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 $E=\frac{\sigma}{2\epsilon_0}\hat{n}$?
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3$\begingroup$ Factors of two usually have a physical reason. My electrostatics lessons were a very long time ago, but in a plate capacitor the field is generated by the field of the charges on both plates and my guess is that it works out to be $E=\sigma/\epsilon$ for the total charge. Take the second capacitor plate away and by symmetry you get a factor of two less for the field of just one plate, but I would not be surprised if I just made a total fool out of myself. $\endgroup$– CuriousOneJan 11, 2016 at 5:22
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1$\begingroup$ Where did you see this? I can only remember seeing it l like this: hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html $\endgroup$– user1717828Jan 11, 2016 at 5:51
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$\begingroup$ It's on a problem where there is a hole drilled into the spherical shell, and the problem is asking what the field on the surface of the shell at the center of the hole, when the hole is infinitesimal. They cite Source: E.M. Purcell, Electricity and Magnetism, 2nd edition (McGraw-Hill, New York, 1985) $\endgroup$– LoonuhJan 11, 2016 at 5:54
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2$\begingroup$ I think It is correct to think of it like the piecewise-mean value theorem of the functions $\frac{1}{2}[f(x_+)+f(x_-)]$ $\endgroup$– CourageJan 11, 2016 at 5:59
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$\begingroup$ This would probably answer it: physics.stackexchange.com/a/62773/74688 $\endgroup$– CourageJan 11, 2016 at 6:14
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3 Answers
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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.
answered Jan 11, 2016 at 7:22
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1$\begingroup$ For details, please refer to Griffith, Introduction to Electrodynamics. $\endgroup$ Jan 11, 2016 at 17:37
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$\begingroup$ I mean, does he actually make this argument and say the E-field at the surface is undefined? $\endgroup$– LoonuhJan 11, 2016 at 17:40
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$\begingroup$ Just checked the book again. He didn't say anything about $E_\text{disk}$ on the surface. In my opinion, it should be undefined. $\endgroup$ Jan 11, 2016 at 17:48
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$\begingroup$ It is in fact, not undefined, it is $E=\frac{\sigma}{2\epsilon_0}$. To my amazement, I have discerned that this field is coming entirely from the charge sitting exactly opposite to the point of interest. It's bonkers. See my answer below. $\endgroup$– LoonuhJan 12, 2016 at 2:38
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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 $E = \frac{\sigma}{\epsilon_0}$.
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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
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1$\begingroup$ Hello Bassem Elkhatib and welcome to Physics SE. Your answer needs editing to make it sensible. At the moment there's punctuation and grammatical issues and (personally) is not easily understandable. I would propose an edit but I am unable to understand the meaning of your answers. I suggest you try and edit it. $\endgroup$– ZaellixAJan 16, 2022 at 21:27