I have rendered an image, where two positive charges with equal electrical charges are present: my application's output I was told that the picture is not correct, because the field lines are getting "concentrated", so if I were to look at this from very far away, I would never see an (almost) perfectly symmetrical picture - it would never be (almost) the same as a single charge with twice the electrical charge.

Here's a picture which (probably) meets such a criteria: (source) picture from someone else

Are both of these images correct - do both of these images correctly show (some of) the field lines of the two charges? Why (not)?

My guess is that both of these are correct, but in the second image only specially selected field lines are drawn, so the outcome contains such evenly distributed field lines, but I have no proof which backs this theory up.

A better question would be whether the density of the field lines should be proportional to the field-strength if the field lines start out from the charges at regular angle intervals.

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    $\begingroup$ Your first image does look incorrect - the concentration on field lines along the separatrix suggests that the field-strength maximum there is much stronger than it really is. However, without knowing exactly what you've done, it's impossible to tell exactly what's gone wrong. $\endgroup$ – Emilio Pisanty Nov 22 '17 at 20:37
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    $\begingroup$ Also, it's important to note that the lower image is only an accurate representation of two line charges coming out of the page (cf. the last section here). In 3D, field lines come out of the page, and if you want to represent the field strength of point charges in 3D accurately, then you need to drop field lines out halfway through to mimic that effect, and to represent the fact that the field strength drops as $1/r^2$, instead of the $1/r$ that your diagram implies. $\endgroup$ – Emilio Pisanty Nov 22 '17 at 20:40
  • $\begingroup$ @EmilioPisanty Thank you very much for your input. Let's say that I am drawing line charges, not point charges. It is my understading that the field lines' density only represent the strength of the electric field by convention, so the real issue with my image is that I start the lines evenly distributed (in the middle of the charge, the field lines close the same angles), but this even distribution doesn't "continue" - the distribution is uneven, despite it starting as even. $\endgroup$ – Trigary Nov 23 '17 at 9:14
  • $\begingroup$ No, it's not just convention - if you draw your diagram correctly, and in the limit of high density of lines, the mapping to the field strength is exact; this is in direct correspondence with the Gauss law and it's explained in depth in my link above. That said, it's unclear what you mean by "uneven", so you should clarify that - you have an inhomogeneous $\mathbf E(\mathbf r)$, so you don't expect any "evenness" as a whole. $\endgroup$ – Emilio Pisanty Nov 23 '17 at 9:31
  • $\begingroup$ On the other hand, if you draw a circle of radius $R$ that's bigger than the separation $d$ of the charges (say, $R>2d$ or so), and you start off your field lines at regular angle intervals out of the line charges, then yes, you expect the number of lines that reach each fixed-length arc of that circle to be extremely uniform. And, moreover, the density of lines that reaches that circle has its minimum at the separatrix, where your plot shows a local maximum. $\endgroup$ – Emilio Pisanty Nov 23 '17 at 9:36

Let's start with the easy one:

A better question would be whether the density of the field lines should be proportional to the field-strength if the field lines start out from the charges at regular angle intervals.

The answer is yes, but: for a 2D plot, when correctly plotted, this is only true for the 2D equivalent of the point charge, i.e. for (combinations of) line charges whose field goes down as $1/r$. This is explained in more depth in the last section of this old answer of mine, but the short of it is that for lines that expand out (or converge in) radially at equal angular intervals, the number of field lines per unit length that cross a given circle of radius $R$ (itself proportional to the field strength) goes down as $1/R$, which corresponds to line charges, not to point charges in 3D.

As for what the field lines should look like, here's a more accurate authoritative plot for the (correctly spaced) field lines of two identical line charges coming out of the page, at a fixed distance but plotted over increasing ranges:

Mathematica source at Import["http://halirutan.github.io/Mathematica-SE-Tools/decode.m"]["http://i.stack.imgur.com/2fN3I.png"]

Some salient points in these plots:

  • The field lines approach each individual charge at a constant angular separation, as they should.
  • When seen at the furthest-out zoom level, the system looks quite a bit like a single line charge of double the charge density, and its field lines reflect that.
  • To drill down on that: this requires the field lines to hit the outer circle at an increasingly even separation as the outer circle's radius increases.
  • At small zoom levels, however, the field is stronger at the horizontal axis than it is at the vertical ends of the outer circle. (This is because, at the vertical end, the horizontal components of the field cancel out, and the vertical components are not as big.) This means that the line density is larger at the horizontal ends than at the vertical ends.
  • In fact, the line density has a local minimum at the vertical ends. This is entirely inconsistent with your first plot, and a surefire way to diagnose it as a buggy plot. (As to what the bug is, it's impossible to tell without seeing the code.)

And finally, an important feature of streamline plots in vacuum in two dimensions: the electrostatic potential there is a harmonic function (i.e. it obeys the Laplace equation $\nabla^2 \varphi=0$), and this means that the streamline plots of its gradient are much easier to find if you see it as a real-valued function of the complex plane, $\varphi:\mathbb C\to \mathbb R$ and find a harmonic conjugate $\chi$ for it, which then acts as its stream function.

Then, because of the nice properties of analytical functions, you know that the streamlines of $\nabla\varphi$ are orthogonal to the gradient $\nabla\chi$, i.e. that $\chi$ is constant over the streamlines. This means that (i) the streamlines are much easier to plot, by simply doing a contour plot of $\chi$, and (ii) that the correct separation in the streamlines can be strictly enforced by simply plotting those contour lines of $\chi$ at regular intervals.

This is the technique used to produce the plots above: here you know that the potential essentially has the form $$ \varphi(x,y) = \ln\left(\sqrt{(x-1)^2+y^2}\right)+\ln\left(\sqrt{(x+1)^2+y^2}\right), $$ but that's much easier to express as $$ \varphi(x,y) = \mathrm{Re}\mathopen{}\bigg[ \ln(x-1+iy) + \ln(x+1+iy)\bigg], $$ which then immediately gives you the stream function as $$ \chi(x,y) = \mathrm{Im}\mathopen{}\bigg[ \ln(x-1+iy) + \ln(x+1+iy)\bigg]. $$ Because of the complex logarithms you might have to wrangle with branch cuts and whatnot, but even then, this is way easier (if and when it is available) than the grind of numerically solving ODEs to get the streamlines, where in the general case it is not easy to enforce the correct streamline spacing.

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