Obtaining Surface Charge Density from the Potential

Question

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. $(*)$.

Answer 1

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 $z \to +0$.

Answer 2

The quantity $\frac{dV}{dn}$ is defined by Griffiths to be equal to $\frac{dV_{above}}{dn}-\frac{dV_{below}}{dn}$. This is equivalent to the following (valid) definition of the derivative of a function:

$$f'(x)=\lim_{h\to 0}\frac{f(x+h/2)-f(x-h/2)}{h}$$

Regarding your specific problem, the trick is that the potential of the image-charge setup and the potential of a sheet of charge are not identical everywhere. They're only identical in the region you care about, which is outside the conductor. The image-charge form of the problem has no discontinuity in the derivative of the potential at zero, whereas the original conducting charged slab definitely does have such a discontinuity. But, for this problem, we are only given (and only care about) what the potential looks like above the conductor, not on it. As such, the boundary of our problem is actually an open set, i.e. we care about the region $(0,\infty)$ in $z$. Since the conductor is outside the boundary of our problem, we can substitute the conductor with anything else that gives us a potential that looks the same for $z>0$.

As it happens, a very simple substitute exists: an image charge. This substitution (which gives us identical conditions for $z>0$) is useful because it takes away the discontinuity in potential that we had before. Therefore, it's possible now to take the derivative of this potential at zero, neatly sidestepping our earlier discontinuity because we have removed all of the charges that exist at the surface using the method of images. As such, your problem doesn't have anything to do with taking the derivative of a configuration with a discontinuous electric field, because you solve it by removing the discontinuity.