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Horizontal field line

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  1. Sep 15, 2016 #1
    1. The problem statement, all variables and given/known data
    rOgE8.png
    cSlqo.png

    2. Relevant equations
    The z component of the field:

    $$E_z = \frac{-Qh}{2\pi\varepsilon_0 (r^2+h^2)^{\frac{3}{2}}}$$

    3. The attempt at a solution
    I tried to choose a cylinder for my Gaussian surface such that the radius of it matches with the distance d I am trying to find and its height is such that all field lines are within the interior of the cylinder.
    I know that the electric field in the z component in this case is:
    \begin{equation}
    \frac{-Qh}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}
    \end{equation}
    Thus, if I integrate:
    \begin{equation}
    \int \frac{-Qh}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}dA=\int_0^d \frac{-Qh(2 \pi r dr)}{2 \pi \varepsilon_o (r^2+h^2)^{\frac{3}{2}}}=\int_0^d \frac{-Qh(r dr)}{\varepsilon_o (r^2+h^2)^{\frac{3}{2}}}=\frac{Q}{\varepsilon_o}
    \end{equation}
    So:
    \begin{equation}
    \int_0^d \frac{r dr}{(r^2+h^2)^{\frac{3}{2}}}=-\frac{1}{h}
    \end{equation}
    And this is ultimately:
    \begin{equation}
    \frac{1}{h}-\frac{1}{\sqrt{h^2+d^2}}=-\frac{1}{h}
    \end{equation}
    However, if a solve for d, it yields a complex number
     
  2. jcsd
  3. Sep 15, 2016 #2

    TSny

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    Hello.

    Can you please describe in more detail your cylindrical Gaussian surface? Where are the locations of the top and bottom of the cylinder?

    I believe you will need to consider a different shape of Gaussian surface for this problem.
     
  4. Sep 15, 2016 #3
    Hello, I tried to do something like this
    http://imageshack.com/a/img921/9379/gQAcim.png [Broken]
    I also thought of doing something with the image charge
    http://imageshack.com/a/img924/2415/9XBrlC.png [Broken]
    However, I am not sure what is the charge enclosed in this sphere...
     
    Last edited by a moderator: May 8, 2017
  5. Sep 15, 2016 #4

    TSny

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    If you included more electric field lines in your drawing (coming out of the upper part of the point charge), wouldn't some of those field lines go through the flat top of your cylindrical Gaussian surface, and also wouldn't some of the field lines go through the vertical (curved) side of the cylindrical surface? Thus, there would be some flux through your Gaussian surface that you have not accounted for.

    Also, how does your picture account for the fact that you are interested in the field lines that leave the point charge in a horizontal direction?
     
  6. Sep 15, 2016 #5
    I don't know. What I thought was that I can make the height of the cylinder even larger in order to cover those field lines you mention because it doesn't affect the result. And the horizontal field line starts bending in the instant it is coming out the charge, so... I really don't know.
     
  7. Sep 15, 2016 #6

    TSny

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    There will always be some field lines that go higher than any finite height you choose for the top surface of your cylinder. The field line that leaves the point charge vertically upward will go vertically upward to infinity. Those that leave the charge almost vertically upward will go very far upward before curving back down and meeting the plane very far away from r = 0. If your cylinder has a finite radius, d, then some of the field lines will pass through the vertical side of the cylinder. So, a cylindrical shape Gaussian surface doesn't appear to be very appropriate for this problem.

    Since you are interested in field lines that leave the point charge horizontally, try to picture all of the field lines that leave the charge horizontally. Can you describe roughly the geometrical shape of the surface formed by all of these lines?
     
  8. Sep 15, 2016 #7
    Could it be a paraboloid?
     
  9. Sep 15, 2016 #8

    TSny

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    Probably not a paraboloid, but it will be some sort of "dome". Can you use that dome surface as part of a closed Gaussian surface?
     
  10. Sep 15, 2016 #9
    Well, since the field lines with the image charge resembles ellipses, I could guess that it could be half part of an ellipsoid...
     
  11. Sep 15, 2016 #10

    TSny

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    The exact shape (paraboloid, ellipsoid, etc.) of the dome doesn't matter. Try to use the dome to your advantage.
     
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