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` the electric field at the point P will become

Options

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

Answer

To solve the problem, we will analyze the electric field produced by a uniformly charged plate, first for a non-conducting plate and then for a conducting plate.

### Step-by-Step Solution:

1. **Understanding the Electric Field of a Non-Conducting Plate:**
- For a non-conducting plate with a uniform surface charge density \(\sigma\), the electric field \(E\) at a point close to the center of the plate is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
- Here, \(\epsilon_0\) is the permittivity of free space.

2. **Given Information:**
- The electric field \(E\) at point \(P\) close to the center of the plastic plate is given as \(10 \, \text{V/m}\).
- Therefore, we can write:
\[ \frac{\sigma}{2 \epsilon_0} = 10 \, \text{V/m} \]

3. **Calculating Surface Charge Density \(\sigma\):**
- Rearranging the equation gives:
\[ \sigma = 20 \epsilon_0 \]

4. **Replacing the Plastic Plate with a Conducting Plate:**
- When the plastic plate is replaced by a conducting plate (copper), the charge \(Q\) will distribute itself on the outer surface of the conductor.
- For a conducting plate, the electric field inside the conductor is zero, and the electric field just outside the surface of the conductor is given by:
\[ E = \frac{\sigma}{\epsilon_0} \]

5. **Calculating the New Electric Field at Point \(P\):**
- Since the charge \(Q\) is the same and distributed on both sides of the conducting plate, the electric field at point \(P\) (which is outside the plate) will be:
\[ E = \frac{\sigma}{\epsilon_0} = \frac{20 \epsilon_0}{\epsilon_0} = 20 \, \text{V/m} \]

6. **Conclusion:**
- The electric field at point \(P\) when the plastic plate is replaced by a copper plate is \(20 \, \text{V/m}\).

### Final Answer:
The electric field at point \(P\) will become \(20 \, \text{V/m}\).

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