
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!

Difficulties with dipoles & point charges

  1. Feb 11, 2006 #1
    Difficulties with dipoles & point charges....

    Help would be appreciated a lot with the following problems:

    ~ A positive point charge +Q is at the origin, and a dipole of moment p is a distance r away (r>>L) and in the radial direction as shown below:


    a) Show that the force exerted on the dipole by the point charge is attractive and has a magnitude ~ 2kQp/r3 (see previous problem).

    [the previous problem was: an electric dipole consists of two charges +q and -q separated by a very small distance 2a. Its center is on the x axis at x=x1 and it points along the x axis in the positive x direction. The dipole is in a nonuniform electric field, which is also in the x direction, given by E=Cxi where C is a constant.

    - Find the force on the positive charge and that on the negative charge and show that the net force on the dipole is Cpi
    - Show that, in general if a dipole of moment p lies along the x axis in an electric field in the x direction the net force on the dipole is given approximately by (dEx/dx)pi].

    I know the force on the negative side of the dipole is -kQq/r2 and the force on the positve side of the dipole is kQq/(r+L)2, but this leads to nothing..... I can derive an expression for the electric field of the dipole as k2p/r3 and conclude the force of the dipole on the point charge= the force on the dipole of the point charge (Newtons 3rd law)= QE, therefore F= Qk2p/r3, but that would ruin the second question below (because I need to use the formula derived in this question in order to prove E= k2p/r3 for a dipole. So how should I tackle this problem?

    b) Now assume that the dipole is centered at the origin and that a point charge Q is a distance r away along the line of the dipole. Using Newton's third law and your result for part (a), show that at the location of the positive point charge the electric field E due to the dipole is toward the dipole and has a magnitude of ~k2p/r3.

    I think I can handle this one if I get part (a) correct

    ~ Two neutral polar molecules attract each other. Suppose that each molecule has a dipole moment p and that these dipoles are aligned along the x-axis and separated by a distance d. Derive an expression for the force of attraction in terms of p and d.

    I do need some help with this one. First of all, if a line like this - is a symbol for the dipole, are the two aligned one after another (like - -) or parallel to each other (like =)? Secondly, I don't really know how to treat a system of dipoles if the distance isn't big (in the previous question it was) so I would really appreciate a couple of hints to give me something to start off with.... I totally don't have a clue.....

    PS Sorry for the big text! :blushing:
  2. jcsd
  3. Feb 12, 2006 #2


    User Avatar
    Homework Helper
    Gold Member

    I guess you need to use a binomial approximation.

    If you take the distance to the center of the dipole as 'r', then the distances to each charge will be r - L/2 and r+L/2, so that your field will be

    [tex] \frac{1}{4\pi\epsilon_0} [\frac{-q}{(r-L/2)^2} + \frac{q}{(r+L/2)^2}] [/tex]

    Now take out an r from the denominator so that you get
    [tex] \frac{1}{r^2} \frac{-q}{(1-\frac{L}{2r})^2)} [/tex]
    since L/R is <<1, you can expand the deonimator using the binomial exapnsion and ignore higher order terms. Do this for the second term also and this should give you the answer.
  4. Feb 12, 2006 #3
    Thank you very very much! :D I managed to solve the problem!!! :)
    Yet I'm still puzzeling on this one:

  5. Feb 12, 2006 #4


    User Avatar
    Homework Helper
    Gold Member

    For the second question, I would still assume that d>>L (In fact, this assumption is valid in most situations, because the inter-molecular distances are very small).
    I know the field due to the dipole along the axis. From that, I can calculate the force on each charge of the second dipole to find the force of attraction.
    Do the Binomial trick again to get it in terms of p and d.
    Last edited: Feb 12, 2006

Similar Discussions: Difficulties with dipoles & point charges
  1. Point charges (Replies: 8)

  2. Point charges (Replies: 4)