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Charge on a point in three different locations with a thin semicircular rod

  1. Jan 22, 2013 #1
    1. The problem statement, all variables and given/known data
    A thin semicircular rod is broken into two halves, the top half has a total charge +Q uniformly distributed along it, and the bottom half has a total charge -Q uniformly distributed along it.
    http://imgur.com/tObt2V0

    2. Relevant equations
    Indicate the direction of the net electric force on a positive test charge placed in turn at points A, B, C.


    3. The attempt at a solution
    Had A's force vector pointing down and to the right, same with b and C pointing down and to the left but I am not sure about my answers.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 22, 2013 #2

    haruspex

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    Think about the forces in the horizontal and vertical directions separately. E.g. for A, compare the horizontal forces (i.e. along the line AB) exerted by the two quadrants.
     
  4. Jan 22, 2013 #3
    I can't really visualize it but my guess is that the horizontal component of +Q would be to the left and the -Q would have a horizontal component equal and opposite so to the right. The vertical component of +Q on the charges would be down, same with the -Q charge.
     
  5. Jan 22, 2013 #4

    haruspex

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    Exactly so. So the resultant force is in which direction?
     
  6. Jan 22, 2013 #5
    Down and to the right. Would it be the same for both A and B. And C would be down and to the right?
     
  7. Jan 22, 2013 #6
    or is the resultant straight down?
     
  8. Jan 22, 2013 #7

    fgb

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    If the horizontal components are opposite (and equal in magnitude, since all distances are the same for the + and for the - quadrant), don't they cancel each other?
     
  9. Jan 22, 2013 #8
    yeah thats what im thinking
     
  10. Jan 22, 2013 #9

    fgb

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    Yeah, so the resultant is straight down, like you said :)
     
  11. Jan 22, 2013 #10
    So all of the points; A,B, and C would have the same vector pointing down?
     
  12. Jan 22, 2013 #11

    fgb

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    The magnitude of said vector would change according to distance from the point to the quadrants (i.e., the vector is smaller for point A than for point B), but it is indeed straight down for three points :)
     
  13. Jan 22, 2013 #12
    Thank you very much.
     
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