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 carrying the same charge `Q` 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 both the plastic and copper plates when they carry the same charge \( Q \).

### Step 1: Understand the Electric Field from the Plastic Plate
The electric field \( E \) at point \( P \) close to the center of a large non-conducting plate with uniform charge density \( \sigma \) is given by the formula:
\[ E = \frac{\sigma}{2 \epsilon_0} \]
where \( \epsilon_0 \) is the permittivity of free space.

Given that the electric field at point \( P \) is \( 10 \, \text{V/m} \), we can write:
\[ \frac{\sigma}{2 \epsilon_0} = 10 \]
From this, we can express the surface charge density \( \sigma \):
\[ \sigma = 20 \epsilon_0 \]

### Step 2: Analyze the Copper Plate
When the plastic plate is replaced by a copper plate (which is a conductor), the charge \( Q \) will redistribute itself on the surface of the conductor. For a conductor, the electric field inside the conductor is zero, and the charge resides on the outer surface.

The surface charge density on the outer surface of the copper plate is:
\[ \sigma' = \frac{Q}{A} \]
where \( A \) is the area of the plate.

### Step 3: Calculate the Electric Field from the Copper Plate
For a conducting plate, the electric field outside the plate (at point \( P \)) is given by:
\[ E' = \frac{\sigma'}{\epsilon_0} \]
Since the charge \( Q \) is uniformly distributed, the surface charge density \( \sigma' \) can be expressed as:
\[ \sigma' = \frac{Q}{A} \]

Thus, the electric field at point \( P \) becomes:
\[ E' = \frac{Q/A}{\epsilon_0} \]

### Step 4: Relate the Electric Fields
From the previous steps, we have:
1. For the plastic plate: \( E = \frac{\sigma}{2 \epsilon_0} = 10 \, \text{V/m} \)
2. For the copper plate: \( E' = \frac{Q/A}{\epsilon_0} \)

Now, we know that:
\[ \sigma = 20 \epsilon_0 \]
Thus, substituting \( \sigma \) into the equation for the electric field of the copper plate, we can find:
\[ E' = \frac{Q/A}{\epsilon_0} = 2 \cdot \frac{\sigma}{2 \epsilon_0} = 2 \cdot 10 = 20 \, \text{V/m} \]

### Final Answer
The electric field at point \( P \) when the plastic plate is replaced by a copper plate will be:
\[ \boxed{20 \, \text{V/m}} \]