3
$\begingroup$

In book (Halliday Resnick Krane, 2nd Part, fifth edition), it's written that when you you put some charge in an isolated conductor, then within around $10^{-9}$ seconds the charges all go to the surface of the conductor, and there's no charge in the inside of the conductor. Even if you cut a hole from inside of the the conductor, then also there would be no net charge in the surface around the hole in the conductor.

The reasoning the apply is using Gauss law and somehow concluding that "the electric field in the conductor must zero everywhere inside it otherwise there would be some field and the particles would not be stationary".

I find this argument to be wrong. Suppose you take an isolated sphere, and in the center of the sphere, you put a negative charge of magnitude $-q$, and construct an hexagon with arbitrary distance centered at the negative charge, and put some positive charge on equal magnitude $+Q$ each vertice of the hexagon so that the entire system is in equilibrim (that such value of positive charge for which the system exists at equilibrium follows by "applying" intermediate value theorem on $Q = 0$ and $Q = \infty$).

Picutre as requested (red is the negative charge of magnitude $-q$ and blue are the positive charges of magnitude $Q$ such that the configuration is stable): enter image description here

But then there's some charge at the inside of the conductor, so the net field is not zero, but the particles are stationary too! What's wrong with my arguement?

| cite | improve this question | |
$\endgroup$
  • $\begingroup$ You can't create a stable configuration. en.wikipedia.org/wiki/Earnshaw%27s_theorem $\endgroup$ – BowlOfRed May 17 '18 at 6:36
  • $\begingroup$ @BowlOfRed I mean the configuration isn't stable and stationary, but it's unstable but stationary. $\endgroup$ – katana_0 May 17 '18 at 11:04
  • $\begingroup$ If it's unstable, then the negative charges will be attracted to the positive charges and combine. They won't remain separated. $\endgroup$ – BowlOfRed May 17 '18 at 16:44
0
$\begingroup$

Your argument seems to be based on fixed point charges and a disregard of the other mobile charges in the conductor. Charges in conductors are assumed to be mobile under the influence of an electric field. The redistribution of the mobile charges of a conductor due to an electric field causes the field inside an conductor to become zero.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Do I conclude that at a distance arbitrarily close to one of these charges the field is zero? $\endgroup$ – garyp May 16 '18 at 13:26
  • $\begingroup$ @garyp - If you want to know the details of the screening of the electric field of point charges in a metal or in a semiconductor by the mobile charge carriers (electrons), you have to consider the Thomas-Fermi or Debye screening. See en.wikipedia.org/wiki/Thomas–Fermi_screening en.wikipedia.org/wiki/Debye_length . The corresponding screening lengths are usually very small. $\endgroup$ – freecharly May 16 '18 at 13:42
  • $\begingroup$ I don't understand what you're trying to say, but in my argument I don't assume that the point charges are rigid in the ball - they're free to move, but they won't because the net force acting on them is zero. $\endgroup$ – katana_0 May 16 '18 at 15:43
  • $\begingroup$ @AlexKChen - The net force on any non-dispersible charge (mobile or immobile) you put into a conductor is zero. But not for the reason you a suggesting, i.e., the other point charges you put there. The real reason is that the electric field of these point charges is shielded by the mobile charges of the conductor so that the electric fields of such charges cannot penetrate the conductor beyond a very short screening length (Thomas-Fermi or Debye length). $\endgroup$ – freecharly May 16 '18 at 15:57
  • $\begingroup$ @AlexKChen - See also en.wikipedia.org/wiki/Lindhard_theory $\endgroup$ – freecharly May 16 '18 at 16:02
0
$\begingroup$

The book says:

when you put some charge in an isolated conductor, then within around $10^{-9}$ seconds the charges all go to the surface of the conductor, and there's no charge in the inside of the conductor.

This is pretty much what is going to happen, if you put your charges inside a conductor.

As you are saying, these group of charges will create a non-zero field. This field will get free electrons inside the conductor in motion and, within a short period of time, perhaps, on the order of nanoseconds, the field inside the conductor will be gone - canceled by free electrons assuming their new positions.

If the net charge of the group is not zero, it'll end up, in a form of extra electrons or positive ions, on the external surface of the conductor. For instance, if the net charge of the group is -1 microcoulomb, about $6.4\times10^{12}$ electrons will be pushed out and end up distributed along the surface of the conductor.

| cite | improve this answer | |
$\endgroup$

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.