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Electric Field

  1. Aug 26, 2005 #1

    My Question:

    What are the magnitude and direction of the electric field at a midway between a -8.0 microcolumb and a 7.0 microcolumb charge 8.0 cm apart? Assume no other charges are nearby.




    F=((9.0 X 10^9 N m^2/C)(-8.0 X10^-16 C)(7.0 X10^-6C))/(.08m)^2

    F=-78.75 or 78.75 N


    E=4.92 x 10^20 N/C

    Is this correct or should I look at findin the electric field using Q1 and Q2 seperately then adding them together to find a net elecrical field.

    Thank you :smile:
    Last edited: Aug 26, 2005
  2. jcsd
  3. Aug 26, 2005 #2

    Doc Al

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    Staff: Mentor

    You need to find the field at the midpoint due to each charge, then add those fields. Don't forget: Direction matters.
    Last edited: Aug 26, 2005
  4. Aug 27, 2005 #3


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    Homework Helper

    In E=F/q , the q has to be the same *test charge* that the Force acts on!
    That is, in all practical situations, the test charge q cancels.
    The part of the E-field at test location t, contributed by charge Q1, is
    E_1t = k Q1/(r_1t)^2 (away),
    where r_1t = distance from Q1 to the test location (.04m in your case).
  5. Aug 29, 2005 #4
    Oh... okay thanks :smile:
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