0
$\begingroup$

In freshman textbooks like Halliday, Resnick and Krane, We derive the electric field on the surface of a conductor by taking a Gaussian surface of infinitesimally small length and infinitesimally small cross-sectional area $dA$, with one face inside the conductor, and then applying Gauss' Law: \begin{align}E=\frac{\sigma}{\epsilon_0}\end{align} My question is that if we do not limit the length of the Gaussian surface, and take it to be a long Gaussian Surface, but of same cross-sectional area $dA$, with one face inside the conductor, then, as the electric field would be constant along the cross section, we would get: \begin{align}E=\frac{\sigma}{\epsilon_0}\end{align} for all locations, which is an incorrect result. enter image description here

The progress I have made so far is deducing that the contribution to the electric flux from the other parts of the surface of the conductor would be zero in the case of infinitesimal length, as they would be perpendicular to the lateral surface of Gaussian Cylinder, and their resultant would both enter and exit the lateral surface; moreover, as the area of the lateral surface is $dAdl$, the area would be of a smaller order than $dA$ anyway, thus making the flux zero. However, this should also be true for the case of a long cylindrical Gaussian surface. This seems to strengthen the argument that leads to the incorrect result.

$\endgroup$

1 Answer 1

1
$\begingroup$

If I understand your problem correctly, you're trying to prove what is sometimes called Coulomb's theorem.

If the conducting surface is an infinite plane, stuying its symmetries shows that E is perpendicular to it everywhere in space, so you can take the Gauss cylinder as tall as you want.

Otherwise, E has no reason to be normal to the surface at all, so its flux through the cylinder is hard to compute (often impossible without numerical tools). Hence the bad result if you assume that E is normal to the surface.

However, if you reduce the cylinder to an infinitesimal height, you remain in a small neighborhood of the surface, where the field coming from that point of the surface dominates all the others sources. Then you can locally assume that the field is normal to the surface and get the correct result, but valid only in the immediate vicinity of the surface.

$\endgroup$
7
  • $\begingroup$ Yes that is correct. However, even if we take it from the centre of a uniform charge distribution conducting disc, then the flux would be parallel to the particular area in consideration here, yet the answer would not be correct. $\endgroup$ May 28, 2022 at 13:26
  • $\begingroup$ I'm not sure I understand your comment. A flux isn't a vector, so it can't be parallel to anything. If you're talking about the field, its direction depends not only on the uniformity (or not) of the distribution, but also on whether it's infinite or not (the symmetries won't be the same). $\endgroup$
    – Miyase
    May 28, 2022 at 16:13
  • $\begingroup$ I meant electric field; sorry for the error. Yes, you are right, but consider a Gaussian Cylinder of infinitesimal area at the centre of the disk. The electric field would be parallel to the cross-sectional area at the centre at least, right? $\endgroup$ May 28, 2022 at 16:26
  • $\begingroup$ Infinitely close to the charged surface, yes. Further than that? No, unless there is some rotational invariance to force the field to be normal at the center. $\endgroup$
    – Miyase
    May 28, 2022 at 18:30
  • $\begingroup$ For a circular disk there is rotational invariance, yet the result does not match. Sorry, I had not considered that a disc could be non-circular. $\endgroup$ May 29, 2022 at 3:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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