-3

When you have a point charge, q, eight field lines are drawn to represent this element of charge - Why eight? Is this something just selected as a standard because electric field lines are infinite theoretically?

peterh
  • 8,208
AAM
  • 3

3 Answers3

1

There are no field lines, not 8, not anything.

The field around the point charge is a vector field. It means, we imagine a vector in all (infinite many) points whole space. It points to the direction of the force of 1 unit charge, and its length is the strength of this force.

The problem is that you can't show it easily understable in a school textbook.

So, in the reality you can imagine infinite many vectors.

The likely reason of that you found mainly 8 lines is, that it wasn't important how many is actually shown in the textbook. Because all of them are equally false, but there is no better way. Thus, it was the easiest to draw with the image editor software of the book author.

peterh
  • 8,208
  • After my fix I got a down - I think my fix made to post significantly better, but if it didn't, I am open to edit suggestions. – peterh Jun 30 '17 at 20:47
  • "There are no field lines" is a rather misleading statement, I should think, without appropriate qualifiers. Field lines do exist - as solutions of a certain differential equation derived from the vector field. If what you mean is that field lines have no physical existence because they are just a mathematical tool for understanding electrostatic forces, then the same applies to the electric field when seen as a vector field, does that mean that it does not "exist"? – Emilio Pisanty Jun 30 '17 at 20:51
  • (And, in addition, there are some pretty severe jagged lines in the grammar that it would be good to have fixed.) – Emilio Pisanty Jun 30 '17 at 20:52
1

The convention is that the number of field lines that start or end on a charge are proportional to the magnitude of the charge. This only becomes significant when there is more than one charge present, because this isn't done in an absolute sense, but a relative sense; so a charge $2q$ should have twice as many lines as one with $-q$ in the same diagram.

Other than that, you just need enough lines to highlight the significant points of the electric field (overall shape, direction of field line bending, etc).

Sean E. Lake
  • 22,482
0

I think it's important to realize that electric field lines are a tool for visualization. The electric field really takes on a vector at every point in space, but we can't really put this on a piece of paper visually. With those in mind, keep the following in mind:

  • The electric field takes on a form $\vec{E} = \frac{q\hat{r}}{r^2}$. From this, it follows that the field has a magnitude based on $\frac{q}{r^2}$ and a direction based on $\hat{r}$. The $r^{-2}$ dependence of the field strength falls off as the area of a circle, so the number of lines should be proportional to $q$. Whatever you draw, a charge of $2q$ should have twice as many lines as a charge of $q$. The sign of the charge combined with $\hat{r}$ determines direction. A positive charge will have arrows pointing out of the point charge and a negative charge will have arrows pointing into the charge. Griffiths' section 2.2.1 probably gives a better explanation of this than I just gave.
  • The number of lines should make a decent representation of the field. I think the reason 8 lines is used a lot is so there is a line for both the positive and negative of each axis in two dimensions and a line to separate each of them. This gives a decent visualization of what happens with a single point charge, but more complicated configurations may require something different.

The big point to take away is that field lines are pretty arbitrary and should only be used as a visual aid, but you should have enough lines for it to actually work as a visual aid.

Edit: After thinking about my reasoning with the $r^{-2}$ dependence, I'm having doubts about the 2D picture providing a true representation. I think it may actually better represent a field that has an $r^{-1}$ dependence and you'd need a 3D drawing to accurately represent an $r^{-2}$ decreasing field with field lines. I don't think it should matter as long as you aren't planning to do measurements based just on the drawing.

  • 1
    "After thinking about my reasoning with the $r^{−2}$ dependence, I'm having doubts about the 2D picture providing a true representation." That's correct. We usually sweep that detail under the rug for the sake of simplicity and because the field line technique is really more effective at representing qualitative features of the field than the quantitative ones. – Sean E. Lake Jul 01 '17 at 01:28