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I know perfectly well how to derive the magnitude of the electric field near a conductor, $$E = \frac{\sigma}{\varepsilon_0}$$ and near a sheet of charge, $$ E = \frac{\sigma}{2\varepsilon_0} .$$

In fact, I can explain with clarity each step of the derivation and I understand why is one two times larger than the other. But here's what bothers me...

When I try to think about it purely intuitively (whatever the heck that actually means), I find it difficult to accept that a planar charge distribution with the same surface density can produce a different field.

Why should it care whether there's a conductor behind it or not... ?

I repeat, I understand Gauss' law and everything formally required, but I want to understand where my intuition went wrong.

EDIT: Thanks to you people, I developed my own intuition to deal with this problem, and I'm happy with it, you can see it posted as an answer!

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8 Answers 8

up vote 3 down vote accepted

(original poster here)

Thank you all for posting your answers! You all helped me to develop an intuition which I think illustrated what the problem really is, so I think this qualifies as an answer, although it's my own.

The problem with my intuition was that I viewed the conducting surface in the same manner as the sheet of charge, while in reality, it's very different.

Let's think about that conductor... There's a sheet of charge on its surface, or put in different words, there's a conducting material behind the sheet of charge. But although we can view the differential element of the surface as being perfectly flat, which justifies our assumption of there being an infinite surface of charge, we must remember that the conductor itself is finite in dimensions.

Putting it simply, there exists another sheet of charge, it must exist in a conductor with finite dimensions, since it must have another surface on the other side.

It's easiest if you imagine a very, very large conducting ball (have to be careful not to say infinite!)... If you get sufficiently close to it, what you see is a very, very large planar sheet of charge. BUT there's another sheet exactly like that on the other side of the ball, way back there, and it generates the same field.

And since both of those fields are distance-independent, voilá, the resulting electric field is twice the magnitude and my intuition was right after all... the charge doesn't care!

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1  
You can change the large conducting ball for a very wide (or infinite) plate that is thick but finitely so, to the same effect. –  Emilio Pisanty May 7 '13 at 18:24
    
I like that too! –  Schlomo Steinbergerstein May 7 '13 at 22:24

To clarify the setting, we have a surface of a conductor, and we consider its small piece, which is nearly flat and carries almost uniform charge density $\sigma$. Then we consider the electric field in the very vicinity of this piece, and the question is why do we have the two times difference.

I think that the right answer is that the formula for the sheet of charge is derived for a very specific global setting -- when it is infinite and flat and uniformly charged, and, as already mentioned by others, electric field depends on the global setting. So, the charged sheet has nothing to do with our "conducting" situation.

However, your second formula actually helps to understand your first formula. You know that the electric field inside the conductor should be zero, because otherwise it would generate currents that will tend to decrease the field. So the requirement of zero field is more or less just the nature of conductor -- the global setting is such that it satisfies it. Ok. Now consider that small piece of surface $dS$ from the beginning of the answer and look at the electric field in its vicinity: $$ E=E_1+E_0,$$ where $E_1$ is the electric field produced by $dS$ and $E_0$ is the electric field produced by all the other charges. You know that $E$ is $0$ inside the conductor and $E_{out}$ outside. You know that $E_1$ has opposite signs and the same value inside and outside of the conductor, while $E_0$ is continious, so nearly constant in our small area. This is enough to conclude that $E_1=E_0=E_{out}/2$, which is in a perfect agreement with your formulae, because $E_0$ is given by the second formula, while $E_{out}$ is given by the first.

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+1. Only answer here that actually addresses and answers the question precisely. –  santa claus May 7 '13 at 19:27
    
@AlecS, thanks) Especially for the fact that your comment made me reread the answer, revealing a typo. –  Peter Kravchuk May 7 '13 at 19:33

Intuitively, the surface charge on the edge of a conductor only produces a nonzero electric field on one side of itself, whereas the surface charge on an isolated sheet produces an electric field on both sides of itself. The charge on the isolated sheet is filling twice the amount of space (for an appropriate definition of "amount of space") with electric field, so the resulting field will be half as strong.

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This is a great question, and it challenges my own intuition.

First off, I have an intuition that the field at any given point can be found uniquely by summing over contributions from all the charges. But this intuition is wrong in many cases where the charge distribution extends over an infinite region of space. You get problems with the fact that the resulting integral is not absolutely convergent. A nice example is discussed here: http://scienceblogs.com/builtonfacts/2011/05/17/gauss-law-proved-wrong/ (see the comment by Adam Jermyn).

This intuition is closely related to a feeling that the Poisson equation must have unique solutions. Well, there are various uniqueness theorems for solutions to the Poisson equation for various types of boundary conditions, but there isn't any such theorem that covers the present case, because the boundary conditions are not given in one of those forms (e.g., they're not given by defining the potential on a bounded surface).

In real life, the result must depend on the details that are eliminated in the idealization of an infinite sheet.

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Thanks, that helped a lot, I've developed what I think is a good intuition, I'll soon post my own answer, everyone's been helpful! –  Schlomo Steinbergerstein Apr 21 '13 at 22:14

The answer is simple. It is all in the definition of sigma. In the case of a non-conducting sheet sigma means entire charge in a given area of the sheet meaning both surfaces and everything between them. In the conducting case it is just easier to think of sigma as being the charge on one surface not the sum of both as in the non-conducting case. Therefore, the conducting case looks twice as big simply because sigma is defined as half what it was before. Conceptually imagine for the non-conducting sheet defining sigma as the charge contained only in the upper half of the sheet. This redefinition of sigma will then give you the same answer as for the conductor.

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The electric field inside a conductor should be zero. But this is not necessary flux entering it should be equal to flux leaving. Thats why we get this answer.

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Great question! For the purpose of intuition, I think the crucial issue here is the fact that the electric field at a point is a "non-local" quantity; it is not just determined by charges in the immediate neighborhood of a given point. Here is why I think this is relevant to your question:

As you're probably aware, the crucial distinction between the two cases you mention is that when there's a conductor behind the sheet of charge, the electric field behind the sheet is zero since in the context of electrostatics, the electric field inside of a conductor vanishes.

The reason why the electric field is zero in the conductor is precisely because all of the electric charges on the surface conspire to distribute themselves in precisely the right way to make this happen. In particular, if the charges were just concentrated at some small patch on the surface, this clearly wouldn't be the case.

When you're at a point just outside of a conductor, the application of Gauss's law to get the right expression depends crucially on using the non-local fact that the electric field just inside the conductor is zero; you're therefore effectively considering the entire distribution of surface charge on the conductor, not just the small patch of charge right next to you.

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My opinion is somewhat different from the books' statements. In my opinion, the assumption that the electric field inside the conductor is zero is in fact a conclusion by assuming a Gaussian surface is placed inside the conductor. However, in this case, we can see that, all that is enclosed by the Gaussian surface is an infinite thin plate of charge, from which the electric field is caused is all that we should pay attention to. By considering both sides of the conductor's surface as two parallel placed infinite thin plates, we can find that on both sides of the conductor, the electric field is actually the superposition of the fields generated by the two thin plates, which is also $E=\sigma/\epsilon_0$, the same as the book says.

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I'm not downvoting, but this is specifically not what the OP wanted. –  santa claus May 7 '13 at 19:28

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