It has been pointed out to me that the Electric field exactly on the surface of the conductor is conventionally taken to be $E=\frac{\sigma}{2\epsilon_0}$; does this come from taking the midpoint of the $E$-field magnitudes before and after the location of the discontinuity (namely, the average of $E=0\hat{n}$ and E=$\frac{\sigma}{\epsilon_0}\hat{n}$? Similar to how the Heavyside function evaluated at 0 is sometimes taken to be 1/2 by convention? Is there any reason other than convention to assign the surface $E$-field as $E=\frac{\sigma}{2\epsilon_0}\hat{n}$?
electrostatics – Electric Field at the surface of a conductor: detailed insights
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Related Solutions
Consider a point charge $q$ at $(0,0,0)$, the potential at $\bf{r}$ is given by $V(\bf{r})$ $= \frac{q}{4\pi\epsilon_0r}$. If you consider a path through $(0,0,0)$, you encounter a discontinuity in electric field and the direction of field changes. Is the potential for the field of point charge discontinuous at the location of the point charge?
If you specify the potential as a function from real numbers to real numbers then the function is discontinuous at zero by virtue of the fact that it is not defined at zero. I.e., the definition of continuity $$ \lim_{r\to 0}V(r)\to V(0) $$ does not hold because V(0) is not defined (it is not a real number).
I'm facing this confusion because, in the book by Griffiths, it is mentioned that the electric field is discontinuous at the surface of charge, with charge density $\sigma$ i.e $E_{above}-E_{below}=\frac{\sigma}{\epsilon_0}\hat{n}$. It is also mentioned that the potential is continuous across a boundary like that. Is my analogy between point and surface charge wrong?
The analogy seems somewhat imperfect to me because both $E$ and $V$ go to infinity in the first example, but neither does in the second example. If E remains finite, phi will be continuous (as discussed below).
Also, is potential for an electric field continuous everywhere? Or is it continuous across field discontinuities?
Typically $\phi$ is continuous since, for example, if $\phi$ had a discontinuity $\Delta\phi$ at some point $x_0$ then $$ \Delta\phi=\lim_{\delta\to 0}\phi(x_0+\delta)-\phi(x_0)=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}\frac{d\phi}{dx}dx=\lim_{\delta\to 0}\int_{x_0}^{x_0+\delta}E_j(x)dx =\lim_{\delta\to 0}E_j(x_0)\delta\;, $$ which is zero unless E(x_0) blows up. I.e., for finite electric fields (even discontinuous finite electric fields) the potential remains continuous.
But, remember, in general, the potential does not have to be continuous everywhere (as evidenced by the case mentioned above of a point charge). This is because the governing equation for $\phi$ is $$ \nabla^2 \phi=-\rho/\epsilon_0\;, $$ where $\rho$ is the charge density. And, I'll bet that you can come up with some wacky charge densities (a single point charge is one case) where $\phi$ is discontinuous.
The boundary conditions by themselves can't tell you anything about a conductor. The boundary conditions can't even tell which side of the surface has the conductor!
One way to model a conductor is as an Ohmic conductor where there is a constant $\sigma$ (different than the surface charge density listed in your boundary conditions) and then you assert the Ohmic condition:
$$\vec J=\sigma \vec E$$ and then you can take the divergence of both sides and get $$\sigma\frac{\rho}{\epsilon_0}=\sigma\vec \nabla \cdot \vec E = \vec \nabla \cdot \vec J$$
Where we used the Maxwell equation $\dfrac{\rho}{\epsilon_0}=\vec \nabla \cdot \vec E$ and we can also take the divergence of $$\vec \nabla \times \vec B=\mu_0\vec J+\mu_0\epsilon_0\frac{\partial \vec E}{\partial t}$$ to get the continuity equation
$$\vec \nabla \cdot \vec J=-\epsilon_0\vec \nabla \cdot\frac{\partial \vec E}{\partial t}=-\frac{\partial \rho}{\partial t}.$$
This means we have $$\frac{\partial \rho}{\partial t}=-\vec \nabla \cdot \vec J=-\frac{\sigma}{\epsilon_0}\rho.$$
So you might have started with an initial charge density but every place inside the conductor it exponentially decays over time.
And this relates to the initial value formulation of Electrodynamics. You start with an actual physical electromagnetic field at a moment and then it evolves according to $$\frac{\partial \vec B}{\partial t}=-\vec\nabla \times \vec E, \text{ and}$$
$$\frac{\partial \vec E}{\partial t}=\frac{1}{\epsilon_0}\left(\frac{1}{\mu_0}\vec\nabla \times \vec B-\vec J\right)$$
So the fields at a later time are a consequence of the fields at an earlier time (and the current) and the evolution equations above.
So for an Ohmic material we know $\vec J$ so we can evolve the fields because the evolution equations are just Maxwell solved for the time rates of change.
So you had an initial charge distribution and an initial electric field. They might have been zero, they might have been nonzero.
I've seen many "conceptual" arguments that if there was a field the charges would move and produce a field canceling this one out.
If you think of statics as the long time limit of dynamics then you don't have to go conceptual. An Ohmic material literally does have a nonzero current where there is a nonzero electric field. But that current causes the charge to go away, you can imagine places where the charge density is positive and negative initially and the initial electric field streamlines could connect some of those and/or it can connect those places to the surface. And since the current points the same way we can see that this exponentially decreasing charge density is becasue opposite charge densities are canceling each other as charge flows or the charge imbalance is moving to the surface, thus increasing the charge density of the surface over time.
The charge density on the surface can change in a different way than exponentially decreasing over time. Why? Because $\sigma$ (from the Ohmic condition) is not constant across the surface on the boundary of the conductor. In fact, the boundary of the conductor might have a vacuum with $\vec J=\vec 0$ on the other side.
Can we argue that the electric field is zero inside? Yes and no. On the one hand if we assert that part of statics is $\vec J=\vec 0$ then we have $\vec E=\vec 0$ right away. But that's more like assuming. But if you allow $\vec J\neq \vec 0$ then your conductor could have a zero charge density everywhere but have a steady current as long as the boundary of the conductor is supplied with the current it needs for that steady current.
It's totally a valid solution of Maxwell to have a cylindrical infinite wire pointing in the $\hat z$ direction with a uniform nonzero $\vec J$ pointing in the $\hat z$ direction inside the cylindrical infinite wire.
So whenever you have a counterexample you know you need to strengthening your hypothesis. That situation can have a static unchanging electric field, but it has a nonzero current.
Best Answer
The E field exactly on the surface in fact should be undefined, because there are surface charges.
But the E field is well-defined if you remove a small disk from the surface.
Let's call the E field due to the disk be $E_\text{disk}$ and the E field due to the other surface charges be $E_\text{other}$.
Then just above the surface
$$E_\text{disk}=\sigma/2\epsilon_0$$
Just below the surface
$$E_\text{disk}=-\sigma/2\epsilon_0$$
And on the surface, $E_\text{disk}$ is undefined.
Now it is clear that $E_\text{other}$ is smooth across the surface and well-defined on the surface.
And because just above the surface
$$E_\text{other}+E_\text{disk}=\sigma/\epsilon_0$$
and just below the surface
$$E_\text{other}+E_\text{disk}=0$$
it can deduced that
$$E_\text{other}=\sigma/2\epsilon_0$$
$E_\text{other}$ on the surface is hence $\sigma/2\epsilon_0$.
So the "E field on the surface" is in fact $E_\text{other}$, viz., the E field at the surface if you remove a small disk of surface charges from the surface, and is well-defined. It is also the E field experienced by that small disk of surface charges.