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Force on Electron Due to Two Charged Objects in an Electric Field

  1. Feb 14, 2014 #1
    1. The problem statement, all variables and given/known data
    Two charged objects are separated by a distance of L=0.4650 m as shown in the diagram.
    Object Q1 has a charge of +5.250 nC. Object Q2 has a charge of +2.960 nC.

    A) What is the magnitude and direction of the electric field at Point A, which is located a distance d=0.09300 m to the right of Object Q1?

    B) If you were to place an electron at Point A, what would be the magnitude and direction of the force on the electron?

    2. Relevant equations
    E= k (q/r^2)

    F= (k*Q1*Q2)/ (r^2)

    3. The attempt at a solution

    A) [ (8.99E9) ((5.25E-9)/(.09300^2) ] - [ (8.99E9) ((2.96E-9)/(.372^2) ]=
    = 5265 N/C = 5.265E3 N/C to the right THIS ANSWER IS CORRECT

    B) The charge on an electron is -1.6E-19. I am also assuming that i will need Coulombs Law to solve.
    [(8.99E9)(5.25E-9)(2.96E-9)] / (.4650) =.64610E-7 to the left
    This is not the right answer and I'm not sure why. AM i even using the right formula?

    Any help would be greatly appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Feb 14, 2014 #2

    gneill

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    Staff: Mentor

    You could use Coulomb's law and work directly with the charge values (you'd need to consider all the charges involved). But since you've already determined the electric field at point A you can use a different formula which relates the force on a charge due to the electric field...
     
  4. Feb 14, 2014 #3
    The only equation I have that relates force and charge is E= F/q
    (5.265E3)= F/(-1.6E-19)
    (5.265E3)(-1.6E-19)= F
    F= 8.42E-16 N to the left

    Was that the equation you were talking about?
     
  5. Feb 14, 2014 #4

    gneill

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    Staff: Mentor

    Yup. Most often you'll see it in the arrangement F = qE. It's analogous to the formula for force due to a gravitational field: F = mg. Makes it easy to remember :smile:
     
  6. Feb 14, 2014 #5
    Oh, OK! That makes sense. Thank you for your help!
     
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