# I Dipole: Which field lines go to infinity?

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1. Sep 26, 2016

### greypilgrim

Hi.

An electric dipole field (two opposite point charges separated by some distance) has fields lines from the positive to the negative charge, but also field lines reaching to and coming from infinity. Starting from the positive charge, is there a way to compute the opening angle of the cone that contains all and only the starting directions of field lines that stop at the negative charge?

2. Sep 26, 2016

### Staff: Mentor

All field lines terminate on the negative charge except for the single line directed exactly away from it.

3. Sep 26, 2016

### greypilgrim

How so? There's certainly not only one field line connecting the positive and the negative charge.

Or do you mean all do except the ones that start/stop exactly facing away the opposite charge?

4. Sep 26, 2016

### Staff: Mentor

Yes, this is correct.

5. Sep 26, 2016

### greypilgrim

I briefly had this suspicion, but couldn't imagine that if I had only one charge in the universe and introduced a second opposite one VERY far away it would change the picture that much.

Is there a simple way to see/prove your statement?

6. Sep 26, 2016

### Philip Wood

This may be too hand-waving for you, but I'd go back to the definition of a field line: a line whose direction at every point along it gives the direction of the (resultant) field at that point. For your dipole, the resultant field strength at a point is found by vector addition of the fields due to the two charges. If you choose a point somewhere near the negative charge you can do an informal vector addition without calculation and determine in a a couple of seconds the rough direction of the field, which is the direction in which the field line is heading at that point. Then take a point in the direction in which the line is heading and do the same thing. You'll soon see what the lines are doing!

7. Sep 26, 2016

### greypilgrim

Well yeah, that's too hand-wavy. Why is the influence of the second charge enough? Would the same happen if the distance dependance was $\frac{1}{r^3}$?

8. Sep 26, 2016

### Staff: Mentor

Well, you cannot have a second charge "VERY far away" when you are asking about the behavior at infinity. At infinity any finite distance will look like an idealized point dipole.

9. Sep 26, 2016

### greypilgrim

But why is this true for an idealized point dipole?

10. Sep 26, 2016

### Khashishi

It's just a silly picture. Dipoles don't have lines coming out from them. Don't worry about it.

11. Sep 26, 2016

### ZapperZ

Staff Emeritus
Why can't you apply Gauss's Law here? Enclose the dipole inside a closed Gaussian surface. What is the net charge inside the enclosed volume? And consequently, what does this imply about the net E-field crossing this surface?

Zz.

12. Sep 30, 2016

### greypilgrim

Well it says only that the positive and the negative part of the surface integral cancel (which they would anyway in this symmetric situation) not that they are zero individually.