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Energy from a dipole

  1. Feb 21, 2016 #1
    1. The problem statement, all variables and given/known data

    An electric dipole with charge ±q is separated by distance d. This dipole is enclosed in a spherical space of radius r = a such that the center of the dipole is located at the origin and the entire dipole in encased in the space. In other words the charges are at ±d/2 and a > d/2 Find the total energy everywhere outside the sphere.

    2. Relevant equations

    WE = ∫∫∫ 1/2 ε0 E2 dv

    Edipole = Qd cosθ ar / (2 π ε0 r3) + Qd sinθ aθ / (4 π ε0 r3)

    Where ar and aθ are unit vectors.

    3. The attempt at a solution

    I know I must do the above integral twice. I need to first do this to ∞ and again with the spherical space. My answer will be the difference between the two. I am having issue because this was something that is not in the text. We only covered briefly in lecture.

    I think I have a problem with my approach. I first want to find the energy to infinity. So I do this

    ε0/2 ∫∫∫ Qd cosθ ar / (2 π ε0 r3) + Qd sinθ aθ / (4 π ε0 r3) dr from r=0 to ∞ dθ from θ=0 to π dφ from φ=0 to 2π

    The result of this blows up due to the 0 in the limits of the first integral. Not sure what to do about this. I suppose I could set my limits from r = a to infinity. I guess this would result in my answer provided my equations listed in the equation section are valid. This leaves a huge question in my understanding however. What if I were asked for the energy in just the sphere? Then my limits would have to have the 0.
     
    Last edited: Feb 21, 2016
  2. jcsd
  3. Feb 21, 2016 #2
    i think first you should calculate field E due to a dipole at any point (r,theta) , and go ahead with integration for the field energy as your expression seems to be 'integration of energy density over whole space - let us see !
     
  4. Feb 21, 2016 #3
    Isn't the expression Qd cosθ ar / (2 π ε0 r3) + Qd sinθ aθ / (4 π ε0 r3) valid for the E field at any r or θ in spherical coordinate system?
     
  5. Feb 21, 2016 #4

    collinsmark

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    I think you've stumbled upon a curious and perhaps "embarrassing" (David J. Griffiths calls it "embarrassing" in his book, Introduction to Electrodynamics) limitation in our (humans) current understanding of electrostatics.

    It's not limited to dipoles either. Ask yourself instead, how much energy does it take to make a single point particle like an electron? [integrate [itex] \frac{1}{2} \varepsilon_0 \left( \frac{q}{4 \pi \varepsilon_0 r^2} \right)^2 [/itex] over all space] You'll also find that even that takes an infinite amount of energy.
     
  6. Feb 21, 2016 #5
    i can not confirm as the expression looks similar but you have magnitude of E as its E^2 in the integrand so take care otherwise your calculation will get affected- i remember some factor sqrt(1 + 3 cos^2 theta) in the numerator - try to do full calculation
     
  7. Feb 21, 2016 #6
    Good point. That goes to infinity as well. This is probably why my professor asked us to do this problem. It is just something he scribbled on the board as we were packing up.
     
  8. Feb 21, 2016 #7
    Right. I left the ^2 off as a simple copy and paste error. This still does not change the fact that I will have division by zero once the integration is carried out.
     
  9. Feb 21, 2016 #8

    collinsmark

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    Yeah, I don't think you can get around that.

    That's probably why the question only asked you to find the energy outside of the sphere. You'll end up with nonsensical answers inside.
     
  10. Feb 21, 2016 #9
    so a good 'learning exercise' given by your teacher - now it will never leave your 'thumb rule book'.
     
  11. Feb 21, 2016 #10
    So does this equation look ok then;

    ε0/2 ∫∫∫ [ Qd cosθ ar / (2 π ε0 r3) + Qd sinθ aθ / (4 π ε0 r3) ] 2 dr from r=a to ∞ dθ from θ=0 to π dφ from φ=0 to 2π
     
  12. Feb 21, 2016 #11
    indeed
     
  13. Feb 21, 2016 #12

    collinsmark

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    Assuming that [itex] \vec E = \frac{Qd}{2 \pi \epsilon_0 r^3}\cos \theta \hat{a_r} + \frac{Qd}{4 \pi \epsilon_0 r^3}\sin \theta \hat{a_\theta} [/itex] (which I haven't verified), then that looks about right.

    You can distribute the squaring operation into the individual components by invoking the Pythagorean theorem.

    [Edit: Oh, and your differentials shouldn't be just [itex] dr [/itex], but rather something a bit more complicated noting that [itex] d \vec r = (dr) \hat {a_r} + (r d \theta) \hat {a_\theta} + (r \sin \theta d \phi) \hat {a_\phi} [/itex], if that's not what you already meant by "dr".]
     
    Last edited: Feb 21, 2016
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