surface**:
- Substituting the expression for \( \sigma \):
\[ E = \frac{Q}{\epsilon_0 A} \]

5. **Comparing Electric Fields**:
- From the insulating surface, we found:
\[ E_{\text{insulating}} = \frac{Q}{2 \epsilon_0 A} \]
- From the conducting surface, we found:
\[ E_{\text{conducting}} = \frac{Q}{\epsilon_0 A} \]
- Thus, the electric field intensity at point \( X \) near the center of the conducting plate is twice that of the insulating surface:
\[ E_{\text{conducting}} = 2 E_{\text{insulating}} \]

### Final Answer:
If the charge \( Q \) is uniformly distributed over a metal plate instead of an insulating surface, the electric field intensity at point \( X \) will become:
\[ E = 2E \]