I'm thoroughly confused by conductors. Let's say you have a round(ish), perfect conductor. Solid and uncharged. Some distance from it, we place a point charge $q$, giving rise to a field $\vec{E}$. Now, the electrons inside the conductor will move around to create an internal field which cancels $\vec{E}$ inside the conductor. Will they not have to do this by going to the surface, creating a negative surface charge? But then, will there not be a positive charge density at least somewhere inside the conductor, since it was neutral to begin with?

Consider the same set up, except this time the conductor is hollow inside, but closed, and with some thickness to its walls. A Faraday cage, I believe. The way I understand it, the "minuses" will all go to the outer surface, while the "pluses" will be scared, by $\vec{E}$, onto the inner surface. This spreading out creates a field which cancels $\vec{E}$ inside the walls. But then the inner wall (inner surface) is full of positive charges, which in my opinion should create a positive field inside the cavity.

I realize that I would have a hard time drawing the field lines inside this cavity; because, where would they end up? But this merely adds to my confusion. If the conductor was perfectly spherical, I think I would understand why there was no field in the cavity: same reason there would be no gravitational force inside a hollow, spherical planet. But what if the conductor has uneven inner surface?

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    $\begingroup$ @Benjamin It's good of you to welcome people to the site, but if you're not already doing so, I'd encourage you to also offer feedback on the post you're commenting on. (Since that is the intended purpose of comments.) Not that there was anything in particular you should have said here, just something to keep in mind for the future. $\endgroup$
    – David Z
    Apr 4, 2016 at 7:46

1 Answer 1


The electrons follow the field as far as they may and as far as there is any field. Be aware, that there is no such thing as "positive field" as you wrote. The field is a vector and has a direction.

  1. Consider first an easier setup: if the external charge would be inside the conductor, the electrons would come and surround it and cancel it completely, there will be no field in the rest of the conductor. There can't be, if there were, the electrons would move until this field is gone.

    And yes, of course there would be some electrons missing now, and the rest of the conductor would be charged positively. But it's no problem, this does not imply a field. Don't think about the distribution of these charges yet, just: it has to be such as to cancel all fields.

    This is a clever mechanism of nature, just like the "invisible hand" in economics: every electron checks whether the field is zero and, if not, contributes by its movement to making it zero!

    The net charge of the whole conductor is still equal to the external charge.

  2. Now consider the external charge outside. The electrons will try the old trick to surround it. But they can't get there, the surface is in the way, which holds them back by acting with a force on them.

    But the surface can not hinder them to move inside the conductor. So inside the conductor the invisible hand is still making the field zero, while on the surface the tangential component of the field is zero. Any component of the field that is orthogonal to the surface remains:

    1. it doesn't have to be made zero, since the surface cancels its impact
    2. it cannot be made zero, since the electrons may not move in this direction.

    $ $

    The field inside the conductor is still zero, and the net charge is still zero. This is no problem, the charge density near the external charge is of opposite sign and rather big, since there is more outside field, and the charge density represents the jump in the field; and the charge density on the other side is of the same sign as the external charge and rather smaller (but on a bigger surface, so the net charge is zero)

    Note: the field outside the conductor is very complicated.

  3. Now cut out some hole inside the conductor. Nothing changes. There was neither field no charges there, and neither is there now after the cutting. This may be not intuitive but is really simple.

    By the way, if you now put another charge inside the hole, there will be some charges needed on the hole surface to cancel it out. They will be missing outside, and there will be a net charge of the whole system as seen from outside. Actually quite the same as in the first case.

  4. I talked about electrons, but there is nothing changed if the outside charge is negative and has to be cancelled by positive charges. Replace "electrons" by "absence of electrons" then. This is less intuitive but completely correct.

  5. Normally you will find arguments for all this which use Gauss' law. This is another viewpoint, it is shorter but less intuitive, but it gives some extra insights.

    But I cannot explain it without actually using it.

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    $\begingroup$ Thanks a lot for this very well thought out explanation, Ilja. Let's see if I get it: "...of course there would be some electrons missing now, and the rest of the conductor would be charged positively." According to Gauss's th., $\rho_v=0$ inside conductor. But if I understand your argument correctly, other electrons will now rush to this new positive area, until the only area left for the "absence of electrons" is the surface - which will then end up having a total charge equal to the point charge hidden inside it? "The field inside the conductor is still zero, and the net charge is still zer $\endgroup$ Apr 4, 2016 at 14:01
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    $\begingroup$ If I get your comment right: Gauss' theorem applied to a surface just below the surface of the conductor gives you a total charge of zero inside it, since there is no field. Since the whole charge is equal to the external charge, the same amount as the external charge has to be on the surface. $\endgroup$
    – Ilja
    Apr 4, 2016 at 14:10

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