Electric dipole

  • Thread starter Rasalhague
  • Start date
  • #1
1,384
2
In lecture 2 of http://oyc.yale.edu/physics/physics/fundamentals-of-physics-ii/content/class-sessions [Broken], around the 50 minute mark, the professor calculates the value of an electric field at a point (x,0,0) on the x-axis of a cartesian coordinate system. (The video can be rather slow loading, so hopefully my description will be clear enough to stand alone.)

This electric field is due to two equal point charges: one positive, one negative. The positive charge is positioned on the x-axis at (a,0,0), where a > 0, and there's an equal negative charge at -a. He begins by giving the sum of the fields generated by each charge as

[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{1}{(x-a)^2}-\frac{1}{(x+a)^2} \right ).[/tex]

Am I right in thinking that this formula is only valid when x > a? In that case, the first term will be positive and the second negative, as I think they should be. But if x < -a, then, E-(x,0,0), the component of the electric field vector at (x,0,0) which is due only to the negative charge should point in the positive x direction (attracting a positive test charge), while E+(x,0,0), the component due only to the positive charge will point in the negative x direction (repelling a positive test charge), shouldn't it? And when (x,0,0) is between the two charges, I think, both components will point in the negative x direction.

I began like this, without making any assumptions as to where on the x-axis (x,0,0) is:

[tex]\frac{q \mathbf{i}}{4\pi\epsilon_0}\left ( \frac{x-a}{\left | x-a \right |^3}-\frac{x+a}{\left | x+a \right |^3} \right ).[/tex]

His formula simplifies to

[tex]\frac{qax\mathbf{i}}{\pi\epsilon_0(x^2-a^2)^2}.[/tex]

I guess if a similar simplification is possible in the more general case, the three special cases of where (x,0,0) can be in relation to (a,0,0) and (-a,0,0) need to be checked.
 
Last edited by a moderator:

Answers and Replies

  • #2
jtbell
Mentor
15,663
3,733
Am I right in thinking that this formula is only valid when x > a?
Yes. In practice, this is the case that we are generally interested in. Usually we end up considering situations in which a is very small, we look at the field only in the region surrounding the dipole (and "outside" of it) and we speak only of the dipole moment p = 2qa.
 
  • #3
1,384
2
Thanks, jtbell!
 

Related Threads on Electric dipole

  • Last Post
Replies
2
Views
659
Replies
2
Views
640
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
7
Views
6K
Replies
2
Views
465
  • Last Post
Replies
1
Views
2K
Replies
4
Views
5K
Replies
2
Views
10K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
3
Views
2K
Top