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Midpoint of an electric field

  1. Sep 21, 2013 #1
    1. The problem statement, all variables and given/known data
    What is the electric field at a point midway between a -8.50μC and a +6.38μC charge 7.79 cm apart? Take the direction towards the positive charge to be positive. Assume no other charges are nearby.


    2. Relevant equations
    E(net) = E(1) + E(2)

    E = (k*q)/r^2


    3. The attempt at a solution
    E(1) = (8.99 x10^-6 * 6.38x10^-6)/ (0.03895)^2
    =3.78x10^7

    E(2) = (8.99x10^-6 * -8.5x10^-6)/ (0.03895)^2
    = -5.037x10^7

    E(net) = 3.78x10^7 - 5.037x10^7
    =-1.257x10^7 [N/C]

    This is not one of the answer choices, and I am not sure what I'm doing wrong. I thought I understood how to do the problem, so it might just be a complete misunderstanding of what the question is asking? I'm not sure...
     
  2. jcsd
  3. Sep 21, 2013 #2
    Oh, I added the two final numbers together:

    3.78x10^7 + 5.037x10^7 = 8.82x10^7

    and got the right answer. But I don't understand - why would you add the negative charge?
     
  4. Sep 21, 2013 #3

    haruspex

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    A negative charge on the left exerts a field from right to left. A positive charge on the right does likewise, so the field strengths add.
     
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