This is a simple question that occurred to me while thinking about electrostatics. Let's consider a positively charged isolated conductor in equilibrium. In general, the surface charge density varies over the surface, peaking at regions of sharp curvature. This is fairly intuitive, but it becomes less intuitive when we allow the conductor to be concave, with any possible shape for its perimeter.

If we allow an arbitrary (connected) shape, is it possible for some region to end up with negative charge even if the entire conductor has positive charge? I suspect not, but I can't prove it can't happen.

  • $\begingroup$ I assume that the conductor is the only charged object we are allowed, yes? If we can put other charges outside the conductor, then it should be fairly easy to produce a negative charge on the surface by putting a positive charge very close to it. $\endgroup$ – Ricky Tensor Sep 7 '18 at 4:59
  • $\begingroup$ @RickyTensor Yes, just the conductor by itself. $\endgroup$ – knzhou Sep 7 '18 at 5:09

Our body is positively charged, so it's potential is positive.

Suppose there is a place on it's surface with a negative charge density. Electric field direction is towards the surface near the spot.

Consider some electric field line that ends on our spot. Where could it originate?

There are only two options: it originated either from some other point of the conductor or from the infinity. (*)

The first option would mean that two points of the conductor have different potentials, so it's impossible.

Second option would mean that the potential of ending point of the electric line field is negative. But it must be positive.

So there are no negatively charged regions on the surface of our body.

(*) I understand that the proof is not rigorous here, but I am quite sure I can improve it.

  • $\begingroup$ Third option; a charge can be inside a conductor but still insulated from the conductor. $\endgroup$ – Whit3rd Sep 7 '18 at 7:05
  • $\begingroup$ @Whit3rd If our body is contained inside some cavity inside a conductor the proof remains the same, only need to replace the word "infinity" with word "the conductor". Actually the proof can be adopted even if any number of not charged conductors are allowed around our body. Instead of one electric field line we would need to consider several: first ends on our charged body and originates from some uncharged body, second one ends on the uncharged body and originates from either infinity or some other body and so on. $\endgroup$ – lesnik Sep 7 '18 at 7:19

I am not sure I fully understand the question, but it seems to me that it is "possible for some region to end up with negative charge even if the entire conductor has positive charge".

Let us place a perfectly conducting cone and a perfectly conducting sphere of a large radius in such a way that the cone and the sphere do not overlap, the center of the sphere is on the extension of the axis of the cone, and the tip of the cone is close to the surface of the sphere. Let us connect the sphere and the cone by a conductor that is placed far from the tip of the cone and put positive charge on the resulting conducting system including the cone, the sphere, and the conductor. I would think that there will be high electric field near the tip of the cone that would repel positive charges on the sphere in the vicinity of the tip, so there will probably be a negative charge density on the sphere near the edge of the cone.

  • $\begingroup$ Can you prove that? It seems much more plausible to me that there would simply be almost zero charge on the tip of the cone and the part of the sphere near it. $\endgroup$ – knzhou Sep 7 '18 at 6:32
  • $\begingroup$ It is tempting to say the positive and negative charge must remain because “they attract each other”. However, this is unambiguously wrong. $\endgroup$ – knzhou Sep 7 '18 at 6:33
  • $\begingroup$ @knzhou : "It seems much more plausible to me that there would simply be almost zero charge on the tip of the cone and the part of the sphere near it." OK, let us assume, for the sake of argument, that the charge is almost zero on the tip of the cone. However, the electric field has a singularity near the tip, so the resulting electric field can still be high. And no, I cannot prove right now that there will be negative charge density somewhere on the sphere. One could just solve numerically the Laplace problem, but I don't have time for that now. $\endgroup$ – akhmeteli Sep 7 '18 at 6:52
  • $\begingroup$ @knzhou : "It is tempting to say the positive and negative charge must remain" I am afraid I don't understand what it means and how this is relevant. $\endgroup$ – akhmeteli Sep 7 '18 at 6:54

This is not possible. The potential on the surface of the conductor has the same positive value everywhere. For a negative surface charge to exist, a field line should connect to a location with higher potential and this does not exist. Only locations with lower potential exist, namely locations that are connected to the conductor and to infinity, where the potential is zero.


Yes, of course a positively charged object can have a negatively charged surface. Consider a hollow sphere of metal,charged A, with an insulated positive-charge, B, inside the hollow. The exterior surface of the sphere will have the sum (A + B) charge, but the interior surface of the sphere will hold an induced (-B) charge.

  • $\begingroup$ No, I meant an isolated conductor; nothing else allowed. It is clearly possible if you put in other stuff. $\endgroup$ – knzhou Sep 7 '18 at 7:06
  • $\begingroup$ @knzhou: Isolated? The only question specification was 'a conductor' meaning a connected conductive object. $\endgroup$ – Whit3rd Sep 7 '18 at 7:12
  • $\begingroup$ Right after I posted the question somebody commented that I should have included "isolated", so I put that in. It's just not an interesting question if it isn't isolated, because there are a million ways to induce negative charge if you allow external positive charge. $\endgroup$ – knzhou Sep 7 '18 at 7:22
  • $\begingroup$ @knzhou OK, how about a million-and-one ways: by heating one end of a wire, the electron density is driven down (Seebeck effect); the cold end picks up excess electrons. It amounts to a voltage gradient inside the conductor (not usually part of an electrostatics problem, but it does happen in real metals). That cold end can have a net negative surface charge. $\endgroup$ – Whit3rd Sep 7 '18 at 10:01

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