# Electric Field Lines

1. Mar 23, 2005

### dspp

Could anyone pls explain to me: what is the similar thing among points in the same Electric Field Line?

(I have 2 particles A and B which are + and -, then I have 2 other ones C and D --> how can I know that C and D are lying in 1 EF line or in 2 different lines?)

Thx so much.

2. Mar 23, 2005

### Chi Meson

One clarification: are you sure you are talking about "electric field lines" and not "equipotential lines" ? And furthermore, what is the situation: is this perhaps in a uniform electric field (as in between plates of a capacitor)?

Or is it a field of a point charge. There is no quantity that is constant for two points on the same field line of a point charge, but the direction would be constant (could that be it?)

3. Mar 23, 2005

### dspp

Sorry coz my question was a bit ambiguous.
Pls have a look at the attachment.
What I got stuck is the difference between C and D if they are lying in 2 separate lines.

What I actually wanna ask is : is there any rule or any equation for each EL line?

Thx so much.

#### Attached Files:

• ###### EL lines.JPG
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4. Mar 23, 2005

### chroot

Staff Emeritus
First, you need to understand what "field lines" are.

Field lines do not really exist. They're just a clever way to visualize a field, but there's nothing physical about a field line.

Think about it this way: at each point in space, you can calculate the electric field, which is a vector. At every point in space, the field has both a direction and a magnitude. You can imagine planting a little vector at each point on a plane, indicating the direction and magnitude of the field at those points.

If you then imagine drawing a line through that field of vectors, so that the vectors are everywhere tangent to the line, you're drawing a field line.

The upshot of this is that you can draw a field line through any arbitrary point. There isn't some finite number of field lines, or some equations which govern them (except for the condition that a field line is everywhere tangent to the field vectors). If you pick any point in the plane, you can draw a field line through that point.

Here's an applet that will help make this clear:

http://www.slcc.edu/schools/hum_sci/physics/tutor/2220/e_fields/java/

First, you can add a positive and negative charge to the plane. Next, select "Draw a field line" and start clicking. A field line will be drawn through the point you chose. With a little practice, you can make a picture that looks just like the one on your assignment.

- Warren

5. Mar 23, 2005

### dspp

However, I still feel unsatisfied with it.
For ex, in my attachment, we have A(+) and B(-). With another article C, how to draw an EL line through C?

From the textbook or other sources I've ever read, they just draw like (1). Why don't we draw like (2)? Can we drwa whatever we want? Coz in 2 cases, the direction of C is still the tangent of the EL line.

#### Attached Files:

• ###### ELlines.JPG
File size:
2.4 KB
Views:
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6. Mar 23, 2005

### chroot

Staff Emeritus
The problem is that a line like (2) is not everywhere tanget to the electric field vectors.

- Warren

7. Mar 23, 2005

### dspp

So there must be some rules about the direction of vectors in Electric Field? Sth like the "continuity of direction", otherwise I could say that line 2 (or any other line) is everywhere tangent in the EF.
And if there is a rule, what is it?

8. Mar 23, 2005

### chroot

Staff Emeritus
The electric field itself (its magnitude and direction at all points in space) can be found by e.g. Coulomb's or Gauss' law; the field lines are just a way to visualize the field.

You could, in principle, draw any arbitrary field line, and then find a distribution of charges that would satisfy it, but that's a difficult problem that isn't normally done.

In the case of line number (2), there are only two charges present, and the field due to those two charges alone cannot look that way.

- Warren

9. Mar 23, 2005

### Original Torch

Field Lines and Coulomb's Law

You can use Coulomb's Law, F= k*q(a)*q(c)/ r^2 to determine the magnitude of the force on q(c) due to charge a, and depending on the sign of both it will either be attractive or repulsive, pointing along the line that joins the two (either inward or outward respectively). If the second charge isn't known, even better, this will give you the field strength and direction due to charge 'a' at that point.

Now, you must do the same thing at the same point in space for the other charge in your system, which you have labelled charge b.

You now have two vectors. Add them using vector addition methods, and this will tell you the direction your field points (and it's magnitude, which you don't seem particularly concerned about at this point).

No for the fun part. Do that very same thing for every point in the space surrounding your two charges. Notice that for small changes in position, the resultant vector direction (generally) doesn't change much. By smoothly connecting all of these vectors such that each is tangent to the connecting line, you have a representation of the electric field.

Torch.

10. Mar 23, 2005

### chroot

Staff Emeritus
Original Torch,

You don't really want to use Coulomb's law directly, since it involves the force between two charges. You should just use the definition of the electric field due to a point charge:

$$E = \frac{kq}{r^2}$$

You can superpose (add) two or more electric fields due to each point charges to find the total field.

- Warren