Suppose a charge Q is uniformly distributed over a large plastic...

For a large non-conducting (plastic) plate with total charge $Q$ and surface area $A$, the electric field at a point close to its center can be found using Gauss's law. Taking a cylindrical Gaussian surface piercing through the plate, the total flux is $2EA = \dfrac{Q}{\epsilon_0} \Rightarrow E = \dfrac{Q}{2A\epsilon_0}$.

Given $E = 10 \text{ V m}^{-1}$.

When the plastic plate is replaced by a conducting (copper) plate of the same dimensions and carrying the same total charge $Q$, the charge distributes equally on its two outer surfaces. The charge on each surface is $\dfrac{Q}{2}$, and the surface charge density on each face is $\sigma' = \dfrac{Q}{2A}$.

The electric field at a point close to a conducting surface is given by $E' = \dfrac{\sigma'}{\epsilon_0}$.

Substituting $\sigma'$, we get $E' = \dfrac{Q}{2A\epsilon_0}$.

This expression is exactly the same as the electric field produced by the plastic plate. Thus, the electric field remains $10 \text{ V m}^{-1}$.