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Why is the electric field from a dipole nonzero? Intuitively, I know there are electric field lines going from the positive to negative charge, so there should be an electric field.

But if I apply Gauss's law, then the enclosed charge should be zero (as the positive and negative charges cancel), meaning the electric field should be zero. Why is this argument invalid?

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Gauss’ Law is a statement about electric flux. A nonzero field can have zero flux.

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  • $\begingroup$ @CharlesFrancis Please write your own answer and do not edit mine to say what you want to say. Edits should be for correcting typos, formatting, etc. $\endgroup$
    – G. Smith
    Mar 31, 2020 at 19:58
  • $\begingroup$ Fair enough. I confess I was surprised to find it where I did, and didn't know the best way forward. $\endgroup$ Mar 31, 2020 at 20:02
  • $\begingroup$ @CharlesFrancis I don’t know what you mean by “find it where I did”. $\endgroup$
    – G. Smith
    Mar 31, 2020 at 20:03
  • $\begingroup$ It was under review as a low quality post, and someone had downvoted. I probably would have done better to comment that the questioner could probably use greater clarification. $\endgroup$ Mar 31, 2020 at 20:05
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    $\begingroup$ I dislike verbosity, and on elementary questions I think the questioner should be led to rethink their misconception, not spoonfed material already in their textbook. If that causes downvotes, I’ll live with them. $\endgroup$
    – G. Smith
    Mar 31, 2020 at 20:15
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Flux is a measure of the electric field through a given surface. The field passes both ways, at different places, through the enclosing surface such that the total flux cancels out, precisely because the enclosed charge is zero. It does not follow that the field is zero at any given point.

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The electric flux does not always determine the electric field. It is only when the field is constant on the Gaussian surface so that one can write $$ \oint \vec E\cdot d\vec S=\vert \vec E\vert S $$ and one use Gauss' law in reverse and in the field. Taking a box as a Gaussian surface, the flux will not be uniform - it will be slightly more positive on a small area right ahead of the positive charge but slightly more negative behind the negative charge - because of the distance from the charges to the particular small area of the Gaussian surface, as exemplified here:

enter image description here

The net flux through the box is $0$ because the patches with overall negative flux exactly make up for the patches with overall positive flux, but as the figure shows this does NOT imply the net field is $0$ on the box, simply because $\vec E\cdot d\vec S$ is not constant on the box.

Figure credit: Young and Freedman's University Physics

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Dipole consists of two charges - one positive charge and one negative charge. Those two charges are not at the same point in space. That means that they give some nonzero electric field in total.

You can apply Gauss law, but Gauss law tells you how much flux goes through your enclosed surface. You cannot always see the magnitude of electric field just using the Gauss law. The reason why we sometimes use Gauss law to determine electric field is because we have some additional information. For example if we have spherical symmetry (which electric dipole does not posses!) then we know that the magnitude of electric field is equal at every point of the sphere and we can use that sphere in the Gauss law to enclose the charged system we are looking at. In that case you can say that $\Phi = E S$ (if electric field is orthogonal to the sphere) where $\Phi$ is total flux which is according to Gauss law equal to $\Phi= \frac{Q}{\varepsilon_0}$. From these two equations you can determine $E$.

In your particular system you do not have spherical symmetry, so you cannot use this kind of reasoning to determine the field $E$ itself.

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The mistake in your reasoning is to conclude that, since the net flux through a closed surface is zero then the electric field everywhere on that surface, and in space, should be zero. This is not true. Since flux contains the dot product of the vector field with the local normal to the surface the field could have some + flux at points and - flux elsewhere on the surface. It is entirely possible for the field lines to cut the the surface in both directions (inside to outside) and (outside to inside) leading to a zero for the sum.

In the case of the dipole you have a + and a - charge of the same magnitude, Q, and you have field lines that leave the +Q and land on the -Q. If you just draw the picture and sketch some field lines you will see that when you draw a circle around both charges as many lines go into it as out of it, therefore ZERO flux over all. Your sketch better be accurate.

Gauss' Law states that the total flux through any closed surface id proportional to the net charge enclosed within that surface, Flux ~ Q_inside. You cannot use this to evaluate the electric field on the surface except in situations with a high degree of symmetry and that might be causing confusion. Many texts use Gauss' law to calculate the Electric field due to a uniform spherical charge distribution or an infinite line of charge. In such cases you are justified making the claim that E is constant on the surface (a surface that respects the symmetry of the source) and puling it our of the flux integral. For the dipole you cannot do this. The fact that Flux = 0 when E != 0 is not a contradiction in this example.

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Why is electric field of dipole nonzero?

Because the positive and the negative charges are separated by some distance.

Now Gauss theorem tells us about the net charge inside a Gaussian surface and the total electric flux crossing that surface. It doesn't tell us whether the charges inside that surface are separated or not.

So though the a Gaussian surface enclosing the electric dipole has zero net charge the electric field in the region isn't zero because there are separated charges inside the enclosed surface. Nevertheless, as stated by Gauss theorem, the net electric flux crossing that Gaussian surface is also $0$.

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