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Electric potential graph analysis

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    o8xszr.jpg


    2. Relevant equations

    ΔU = -q∫E * ds

    3. The attempt at a solution

    Just by looking at the graph I am thinking, if I solve for E using the equation above, the number would be always negative therefore I can conclude that the electric field at any x position would be negative and the graph will look the same it will be just inverted. (-y values)

    I don't know if this is the correct way to look at the problem or if I have to use energy conservation but any help would be much appreciated.

    Thanks!
     
    Last edited: Oct 2, 2013
  2. jcsd
  3. Oct 2, 2013 #2
    Ex = -dV/dx =negative of the slope of V-x graph.

    What is the slope of graph between 0 and 1,1 and 3,3 and 4 ?
     
  4. Oct 2, 2013 #3
    between 0 and 1, the slope is positive, 1 and 3 the slope is zero, 3 and 4 the slope is negative.

    So if I use that equation you posted, between 0 and 1, the electric field will be negative, between 1 and 3 the electric field remains negative and then when x is between 3 and 4 the slope will be positive?
     
  5. Oct 2, 2013 #4
    Recheck your reasoning
     
  6. Oct 2, 2013 #5
    Oh, between 1 and 3 electric field is zero. 0 and 1 V slope is positive therefore efield is negative, 3 and 4 v slope is negative therefore efield is positive.
     
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