Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
    
As I understand it, field lines are no more real than contour lines on a topographic map. If so, discontinuities at boundary conditions may be no more paradoxical than pondering the direction of the 5000 m contour "dot" at the top of a 5000 m mountain. –  RedGrittyBrick Feb 18 '13 at 16:48

2 Answers 2

up vote 3 down vote accepted

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: enter image description here

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.

share|improve this answer
    
Thanks Emilio. The last paragraph is actually what I thought about. That single line of force does not contribute to the flux. I guess one can think of the origin as a point at infinity where the field is zero and the field line asymptotically approachs that point. Am I right? –  richard Feb 18 '13 at 15:41
    
Sort of. You should really formulate the differential equation for the field lines, $\frac{d \mathbf{r}}{d\tau}=\mathbf{E}$, and solve it for this case. You'll find, I think, that there is a well defined line going from each charge to the origin, but it takes infinite "time" $\tau$ to get there. I think it's more to do with the fact that the origin is an unstable saddle point, and therefore very hard to get to. –  Emilio Pisanty Feb 18 '13 at 17:27

This: enter image description here

Which I borrowed from: enter link description here

The lines go off to infinity and never terminate.

EDIT: As per your comment, you seem to be asking about the difference between an electric field of each point charge, and the net electric field of both. If you draw the electric field lines of one point charge Q, then they just point radially outwards. Now if you draw another point charge q on that same picture, then there will be an electric field line that 'goes through' that second point charge q. The relevance of this picture is that this is the electric field of Q and q will feel a force given by $\vec{F}_{on q} = q \vec{E}_{of Q}$. However, if you draw the $\bf{net}$ electric field of 2 equal point charges, then it will look like the picture above.

share|improve this answer
    
No, follow the field line which connects two charges the other lines yes scape to infinity –  richard Feb 18 '13 at 14:54
    
I don't understand your comment. Please elaborate. –  DJBunk Feb 18 '13 at 14:56
2  
I mean there is a field line starting from one of the charges toward the other (which is not sketched in your picture). This line cannot turn to left or right to scape to infinity and it seems it terminates in the middle. Hope this helps. –  richard Feb 18 '13 at 15:01
    
I mean the net electric field. Im saying that the above picture has not all field lines in it. The net electric field is not zero on the connecting line so you can draw a field line which Im concerned about. –  richard Feb 18 '13 at 15:31

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.