Can the formula for a conductor be applied to a grounded infinite conductor plate?

Question

I'm studying about Image method with "Introduction to electrodynamics-Grffiths"

And, this book explains the process finding the induced surface charge density when a point charge $q$ is held a distance $d$ above an infinite grounded conducting plane.

I was able to understand the process of finding the electric potential $V$. But my confusion begins now.

Now that we know the potential, it is a straightforward matter to compute the surface charge $\sigma$ induced on the conductor. According to $\sigma=-\epsilon_{0}\frac{\partial V}{\partial n}$....

I can't understand how $\sigma=-\epsilon_{0}\frac{\partial V}{\partial n}$ can be used in this situation. As i know, it is formula when $V$ is electric potential 'immediately outside of conductor', not 'grounded conducting plane'.

Also, it is written like this in the book:

Because the field inside a conductor is zero, boundary condition requires that the field immediately outside is: $\vec{E}=\frac{\sigma}{\epsilon_0} \hat{n}$. In terms of potential, this yields $\sigma=-\epsilon_{0}\frac{\partial V}{\partial n}$

In summary, I don't know why the formulas that apply for conductors can't be applied to a 'grounded infinite conductor plate'.

Or can a 'grounded infinite conductive plate' be considered $\vec{E}=0$ for the region without charge, like a conductor ?

Answer 1

If you're only interested in what's happening on one side of the plate, you may ignore the other side (assuming infinite conductivity). Infinite conductivity forces $\vec{E}=0$ inside the plate, so there cannot be charge there: charge can only be on the surface. It thus doesn't matter how thick the plate is or what's happening on the far side.