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Homework Help: Dipolemoment of charged beam

  1. Sep 17, 2010 #1
    1. The problem statement, all variables and given/known data
    We have a non-conducting beam with length L and uniform chargedistribution [tex]\lambda[/tex]. Half the beam is negativly charged and the other half positive.

    The problem is to find the dipole moment of the beam.

    I don't know how to proceed for a continous distribution like this, but I suppose it
    is some kind of integral.

    I found this integral at wikipedia, but I don't get the logic and what it reduces to in the
    one-dimentional case. It seems like the dipolemoment is a function of r, but should'nt this be a property of the chargedistrobution alone?

    [tex]\boldsymbol{p}(\boldsymbol{r}) = \int_{V} \rho(\boldsymbol{r_0})\, (\boldsymbol{r_0}-\boldsymbol{r}) \ d^3 \boldsymbol{r_0}[/tex]
     
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  3. Sep 17, 2010 #2

    kuruman

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    Think of ρ(r0)d3r0 as a charge element dq. In one dimension, this becomes λdx0. Does this help?
     
  4. Sep 17, 2010 #3
    humm..
    then the formula reduces to [tex]\int_L \lambda (x_0 - x) dx_0[/tex]. That is all good.
    But what is the intuition behind it? What is x0 and how do one proceed with the calculation?

    How can one think about the dipole moment of a continous charge distrobution like this?
    I have never seen the dipole moment be defined for more then a 3-charge system.
    One example is the H20 molecule. Then we can define two dipolemoments from
    the oxygen atom to each of the hydrogen atoms and the total dipole moment is the
    sum of the individual ones..

    How do this relate to a continous distribution in the problem above and to a system
    of say several positive and negative charges?
     
  5. Sep 17, 2010 #4

    kuruman

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    You can think of this distribution as a continuous distribution of pairs of charges separated by some distance. Say you put two point charges ±dq at ±L/2, that's a dipole; now put another two point charges +dq at (L/2)-ε and -(L/2)+ε where ε is a small distance, that's another dipole. Keeping doing this until you end up at the origin. You end up with a distribution of dipoles and you are looking for the net field produced by all of them in the limit that you smear the positive and negative charges into a continuous distribution. Forget the water molecule. Imagine charging a rod of length L/2 with charge +q and another identical rod with charge -q and butting their end together. That's the visualization of this distribution.

    Variable x0 integrates over the length of the rod. You need to do two integrals, one with x0 running from -L/2 to zero and negative λ and one with x0 running from zero to +L/2 and positive λ.
     
  6. Sep 17, 2010 #5
    So one charge dq only gets paired up with one charge -dq at the opposite end? It does not get paired up with several others as in the H2O? I think I would get the intuition behind the formula if i could derive it from a sum.. Can you help me with that?

    I wont forget the H2O molecule just yet. Originaly I tought that, because of the case with the molecule, you had to pair each positive charge up with all of the negative ones.. as we did in the case of the H2O. Then I got to a sum like
    [tex] \sum_i q_i \sum_j (x_i - x_j)[/tex] where charge number i get paired up with each other negative charge j and multiplied with the distance between them.

    Is what you're saying that the dipole moment is really a vector field and not just one
    vector? Isn't the dipole moment of some thing a property of that thing alone and not dependent on where you are in space?

    I did the calculation btw, but it's late now here in Norway, so I might have gotten it wrong. I got

    [tex] p = -\lambda\frac{L^2}{4} = -\frac{QL}{4} [/tex]

    The x's canceled in my calculation. What is the meaning of x relative to x0?
     
  7. Sep 18, 2010 #6

    kuruman

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    The correct discrete expression derived from the integral expression you originally posted should be

    [tex]\vec{p}(\vec{r})=\sum_{j}q_j(\vec{r_{0j}}-\vec{r})[/tex]
    It is just one vector but it generates a dipole field in space.
    No. The general expression for the dipole moment says that it depends on your choice of origin with respect to which you calculate it. In the expression that I posted above, there are two "origins". One is, say O, and one is O'. Origin O' is positioned at r with respect to O and vector r0 is a vector from O' to charge qj in the distribution. You can clearly see that the dipole moment depends on your choice of origin in general. Most definitions of dipole moment that I have seen set r = 0, i.e. O and O' are one and the same. Nevertheless, the dipole moment still depends on your choice of origin given a charge distribution in space.

    A notable exception is two equal and opposite charges separated by some distance d. In this case you have what is also known as a "pure" dipole, one that does not depend on your choice of origin. Note that the particular problem that you posted can be viewed as a continuous distribution of pure dipoles, so the final answer should not depend on your choice of origin. In that case, you might as well line up your beam on the x-axis, with the positive charges on the positive side and the negative charges on the negative side and do the easy integral.
    I think the negative sign should not be there, check it. The x's canceled for the reason I just posted: you have a pure dipole that should not depend on x.
     
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