A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the centre of the plate is `10V m^(-1)`. If the plastic plate is replaced by a copper plate of the same geometrical dimension and carryin the same charge `Q_1` the electric field at the point P will become

Options

A. zero
B. `5 V m^(-1)`
C. `10 V m^(-1)`
D. `20 V m^(-1)`.

Answer

To solve the problem, we need to analyze the electric field produced by a uniformly charged plate and how it behaves when the plate material is changed from plastic to copper.

### Step-by-Step Solution:

1. **Understanding the Electric Field from a Charged Plate**:
The electric field (E) produced by an infinite plane sheet of charge with surface charge density (σ) is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

2. **Given Information**:
- The electric field at point P near the center of the plastic plate is \( E = 10 \, \text{V/m} \).
- The charge \( Q \) is uniformly distributed over the plate.

3. **Surface Charge Density**:
Since the electric field due to the plastic plate is known, we can express the surface charge density \( \sigma \) as:
\[ \sigma = 2 \epsilon_0 E \]
Substituting the given electric field:
\[ \sigma = 2 \epsilon_0 \times 10 \, \text{V/m} \]

4. **Replacing the Plastic Plate with Copper**:
When the plastic plate is replaced with a copper plate, the charge \( Q \) remains the same, and the geometrical dimensions of the plate are unchanged.

5. **Electric Field from a Conducting Plate**:
For a conducting plate, the electric field outside the conductor is also given by the same formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
The key point is that the electric field produced by a uniformly charged conductor (like copper) is the same as that produced by a uniformly charged non-conducting plate (like plastic) when the charge distribution is the same.

6. **Conclusion**:
Since the charge \( Q \) and the dimensions remain the same, the electric field at point P will also remain the same:
\[ E = 10 \, \text{V/m} \]

### Final Answer:
The electric field at point P when the plastic plate is replaced by a copper plate will still be \( 10 \, \text{V/m} \).

---