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. $(*)$.

  • $\begingroup$ You should read the context of equation (#) in Griffiths book (1999 edition) which is equation (2.49) there. He explicitly states in the sentence before introducing this equation that this corresponds to the field immediately outside the conductor because the field inside the conductor is zero. Thus your equation (#) follows from equation (*) in the case of a metal. $\endgroup$
    – freecharly
    Mar 13, 2018 at 5:50

2 Answers 2


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 $z \to +0$.

  • $\begingroup$ I would argue that even in the case of nonzero electric field on both sides, the equations are consistent using this form of the derivative: $f'(x)=\lim_{h\to 0}\frac{f(x+h/2)-f(x-h/2)}{h}$. $\endgroup$ Mar 13, 2018 at 0:43
  • $\begingroup$ I wanted to think that one derivative was zero, but I think there is electric field both above and below the plane. Why do you say there is no electric field below the plane? Additionally, when Griffiths presents his form, he does so generally, without saying one derivative is zero. $\endgroup$
    – Ptheguy
    Mar 13, 2018 at 0:49
  • $\begingroup$ @Ptheguy In the context of a metal surface the electric field inside the metal is zero. Therefore $$\frac{\partial V_{below}}{\partial n}=0$$ The image charge is only a virtual charge to obtain the potential zero condition on the metal surface. The superposition of the field/potential of this virtual charge with the field/potential of the real charge can only be used to find the field in the space of the real charge down to the metal plane. To calculate the surface charge you need thus only the normal derivative of the potential above the surface (corresponding to the electric field). $\endgroup$
    – freecharly
    Mar 13, 2018 at 4:23
  • $\begingroup$ @probably_someone In the usual electrostatic model of a metal surface, there is an infinitely thin surface charge and always a zero electric field below the metal surface. There is definitely no electric field inside the metal. If there was an electric field inside the metal, it would produce a current flow that would lead to a surface charge whose field cancels the original field perfectly. Thus inside a metal there can only be a zero electric field. $\endgroup$
    – freecharly
    Mar 13, 2018 at 4:33
  • $\begingroup$ My point was that this formula does not only apply to metals, but rather is a general boundary condition. $\endgroup$ Mar 13, 2018 at 4:34

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.

  • $\begingroup$ In that sense, shouldn't $dV/dn$ be zero when evaluated at the surface since both the above and below parts are equal at the surface? $dV/dn$ is not zero when evaluated at the surface, however. $\endgroup$
    – Ptheguy
    Mar 13, 2018 at 0:24
  • $\begingroup$ What if there's a discontinuity in the electric field? Then the derivative just below the surface would not be the same as the derivative just above the surface. $\endgroup$ Mar 13, 2018 at 0:25
  • $\begingroup$ Well, we know there is discontinuity in the E field. But that's for just above and just below the surface. Isn't the $E_{above}=E_{below}$ when evaluated AT the surface? $\endgroup$
    – Ptheguy
    Mar 13, 2018 at 0:26
  • $\begingroup$ What do you mean by $E_{above}$ and $E_{below}$ "at the surface"? $E_{above}$ is above the surface, and $E_{below}$ is below the surface. $\endgroup$ Mar 13, 2018 at 0:29
  • $\begingroup$ In the problem I'm working on, $\mathbf{\hat{n}}=\mathbf{\hat{z}}$. So $dV/dn=dV/dz$, and Griffiths evaluates this at $z=0$, which is where the surface resides. Isn't this taking the limit as you approach the surface from BOTH sides? And we know if that limit exists, then the respective limits from either side (one from bottom, one from top) are equal. But if $dV/dn=dV_{above}/dn-dV_{below}/dn$, then it should be zero. $\endgroup$
    – Ptheguy
    Mar 13, 2018 at 0:32

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