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Homework Help: Components of an Electric field due to a dipole

  1. Feb 16, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is: Show that the components of [tex]\vec{E}[/tex] due to a dipole are given at distant points, by Ex=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{3pxz}{(x^2+z^2)^{5/2}}[/tex] and Ez=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{p(2z^2-x^2)}{(x^2+z^2)^(\frac{5}{2})}}[/tex]

    http://physweb.bgu.ac.il/COURSES/PHYSICS2_B/2009A/homework/Homework-2_files/image006.jpg [Broken]

    2. Relevant equations

    E=[tex]\frac{1}{4\pi\epsilon{o}}[/tex] [tex]\frac{Q}{r^2}[/tex]

    3. The attempt at a solution

    I have tried to break the fields of each one into vector components and add the components, however, it got really messy really quickly and after simplifying it a bit i got a ridiculous equation for just the x component, i had no clue where to go and gave up on even try to get the z component.

    Ex=[tex]\frac{q}{4\pi\epsilon{o}}[/tex][tex]\frac{(x^2+(z+\frac{d}{2})^{2})^{\frac{3}{2}}-(x^2+(z-\frac{d}{2})^{2})^{\frac{3}{2}}}{((x^{2}+z^{2})^{2} + (\frac{d^{2}x^{2}}{2}-\frac{d^{2}z^{2}}{2}+\frac{d^4}{16}))^{\frac{3}{2}}}[/tex]
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 17, 2010 #2


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    I haven't checked your expression, but the key thing to notice here is that you are asked to find [itex]E_x[/itex] and [itex]E_z[/itex] at distant points. To me, that means you should assume that the distance to the field point from the center of the dipole is much greater than the length of the dipole, or

    [tex]\frac{d}{\sqrt{x^2+z^2}}\ll 1[/tex]
  4. Feb 17, 2010 #3
    But how does that help?
  5. Feb 17, 2010 #4


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    Well, looking at the denominator of your expression; if you factor out an [tex](x^2+z^2)^3[/itex], you're left with [tex](1+\text{some stuff})^{3/2}[/itex]. It shouldn't take you much effort to show that the "some stuff" is much smaller than one at distant points, and so your denominator is approximately just [tex](x^2+z^2)^3[/itex].
  6. Feb 17, 2010 #5
    Ok. but my biggest problem is then simplifying the numerator, and I tried using binomial expansion for that and you get (x[tex]^{2}[/tex]+z[tex]^{2}[/tex]+dz)[tex]^{3/2}[/tex] - (x[tex]^{2}[/tex]+z[tex]^{2}[/tex]-dz)[tex]^{3/2}[/tex] and a denominator of (x[tex]^{2}[/tex] + z[tex]^{2}[/tex])[tex]^{3}[/tex]

    sorry to ask but what do you do from there?
  7. Feb 17, 2010 #6


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    To be honest, your expression just doesn't look right to me. And I think you are making the calculations more difficult than they need to be by working in Cartesian coordinates.

    If I were you, I'd redo your calculation using polar coordinates, [itex]x=r\cos\theta[/itex] and [itex]z=r\sin\theta[/itex]. The advantage of this is that your "at distant points" approximation simply means that [itex]\frac{d}{r}\ll 1[/itex], and so writing your expression in terms of the dimensionless variable [itex]\eta\equiv\frac{d}{r}[/itex], you'll only need to Taylor expand it around the point [itex]\eta=0[/itex] and keep terms up to first order.
  8. Feb 18, 2010 #7
    actually, doing the whole thing from the electric field is ridiculously stupid. I realized I should start it in Volts so it looks like this

    V = [tex]\frac{q}{4\pi\epsilon}[/tex] ([tex]\frac{1}{r-\frac{l}{2}cos\vartheta}[/tex]-[tex]\frac{1}{r+\frac{l}{2}cos\vartheta}[/tex]) which then simplifies to V=[tex]\frac{q}{4\pi\epsilon}[/tex][tex]\frac{lcos\vartheta}{r^2}[/tex]


    then -[tex]\nabla[/tex]V=E so then you just take the partial derivatives of it in terms of x and z.

    I feel stupid for not thinking of it sooner
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