3
$\begingroup$

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 $E=\frac{\sigma}{2\epsilon_0}\hat{n}$?

$\endgroup$
  • $\begingroup$ Could you elaborate please? $\endgroup$ – Loonuh Jan 11 '16 at 5:15
  • 1
    $\begingroup$ 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. $\endgroup$ – CuriousOne Jan 11 '16 at 5:22
  • 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 $\endgroup$ – user1717828 Jan 11 '16 at 5:51
  • $\begingroup$ 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) $\endgroup$ – Loonuh Jan 11 '16 at 5:54
  • 1
    $\begingroup$ I think It is correct to think of it like the piecewise-mean value theorem of the functions $\frac{1}{2}[f(x_+)+f(x_-)]$ $\endgroup$ – Oswald Jan 11 '16 at 5:59
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Can you cite this conventionally? $\endgroup$ – Loonuh Jan 11 '16 at 17:14
  • $\begingroup$ For details, please refer to Griffith, Introduction to Electrodynamics. $\endgroup$ – velut luna Jan 11 '16 at 17:37
  • $\begingroup$ I mean, does he actually make this argument and say the E-field at the surface is undefined? $\endgroup$ – Loonuh Jan 11 '16 at 17:40
  • $\begingroup$ Just checked the book again. He didn't say anything about $E_\text{disk}$ on the surface. In my opinion, it should be undefined. $\endgroup$ – velut luna Jan 11 '16 at 17:48
  • $\begingroup$ 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. $\endgroup$ – Loonuh Jan 12 '16 at 2:38
0
$\begingroup$

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 $E = \frac{\sigma}{\epsilon_0}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.