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Homework Help: Dipole dipole interactions

  1. Jul 11, 2013 #1
    Hi everyone,

    we haven't discussed this topic in our lecture but this task could be in the final exam so I'd need some help with it.

    1. The problem statement, all variables and given/known data

    Two dipoles with the dipole moment p are given. The distance between these two is d.
    Find the potential aswell as the electric field from one dipol at the place of the other.

    2. Relevant equations


    3. The attempt at a solution

    I have no idea how to do this. We don't have further information such as one dipole has a fixed position or something like that. Can anyone give me a hint on how to solve this?

    Thanks for your help

    I found [tex] \Phi= \frac {1}{4 \pi \epsilon_0} \frac { \vec p \vec r}{r^3} [/tex]

    if r is the distance d with x y z then this is already my solution? But I don't think that's right. Because I think the interaction with the second dipole is missing somehow.
    Last edited: Jul 11, 2013
  2. jcsd
  3. Jul 11, 2013 #2


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    The dipole field of one dipole, at the position of the other dipole, depends on the first dipole only. You got an expression for the potential - that is fine, you can use it.
    Now you just need the field. Hint: How are potential and field related?
  4. Jul 11, 2013 #3
    Notice that the answer depents not only on p and d but rather on [itex]\vec{p} \,and\, \vec{d}[/itex]
  5. Jul 11, 2013 #4
    Thanks for the answer.

    How can I imagine the fact that the potential from the one dipole is independent from the other? I guess I can see it mathematically but how do I interpret it?

    One more question, does the vector r in my formula for the potential represent the distance d?


    to answer your last question: [tex] \vec E(\vec r) = - \nabla \Phi(\vec r)[/tex]
  6. Jul 11, 2013 #5


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    The question basically asks "what is the field at point x, if the dipole at point x is not there". Expressed in that way, I don't see anything worth an interpretation.

    d as vector, right.

  7. Jul 12, 2013 #6
    Ok I think I got it now, but I have further problems. Next task says I have to calculate the moment of force from one dipole on another. Furthermore I have to find the equilibrium.

    my electric field looks like

    [tex] \vec E(\vec d)= \frac{1}{4 \pi \epsilon_0} [ \frac {3 \vec p \vec d}{d^5} \vec d - \frac {\vec p}{d^3}][/tex]

    wiki told me I can write it aswell as:

    [tex] \vec E(\vec d)= \frac{1}{4 \pi \epsilon_0} \frac {p}{r^3} [2 cos \theta - sin\theta ][/tex]

    I need some help how to get to that term I know the definition of the inner product betwenn two vectors but I don't know where that sinus came from.

    formula for the moment of force [tex] \vec M = \vec p \times \vec E = p \cdot E \cdot cos \theta [/tex]

    and to find the equilibrium M=0 and I get that the angle has to be arctan(2).

    Can anyone approve this or is this wrong?

    Thanks for your help
  8. Jul 12, 2013 #7


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    yep. that's right. nice work. Although, you should use the \cdot symbol so that people know that you are taking the dot product.

    There should also be some unit vectors next to the sine and cosine. That might be why you are confused. (hint - you can get to this equation by using dot product with suitable unit vectors on your equation for the electric field).
  9. Jul 13, 2013 #8
    thanks for your answer.

    I actually just forgot to write down the unit vectors (I copied it from wiki), but I tried it with the inner product, but I don't know where the sine comes from.

    I get [tex] \vec p \cdot \vec r = r \cdot p \cdot cos \theta[/tex]

    but on the right side I just have vector p (the second term) I tried to add a unitvector for r but that didn't help me much.

    edit: wow, nevermind, I was so stupid and used cylindrical coordinates instead of spherical. Now I got it.

    But one problem is still unsolved: I have to find the equilibrium. Is my solution right for theta=arctan(2)?
    Last edited: Jul 13, 2013
  10. Jul 13, 2013 #9


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    uh... I don't know why your solution is theta=arctan(2) ?
  11. Jul 14, 2013 #10
    I think that solution is wrong. my attempt was to set M=0 (because I think thats the condition when the task says "equilibrium") and then find theta.
  12. Jul 14, 2013 #11


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    yes, M=0 is what you should look for. And remember that [itex]\vec{E}[/itex] is the electric field due to the first dipole, at the position of the second dipole.
  13. Jul 14, 2013 #12
    thanks for your reply.

    I used:

    [tex] \vec E(\vec{r})

    = \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} \left( 2 \cos(\theta) \cdot \hat{r} + \sin(\theta)\cdot \hat{\theta} \right)[/tex]


    [tex] \vec M = \vec p \times \frac{1}{4 \pi \varepsilon_0} \frac{p}{r^3} \left( 2 \cos(\theta) \cdot \hat{r} + \sin(\theta)\cdot \hat{\theta} \right)[/tex]

    and then [tex] M=p \cdot E \cdot sin \theta [/tex]

    is that right?
  14. Jul 14, 2013 #13


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    I agree with this. (let's call it equation 1)

    But I don't see where this comes from. Oh I understand you now, here you mean theta is the angle between the electric field and the dipole moment. But you should be careful. This theta is not necessarily the same as the theta you used in equation 1. That theta was the angle between the dipole moment and the displacement vector between the two dipoles. It might be best to use two different symbols. Maybe phi for one and theta for the other.

    Anyway, going back to equation 1. It is correct. But I think you were closer to the answer by using the equation you wrote a while back:
    [tex]\vec E(\vec d)= \frac{1}{4 \pi \epsilon_0} [ \frac {3 \vec p \vec d}{d^5} \vec d - \frac {\vec p}{d^3}][/tex]
    From here, try to solve for [itex]\vec{M} = \vec{p} \times \vec{E} = 0 [/itex]
    Last edited: Jul 14, 2013
  15. Jul 14, 2013 #14


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    Isn't all this the wrong answer? You are supposed to calculate the potential energy of one dipole in the field of the other when both dipoles are separated by a distance [itex]\vec{d}[/itex].

    Let be the first dipole located at the origin of the coordinate system. Then it's electric potential is (in Heaviside-Lorentz units)
    [tex]\Phi_1(\vec{r})=\frac{\vec{r} \cdot \vec{p}_1}{4 \pi r^3}.[/tex]
    Now the potential energy of an arbitrary charge distribution [itex]\rho(\vec{r})[/itex] is given by
    [tex]V=\int \mathrm{d}^3 \vec{r} \rho(\vec{r}) \Phi_1(\vec{r}).[/tex]
    For a dipole at [itex]\vec{r}=\vec{d}[/itex] you have to set the charge distribution to
    [tex]\rho(\vec{r})=-\vec{p}_2 \cdot \vec{\nabla} \delta^{(3)}(\vec{r}-\vec{d}).[/tex]
    Now it's easy to prove that the dipole-dipole potential is
    [tex]V=-\vec{p}_2 \cdot \vec{E}_1(\vec{r}_0),[/tex]
    where [itex]\vec{E}_1=-\vec{\nabla} \Phi_1[/itex] is the electric field of the first dipole. Just write this out to find the dipole-dipole potential.
  16. Jul 14, 2013 #15


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    Lindsayyyy just had to find the potential. Not the potential energy. And now she/he? needs to calculate the moment.
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