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Electrostatics: Moving Charge (comprehension problem)

  1. Aug 29, 2009 #1
    I already know what the answer to this problem should be, however I do not understand why that is the case.

    1. The problem statement, all variables and given/known data
    The negative charge (= -1 micro Coulombs) in the figure below goes from y = -5 to y = 5 and is made to follow the dashed line in the vicinity of two equal positive charges (= +5C). What is the work required to move the negative charge along the dashed line?


    2. Relevant equations
    Vb - Va = Wab/q0
    V = kq/r
    F = qE
    E = kq/r2
    W = Fd cosine theta

    3. The attempt at a solution
    Below is a representation of the arrangement of the charges. The horizontal dashes (-) are just space fillers and can be ignored, the vertical ones represent the negative charge's path.

    ------------y=5
    -------------|
    -------------|
    -------------|
    ----+5C----y=0----+5C
    -------------|
    -------------|
    -------------|
    ------------y=-5

    I tried solving it on my own by taking the net displacement as 10 and coming to the conclusion that there would be work. Then I checked the answer and it said there is no net work because the force changes as the charge passes the origin. I don't understand why that is and was hoping someone could explain it to me. The only thing I could think of was that as the charge approaches the origin it is moving toward the positive charges (natural attraction) and as it moves past them it is moving away from the positive charges (against attraction).
     
  2. jcsd
  3. Aug 29, 2009 #2

    kuruman

    User Avatar
    Homework Helper
    Gold Member

    Look at your first equation. What is Vb - Va?
    Another way of seeing it is that the electric force is attractive, but changes sign when the charge crosses the origin. Because of the symmetry, this force does positive work over the top half of the path and negative work of equal absolute value over the bottom half of the path so that the net work is zero. Your thinking along these lines is OK.
     
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