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Field in the plane where a dipole lies

  1. Sep 2, 2015 #1
    Hello, friends! I read that, if a dipole is centred on the origin, with the ##+q>0## charge in ##a>0## and the negative ##-q<0## charge in ##-a##, the field in a point ##(x,y)## of the plane is $$\mathbf{E}=k\frac{3pxy}{(x^2+y^2)^{5/2}}\mathbf{i} +k\frac{p(2y^2-x^2)}{(x^2+y^2)^{5/2}}\mathbf{j}$$ where ##k## is Coulomb's constant.

    If I have correctly calculated, the exact electric field in ##(x,y)## should be ##\mathbf{E}=kq\Big(\frac{x}{(x^2+(y-a)^2)^{3/2}}-\frac{x}{(x^2+(y+a)^2)^{3/2}}\Big)\mathbf{i}## ##+kq\Big(\frac{y-a}{(x^2+(y-a)^2)^{3/2}}-\frac{y+a}{(x^2+(y+a)^2)^{3/2}}\Big)\mathbf{j}##, but I have got serious troubles in finding a way to see that this is desired result. Nevertheless, I suspect that an approximation for ##|x|\gg a## and ##|y|\gg a## represents the formula given by my book, but I cannot find a way to prove the approximation.

    I heartily thank you everybody for any answer.
     
    Last edited: Sep 2, 2015
  2. jcsd
  3. Sep 2, 2015 #2

    blue_leaf77

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    How is the dipole oriented in your book?
     
  4. Sep 2, 2015 #3
    From ##-q## to ##+q##, i.e. as the ##y## axis. Thank you for asking!
     
  5. Sep 2, 2015 #4

    blue_leaf77

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    The electric field due to an electric dipole at a point far away from the dipole is usually easier when computed from the potential and then take its gradient. For example look at this http://phys.columbia.edu/~nicolis/Dipole_electric_field.pdf. In this link the dipole is oriented along z direction, the equation you will be interested in is equation (13). If you express this equation in Cartesian coordinates you will obtain expression identical to the equation in your book only with ##y## replaced by ##z##. Let me know if there is something you still don't understand.
     
  6. Sep 2, 2015 #5
    Thank you for the link! I will read it when I study the concept of potential. For the moment my text has not introduced it yet... :frown:
     
  7. Sep 2, 2015 #6

    blue_leaf77

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    You can instead proceed from the exact expression which you have calculated on your own. The key is to make use of the identity
    $$
    (1+x)^p = 1+px+p(p-1)x^2+\ldots
    $$
    for |p| bigger than or equal to unity. For example let's calculate the x-component of the field, first expand the denominator so that
    $$kq\Big(\frac{x}{(x^2+(y-a)^2)^{3/2}}-\frac{x}{(x^2+(y+a)^2)^{3/2}}\Big) \approx kq\Big(\frac{x}{(x^2+y^2-2ay)^{3/2}}-\frac{x}{(x^2+y^2+2ay)^{3/2}}\Big)
    $$
    where the term quadratic in ##a## has been neglected, then factor out ##(x^2+y^2)##. You will find, for the first term,
    $$
    kq\frac{x}{(x^2+y^2)^{3/2}} \left(1-\frac{2ay}{x^2+y^2}\right)^{-3/2}
    $$
    Finally make use of the fact that ##a## is small and the identity above to approximate the term
    $$
    \left(1-\frac{2ay}{x^2+y^2}\right)^{-3/2}
    $$.
    up to the first order. Do the same way for the second term in the x component of the field.
     
  8. Sep 3, 2015 #7
    Very, very clear! Thank you so much!
     
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