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Yet another electric feild problem

  1. Sep 6, 2004 #1
    This one I'm having a bit of conceptual difficulty with.
    Problem is: three positive point charges, q, are placed in an equal lateral triangle, charges being the vertex's. Length of one side is a. Calculate the electric feild at point P, which is also the the location of the topmost charge. Base of triangle is parrallel to page.

    Hoping everyone understands that description, this is where I'm having problems.

    eq: E = 2 * ((k * q)/(r^2)) * (unit vector, r)

    My natural inclanation is to sum up the E's and treat it like a gravitation problem, however, I don't grasp the effect of having a charge ontop of point P will have on the electric field. Do I really just simply sum up the components of the fields from each charge?
     
  2. jcsd
  3. Sep 6, 2004 #2
    Dink,
    I thought that the electric field should be infinite at point P, because P is the charge itself, and then u can neglect the other fields caused by the other charges, but I hope you will get more posts so as to correct me...
     
  4. Sep 6, 2004 #3
    The question actually asks: What is the magnitude and direction of the field at P due to the two charges at the base? Does that mean to neglect the top charge?
     
  5. Sep 6, 2004 #4

    Doc Al

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    Staff: Mentor

    Yes.
    Find the field at P by finding the contribution from each of the two base charges and adding them. Remember that electric field is a vector and must be treated as such.
     
  6. Sep 6, 2004 #5
    yes, that means to neglect the top charge, because they said find the electric field due to the other 2 charges. However if they didn't say due to the 2 charges, just find the electric field at the charge itself, it would have been infinite.
     
  7. Sep 6, 2004 #6
    Thanks for your replies guys, with this simple elightenment begins the blossoming of understanding, and also 5 more homework points. :D
     
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