Consider the diagram given below. If I put my reference point at infinity, then I get different line integral values for paths in blue and red, and hence different values and yet the integral $E.dl$ is supposed to be be path independent. You can think of the $E$ field as if there were positive and negative charges to the left and right respectively. The red path is along the axis through the midpoint connecting the positive and negative charges.

enter image description here

Edit:The potential in the middle of a dipole is zero. When you calculate the potential in the middle of the dipole, you do the line integral starting from infinity right? If so, then consider the two figures below. enter image description hereenter image description here

Are both of the line integrals zero? The vertical one is obviously, but the other one doesn't seem to be if you draw electric field lines around the dipole. Doesn't that mean that there is a certain sense of ambiguity in defining $r=\infty$ as a reference point?

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    $\begingroup$ The integral $\mathbf{E}\cdot d\mathbf{l}$ should give the same value, independent of path, given the same endpoints. Your red and blue lines do not have the same endpoints. $\endgroup$ – user197851 Feb 18 '19 at 18:43
  • $\begingroup$ But if you just write that the integral is done from r=infinity to r=a where r=a is the point where both the red and blue lines meet, then shouldn't both line integrals be valid? $\endgroup$ – Brain Stroke Patient Feb 18 '19 at 19:50
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    $\begingroup$ It might clarify your question if you could specify it completely in terms of vectors, stating $\mathbf{E}(\mathbf{r})$ as a function of position $\mathbf{r}$. Depending on what this is, it might not be consistent to take a reference potential of zero at "$r=\infty$". You may need to specify your two starting points for the integration as vectors as well. The integrals are line integrals (which you may be able to express in $x$, $y$ and $z$ components of $\mathbf{E}$ and $d\mathbf{l}$). In any case, I think more clarification is needed, in order to understand what is giving you difficulty. $\endgroup$ – user197851 Feb 18 '19 at 20:21
  • $\begingroup$ Edited @LonelyProf $\endgroup$ – Brain Stroke Patient Feb 19 '19 at 13:23

So there are a bunch of different things to consider here. The basic one is, if you complete the line integral off at infinity, you get a closed-loop, and then you get to apply the curl theorem,$$\oint_{\partial D} d\vec\ell\cdot\vec E=\iint_D d\vec A\cdot\big(\nabla\times\vec E\big).$$so the left hand side will only be zero in cases where the right hand side is also zero, and the easiest way to guarantee that is to only consider fields that are curl-free. Your first example does not look curl-free unless we extend the field out to infinity and make it constant everywhere, and in that case we get a non-trivial contribution for the part of the line integral that's off at infinity. So that leads to principle #1, a point at infinity is only well-defined for irrotational fields that decay to zero faster than $1/r$ at infinity. If it's not irrotational then you might have a circle of points at infinity; if it's not decaying to zero then probably all bets are more or less off, and that particular asymptote comes from the fact that the arc length is scaling like $r$, so that at least some fields that decay like $1/r$ have non-decaying contributions over circular arcs of fixed angle, as they are scaled out to infinity.

The second part is, the electric dipole field is in fact curl-free, and it decays to zero sufficiently fast. So, you must find that both line integrals come out the same. Like, if you found out that this weren't true, it would be the case that the principle of superposition has failed: because you know what the electric potential is for each of those charges, and you can just add them together, and get a potential field for the electric dipole. That potential field is clearly zero at infinity, and it is clearly zero in the middle. It is clearly nonzero in the middle of your curve, which means that its gradient the electric field points inward on the inner-part of the curve, but outward on the outer-part of the curve, and this integrates to 0.


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