Electric field at a point close to the centre of a conducting plate

[Pushoam]

Homework Statement

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 10 volt per metre. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become ......

Relevant Equations

For conducting plate
$$ E = \frac { \sigma}{\epsilon_0}$$
For plastic plate
$$ E = \frac { \sigma}{2\epsilon_0}$$

Since the electric field due to a conducting plate is twice the electric field due to a plastic plate having same charge density, the electric field at the point P will be twice in case of conducting plate and hence it is 20 volt per metre.

Is that correct?

 

[TSny]

A plate has two large parallel surfaces. If you somehow start with all of Q spread on one of the surfaces of the conducting plate, does the charge redistribute itself? If so, when electrostatic equilibrium is established, think about $\sigma$ on each of the surfaces of the conducting plate.

 

[Pushoam]

A plate has two large parallel surfaces. If you somehow start with all of Q spread on one of the surfaces of the conducting plate, does the charge redistribute itself? If so, when electrostatic equilibrium is established, think about $\sigma$ on each of the surfaces of the conducting plate.

So, charge density of the conducting plate is half of the charge density of plastic plate hence the electric field will be 10 volt per metre. The charge Q on both plates remain same.

 

[TSny]

So, charge density of the conducting plate is half of the charge density of plastic plate hence the electric field will be 10 volt per metre. The charge Q on both plates remain same.

Yes, I believe this is correct.

The charge density on each of the surfaces of the conducting plate will not be uniform, especially near the edges of the plate. However, near the center of a large conducting plate, the charge density on each surface will be approximately uniform and approximately equal to $\frac Q {2A}$, where $Q$ is the total charge on the plate and $A$ is the area of one of the surfaces.

 

[ChiralSuperfields]

Yes, I believe this is correct.

The charge density on each of the surfaces of the conducting plate will not be uniform, especially near the edges of the plate. However, near the center of a large conducting plate, the charge density on each surface will be approximately uniform and approximately equal to $\frac Q {2A}$, where $Q$ is the total charge on the plate and $A$ is the area of one of the surfaces.

Would you please know where you got $\frac Q {2A}$ from @TSny?

 

[haruspex]

Would you please know where you got $\frac Q {2A}$ from @TSny?

Two surfaces, each of area A.
It's not clear to me whether the charge on the plastic plate is on one surface or both, but for the same total charge the field is the same. Both sides.

 

[ChiralSuperfields]

Two surfaces, each of area A.
It's not clear to me whether the charge on the plastic plate is on one surface or both, but for the same total charge the field is the same. Both sides.

Thank you for your help @haruspex!

 

[Pushoam]

Two surfaces, each of area A.
It's not clear to me whether the charge on the plastic plate is on one surface or both, but for the same total charge the field is the same. Both sides.

The question says uniform distribution. So, I think it is uniformly distributed on both sides.