If electric field lines cannot terminate in empty space, in the simple example of two equal charges what happens to the line starting from one of the charges toward the other in the middle?
[Physics] Electric field lines
electric-fieldselectrostatics
Related Solutions
Yes the additional positive charge to one plate increases electric field behind the other one.
To see this let's look at general Gauss's law:
$$\oint_{S}\vec{E} \,d\vec{A}=Q/\epsilon_0$$
The $S$ is the Gaussian surface which we are free to choose therefore it could include both plates even if the surface goes to infinity with both plates.
The charge Q however could be written as charge density $\sigma$ times surface area or in general:
$$Q = \int_A \sigma \, dA $$
where $\sigma$ could be $\sigma_1$, $\sigma_2$ or $\sigma_1 + \sigma_2$.
Now we are ready to use symmetry of configuration. Lets assume that field is homogeneous at the middle of of both infinities and rewrite the left side of Gauss law as:
$$\oint_{S} \vec{E}\, d\vec{A} = \oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} $$
The right side with the same division:
$$\int_A = \int_{in} +\int_{out}$$
Now we are ready for the main part:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} + \oint_{S_{out}} \vec{E}\, d\vec{A} =\int_{in}\sigma \, dA + \int_{out}\sigma \, dA$$
And the Gauss law says that charges which is outside from closed surface does not affect the flux or $\vec{E} = \vec{E}_{in} + \vec{E}_{out}$ and the $\oint_{out} \vec{E}_{in} \, dA = 0 $ which makes to satisfy:
$$\oint_{S_{out}} \vec{E}\, d\vec{A} = \int_{out}\sigma \, dA$$
And the only thing which is left is always satisfied:
$$\oint_{S_{in}} \vec{E}\, d\vec{A} = \int_{in}\sigma \, dA$$
which for homogeneous field means:
$E_{top} + E_{bottom} = \sigma / \epsilon_0$
where the signs for field is positive if it is directed out of surface.
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 $+2q$.
"…and the rest will go to other charges." I would add here "or go to infinity." Even in the hypothetical situation where the whole universe consists of only these two charges the field lines still make sense If we consider only the two charges to be present that is, isolated 2 charges, what would be the explanation in that case?
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.
Best Answer
There are in fact two field lines that depart each charge headed towards the other. These lines meet at the origin (the mid-point of the two charges), where the field is zero, and vanish there. There are also two other lines, which are born at the origin and depart along the vertical axis. Thus, formally, two lines go in and two lines go out, so no lines actually die in empty space.
These lines are actually a limiting case of lines that leave the point charges at a small angle $\epsilon$ from the intercharge axis; these lines make increasingly close approaches to the origin as $\epsilon\rightarrow0$, and then they shoot off to infinity, increasingly close to the vertical axis.
(If you're sharp, you'll notice there's actually an infinity of such lines, since there's also lines that go off perpendicularly to the screen and at any angle in between. Thus my "two-for-two" argument is not actually quite right. Can you see the limiting behaviour that makes it right?)
Pictures of this were relatively hard to find, but you can see them in this Wolfram web app:
You also have to consider one key point: at the origin, the field is zero, so actually there should be no field lines through it. Or, more formally, the density of field lines should be zero. This comes about in that the angle $\epsilon$ should be really small for the lines to actually approach the origin. You should then plaster the diagram with lines leaving equiangularly at angle $\epsilon$ from each charge, and that will mean a lot of lines on the "outside" of the charges.
Ultimately, though, the lesson is that individual field lines are not that important, and it is the set of lines, equiangularly leaving the charges (in 3D!), that makes a physically relevant diagram. And even then, field line diagrams are only of limited utility in understanding electric fields, mostly because they only incorporate with the utmost difficulty the superposition principle, which is at the real heart of classical electromagnetism.