Let me begin by noting that for a surface with charge density $\sigma$, we know the component of the electric field perpendicular to the surface is discontinuous. This relation is given as
$$\mathbf{E_{above}-E_{below}}=\frac{\sigma}{\epsilon_0}\mathbf{\hat{n}},$$ or equivalently in terms of the potential
$$\nabla V_{above}-\nabla V_{below}=-\frac{\sigma}{\epsilon_0}\mathbf{\hat{n}}$$ $$\tag{*}\frac{\partial V_{above}}{\partial n}-\frac{\partial V_{below}}{\partial n}=-\frac{\sigma}{\epsilon_0},$$
where for the last step we can dot both sides of the first equation by $\mathbf{\hat{n}}$ and define the normal derivative of $V$.
Now, in Griffiths Electrodynamics book, he suggests that the surface charge density of a plate is given as
$$\tag{#}\sigma=-\epsilon_0 \frac{\partial V}{\partial n}.$$
I'm a bit confused because results $(*)$ and $(\#)$ don't look the same to me. Could someone clarify how these two relations are connected, because I think they must be, but can't see it in.
Here is the context of the problem where this shows up. You have a conducting plane that is grounded and resides at the $xy$-plane. There is a positive point charge $q$ a distance $d \mathbf{\hat{z}}$ above this plane. You apply the method of images, replacing this problem with one where there is a mirror charge and no conductor. You obtain the potential V, which is also the potential for the original problem for $z \geq 0$. Next, you want to find $\sigma$ of the plane. I think that since there is an electric field both above and below the plane, we should use Eq. $(*)$. But, Griffiths uses Eq. $(\#)$ and evaluates the derivative at $z=0$, noting that here $\mathbf{\hat{n}}=\mathbf{\hat{z}}$ so the derivative is taken with respect to z. This confuses me because if the derivative exists as we take the limit $z \rightarrow0$, then that means $dV/dz$ evaluated from limit $z\rightarrow0^+$ is equal to that taken with limit $z\rightarrow0^-$. But this contradicts the discontinuity claim in Eq. $(*)$.
Best Answer
Griffiths equation $$\sigma=-\epsilon_0 \frac{\partial V}{\partial n}$$hold for the case of a metal surface charge where the interior electric field is zero. It is equivalent to $$\frac{\partial V_{above}}{\partial n}-\frac{\partial V_{below}}{\partial n}=-\frac{\sigma}{\epsilon_0}$$ where $$\frac{\partial V_{below}}{\partial n}=0$$ Thus your equation (*) also holds in the case considered by Griffiths.
In regard to context of the image charge solution of the potential, you have to consider that this solution holds only for the potential/electric field above the conducting plate $z \ge 0$. Thus for the determination of the surface charge on the conductor, you have to take Griffiths equation (#) $$\sigma=-\epsilon_0 \frac{\partial V}{\partial n}$$ with the normal derivative of this potential solution at $z \to +0$.