I have come across this doubt, if electric field lines originating from a $+2q$ charge terminate on $-q$ charge (both point charges), will all field lines terminate on the negative charge or only half of them will ? That is, are field lines coming from or coming to a charge proportional to magnitude of charge or not?
(ISOLATED point charges)
I have found many books saying it's proportional but I can't seem to find a credible/ authentic source or book .
So please answer with proof and provide a source too if possible. Thanks.
[Physics] Electric Field Lines Proportional To Magnitude Of Charge Or Not
chargeelectric-fieldselectrostatics
Related Solutions
It's helpful that the posted image is sideways, because it already shows the answer: as you rotate the coordinate system, and advance by the same distance, $r$, everything looks exactly the same. This is true if you rotate in the plane of the table, or if you hold the page against a wall and rotate it.
If instead of rotating the coordinate system you were to imagine holding a smooth, uniformly colored sphere, like a ball bearing, you will find that no matter how you rotate it, it will appear the same.
This is spherical symmetry. Since everything is the same, the intensity calculated at one point will be the same as that for any other point. However, the radial vectors will all point in different directions - they all point directly away from the center, each one towards its own point on the surface.
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.
Best Answer
Field lines are a visual aid.
Gauss's law is really "counting" the number of field lines which pass through a surface - the flux (flow) of field lines.
So a closed surface around the $+2q$ charge will have twice the flux, "field lines", going out from the charge $+2q$ as the flux going into the $-q$ charge.
Not all the field lines ending on the $-q$ necessarily originate from the $+2q$ charge although if the charges are close together most of them would have come from the $+2q$ charge with the rest coming from other charges.
In the same way nearly half of the field lines originating from the $+2q$ will end up on the $-q$ charge and the rest will go to other charges or go to infinity.
Update in answer to some comments from @Mr.Momo
My answer above was in the "spirit" of the question ie I wrote about electric field lines which I said are visual aids although Faraday did "count" them.
In terms of electric flux I would say that the electric flux through a closed surface around the $+2q$ is double that passing through a closed surface around the $-q$ charge.
The electric flux passing through a closed surface enclosing the two charges is half that of the electric flux passing through a closed surface around the $+2q$.
I did consider adding "or go to infinity" but did not add the phrase because I do not know what happens at infinity and whether this is only a hypothetical concept.
I do not mind adding it to my answer.
With two isolated charges the flux which does not pass through the surfaces around the two individual charges passes through a closed surface around the two charges.
I think it is worth reading this chapter about how Maxwell applied Faraday's lines of force.