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Shape of field line in an electric dipole

  1. Feb 20, 2008 #1
    1. The problem statement, all variables and given/known data

    In a xy-plane, we have a +q charge in A(0,h) and a -q charge in B(0,-h).
    Let's take a field line passing through A, that has horizontal slope in A. Find the point C(m,0) where this field line intersects the X axis.

    2. Relevant equations

    I think Gauss' law [tex]\epsilon_0 \oint \vec{E} \cdot d\vec{A}=q[/tex]


    3. The attempt at a solution

    Let's take a surface passing through A and following the field lines till the y=0 plane. Apply the Gauss' law to it: the flux is only through the base of this surface, so that we should get
    [tex]\displaystyle q=\epsilon_0 \int d\Phi = \epsilon_0 \int_0^m \frac{2khq}{(x^2+h^2)^{\frac32}} \cdot 2 \pi x dx[/tex]​
    where the second integral is the flux calculated for concentric rings around the origin. This equation, though, has no solutions for m, since it reduces to
    [tex]\displaystyle h(\frac{1}{h}-\frac{1}{\sqrt{h^2+m^2}})=1[/tex]​
    which is clearly impossible...

    Thank you all for your time, and please forgive my bad english and my even worse physics :)
     
  2. jcsd
  3. Feb 20, 2008 #2
    It's early and I think I'm missing something here but I'll get the ball rolling, I'm starting by assuming you got to the integral correctly and had the correct integrand

    First, what happened to k? Second, you're integrating over x in that integral, the only dependency on x is just, well, that single x
     
  4. Feb 20, 2008 #3
    I'll write it in a bit more detail. Let's take the flux across a ring ([tex]x[/tex] to [tex]x+dx[/tex]). The distance of every point of this ring from the two charges is the same, so the electric field must be the same on the whole ring. We get this field by multiplying by 2 the y-component of the field generated by the first charge only (this should be true by symmetry). Putting all together, the field SHOULD be [tex](2) \cdot \frac{kq}{x^2+h^2} \cdot \frac{h}{\sqrt{x^2+h^2}}[/tex], where the first 2 is to take in account the doubling I was talking about, the first fraction is the field generated by the first charge, and the second the cosine of the angle (to get only one the y-component). Then I'm multiplying all by the area of the ring, and that's all for the integrand.
    For k... that's what puzzles me, mainly. Taking what I can out of the integral, I get a [tex]\epsilon_0 4 \pi k[/tex], and everything magically vanishes...

    For the integral, I got [tex]\int \frac{x}{(x^2+h^2)^{\frac32}} = - \frac{1}{\sqrt{h^2+x^2}}[/tex], and I think this should be correct (anyways, if anybody is willing to check...)

    Just one last thing: in the integrand... there isn't only one x... I couldn't get what you meant, sorry

    Thank you very much for your answer, anyway!
     
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