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Where does a field line meet the surface of the conductor?

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data
    In the field of a point charge over a plane, if you follow a field line that starts at the point charge in a horizontal direction, that is, parallel to the plane, where does it meet the surface of the conductor?

    2. Relevant equations
    The problem 'hint' is "You'll need Gauss' law and a simple integration."

    3. The attempt at a solution
    The electric field on the surface of the conductor at a radius R=[tex]\sqrt{r^{2}+h^{2}}[/tex] (h is the height of the pt charge, r is the x component of the radius on the plane), the Electric field due to the point charge is:
    E=[tex]\frac{-2Qh}{(r^{2}+h^{2})^{3/2}}[/tex]. This is given in the book.
    I have no idea where to start...
     
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 12, 2008 #2

    Gokul43201

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    Have you quoted the question exactly as it was given to you? It appears somewhat poorly written. Is this supposed to be understood in the context of a previous problem, for instance?

    Specifically, is the point charge located near an infinite, grounded conducting plane?

    Please write down the original question exactly as provided.
     
  4. Oct 13, 2008 #3
    the plane is grounded and i figured out that the net electric field is directed towards negative Z direction. I have to find out the distance of a point in the plane (x-axis). I thought of using a concept of projectile motion as the trajectory looks parabolic, but I am not quite sure how to get to the final conclusion.
     
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