1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force between electric dipoles along the dipole axis

  1. Oct 3, 2016 #1
    1. The problem statement, all variables and given/known data
    Compute the force between two identical dipoles. See Problem P63 on page 544 to set up the problem. Explain why this result makes sense by comparing it to the force between two point charges and the force between a point charge and a dipole, in terms of the relationship to the distance between the two objects.
    P64 shows the two dipoles that are parallel and in the same orientation, separated by a distance r from their centers and have a distance of s between them. The charges that make up the dipole are -q and +q.
    2. Relevant equations
    E(dipole)=2kqs/r^3

    3. The attempt at a solution
    I thought I would calculate the force on each charge of the dipole on the right by using the equation above and F=Eq. Then I would add the two. For r, I'm confused about weather I should use (r+s/2) and (r-s/2).When I tried this, I couldn't find a way to simplify to the desired answer which I know is 6kq^2s^2/r^4.
    Thanks so much for any help in advance.
     
  2. jcsd
  3. Oct 3, 2016 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Sounds good.
    I think you have a good approach. The "binomial approximation" might be of use.
    https://en.wikipedia.org/wiki/Binomial_approximation
     
  4. Oct 3, 2016 #3
    My teacher just told me to do it differently by calculating four forces using coulomb's law. How do binomial approximations work, I know how to do regular binomial expansions.
     
  5. Oct 3, 2016 #4

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Doing it this way, I think you will need to go beyond the binomial expansion (i.e., beyond first order in the small quantity s/r).

    Let's see your expressions for the forces on each charge of the dipole on the right.
     
  6. Oct 3, 2016 #5
     
  7. Oct 3, 2016 #6
    WIN_20161003_21_23_22_Pro.jpg
     
  8. Oct 3, 2016 #7

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK, this looks correct for the net force on the dipole. You need to approximate this expression using the condition that s << r; or, equivalently, s/r << 1. Can you manipulate each of the last two terms in F so that they are expressed in terms of the small quantity s/r?
     
    Last edited: Oct 3, 2016
  9. Oct 3, 2016 #8
    WIN_20161003_22_07_50_Pro.jpg
     
  10. Oct 3, 2016 #9
    made an error, WIN_20161003_22_10_34_Pro.jpg
     
  11. Oct 3, 2016 #10

    TSny

    User Avatar
    Homework Helper
    Gold Member

    OK. Looks good. Do you know how to approximate ##\frac{1}{(1+s/r)^2}##?
     
  12. Oct 3, 2016 #11
    unfortunately not, does it have something to do with a McLaurin expansion? I couldn't find any clear explanation online.
     
  13. Oct 3, 2016 #12

    TSny

    User Avatar
    Homework Helper
    Gold Member

  14. Oct 3, 2016 #13
  15. Oct 3, 2016 #14

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You're almost there. You have a sign error in the first equation of the second picture. The expansion of ##\frac{1}{(1+s/r)^2}## does not equal the expansion of ##\frac{1}{(1-s/r)^2}##.

    I don't quite follow your cancellations in the second equation of the second picture.
     
  16. Oct 3, 2016 #15
    I'll fix the signs :D. I thought that since s is much smaller than r, the terms with s In them will be negligible.
     
  17. Oct 3, 2016 #16
    Okay so I fixed the signs and just expanded and found that it simplified to the right answer XD XD XD, thank you so soooo soooo much
     
  18. Oct 3, 2016 #17

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Good work!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Force between electric dipoles along the dipole axis
Loading...