# Electric field lines properties

I wanted to ask a question. My question is that are the properties of field lines ( number of field lines leaving/entering a point charge is proportional to the charge, field strength between points can be compared using relative field line density etc.) true for any symmetrically drawn diagram. Like if I had charges $$-q$$ and $$2q$$, then will out of all the field lines drawn symmetrically from $$2q$$ only half of them enter $$-q$$? Or are the field lines drawn in such a way that they they follow the mentioned properties?

Also what is the mathematical idea associated with field lines? For example divergence of a vector field is intuitively 'how much' of a source/sink a point is. Rigorously, it is limit of the flux of vector field per area of surface as the surface approaches zero. I am asking this because I feel that this may help me derive results related to field lines in a well-drawn diagram. So, could somebody tell me what to associate a 'field line' with? Flux? Electric field strength? Something else?

• Are you familiar with Gauss' Law and Gaussian surfaces? – Bob D Jan 18 '19 at 18:59
• In which ways does the previously-linked thread fail to answer the second half of this question? – Emilio Pisanty Jan 18 '19 at 19:58
• If I have time later, I might expand this comment into an answer. If you have a function $f(x) := x^2$, do you need to have a visualization (graph) of it? Even if you did try to turn the function into a picture, there's no reason to have the x and y axes perpendicular, nor do your tick marks on the x axis need to be evenly spaced, nor do the tick marks on the x and y axes have to exactly match. Say you do decide an x,y coordinate plane for the function $f$. What if you have another function $g$. Could you use a different coordinate system for $g$? You could but you'd have to tell the reader – DWade64 Jan 18 '19 at 21:10
• To graph $f$ and $g$ in the same space, we all assume they are drawn with respect to the same coordinate system. Likewise with charges, you don't need a visualization of the vector field. But if you do want one, there is nothing wrong with 4 lines coming out of $q$ and 1 line coming out of $2q$. This would be a visualization, but it wouldn't be a standard one (It's more useful to go with the "standard" visualization - i.e. "fluid flow (literally) convention" of visualization – DWade64 Jan 18 '19 at 21:16
• See similar question and its answers – Thomas Fritsch Sep 22 '19 at 13:12

Now, let us use these properties to answer your question. In the region surrounding the $$2q$$ charge, the field will be twice as strong as and pointing in the opposite direction from (in terms of divergence) the field in the region surrounding the $$-q$$. Field strength corresponds to line density, so $$2q$$ has twice as many field lines diverging from it as $$-q$$ has converging. Since lines can't combine or cross, only half of the lines starting at $$2q$$ make it to $$-q$$. The rest do not end.
This should make sense. Remember, field lines can be compared to flux. If we create a closed (Gaussian) surface around our two particle system, we get that the half of the lines not ending at $$-q$$ leave the closed surface. Tallying up these lines and multiplying by the area of the surface gives us the total flux which, if we recall Gauss' law, must be proportional to the net charge ($$q$$) of our system. Indeed, this is half of the flux we would calculate if our surface only enclosed $$2q$$.