Electric Field "at" the surface of a conductor 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 $E=\frac{\sigma}{2\epsilon_0}\hat{n}$?
 A: 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.
A: I got it. Consider the sphere of charge, or the the plane of charge, that sits directly on top of the conductor. We know that this field has $E = \frac{\sigma}{2\epsilon_0}$ pointing away from it (away and towards the surface of the conductor). We also know that the electric field inside the conductor must be 0; thus the conductor itself must have an electric field of $E = \frac{\sigma}{2\epsilon_0}$ pointing outward to cancel out the incoming field of the charge distribution. Off the surface of the conductor, the electric field of the conductor and that of the charge distribution add via the super position principle to give a field of $E = \frac{\sigma}{\epsilon_0}$. 
A: The electric field just above the surface is defined as force on unit charge  so youbarecnot allowed to take a disc from the surface and find for other charges thats untrue .the same thing is applicaple to all points outside or inside conductor .exactly on the surface the electric field function is dicontinuous  inside 0 and outsid the linit when you approach the surface is proportional to surface charge density
This can be derived easily from Gauss law by takiñ small cylinder part of it outside and the other part inside conductor
