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Electric field on point charge

  1. Sep 26, 2008 #1
    1. The problem statement, all variables and given/known data

    what is the electric field at point A (0,12) if there is charge c of 8*10^-9 coulombs at (16,0) and a charge d of 6*10^-9 coulombs at (-9,0)
    determine x and y components of electric field (see diagram)

    2. Relevant equations

    electric field e = kq/r^2 where k is 9*10^9, q is charge, r is distance from point

    3. The attempt at a solution

    x components:
    e_c = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ]sin(phi)
    = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ] * [ 16 /(sqrt(16^2 + 12^2))]
    = [(9*10^9)(8*10^-9)(16)] / [(16^2 + 12^2)^(3/2)]
    = 0.144 newtons/coulomb

    e_d = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ]sin(theta)
    = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ] * [ 9 /(sqrt(9^2 + 12^2))]
    = [(9*10^9)(6*10^-9)(9)] / [(9^2 + 12^2)^(3/2)]
    = 0.144 newtons/coulomb

    y components:
    e_c = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ]cos(phi)
    = [(9*10^9)(8*10^-9)/(sqrt(16^2 + 12^2)^2 ] * [ 12 /(sqrt(16^2 + 12^2))]
    = [(9*10^9)(8*10^-9)(12)] / [(16^2 + 12^2)^(3/2)]
    = 0.108 newtons/coulomb

    e_d = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ]cos(theta)
    = [(9*10^9)(6*10^-9)/(sqrt(9^2 + 12^2)^2 ] * [ 12 /(sqrt(9^2 + 12^2))]
    = [(9*10^9)(6*10^-9)(12)] / [(9^2 + 12^2)^(3/2)]
    = 0.192 newtons/coulomb

    net electric force components:
    x = 0.144+0.144 = 0.288 newtons/coulomb
    y = 0.108+0.192 = 0.300 newtons/coulomb

    my net components are wrong. why?
     

    Attached Files:

  2. jcsd
  3. Sep 26, 2008 #2

    Nabeshin

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    Science Advisor

    Your work is a little strange here in that you first find the vectors, and then component-ize them. I can't see your picture so I'm thinking you might have gotten an angle wrong.

    You could try simply solving for the compoents without the general vector by saying in the y component both charges are 12 units away and in the x they are 16 and 9 respectively. This should help to minimize mistakes, just make sure to keep track of signs.
     
  4. Sep 27, 2008 #3
    i don't think i got the angles wrong, if you look at the diagram, is the way i calculated them the wrong way? will it affect the final answer?
     
  5. Sep 27, 2008 #4

    hage567

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    Homework Helper

    Are both your components wrong?

    One thing I notice is that you added your x components. I think the x component due to charge C and the x component due to charge D are NOT pointing in the same direction. I don't know if that makes it right but I just thought I'd mention it so you could check it out.
     
  6. Sep 27, 2008 #5
    it seems both my net components are wrong. isn't the problem asking for the net components, therefore wouldn't i just add them regardless of direction?
     
  7. Sep 27, 2008 #6

    hage567

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    Homework Helper

    They are vectors, though. You are trying to find the net electric field at that point due to both charges. If the x component vectors act in opposite directions, you subtract them, they don't add up.
     
  8. Sep 28, 2008 #7
    so i should be subtracting vectors then? how do i know which vector component is the positive one, is it based on magnitude?
     
  9. Sep 28, 2008 #8
    using the vectors at point A -going upward (use A as the origin), the vector e_c is in the negative x-direction and positive y-direction. e_d is in the positive x, positive y. change your signs according to the directions, and that should change your answer for e_net(x). your net electric field in the y-direction is right.
     
  10. Sep 28, 2008 #9
    thanks for clarifying how to get the directions, i didn't even think to look at it that way, really cleared up things for me

    so if i change the direction (signs) of my vectors and re-sum them i should get the answer i am looking for? are my original calculations, ignoring direction, correct?

    so:

    x components:
    e_c = -0.144 newtons/coulomb
    e_d = + 0.144 newtons/coulomb

    y components:
    e_c = + 0.108 newtons/coulomb
    e_d = + 0.192 newtons/coulomb

    is that correct now?

    so net components
    x = -0.144 + 0.144 = 0 newtons/coulomb
    y = +0.108 + 0.192 = 0.300 newtons/coulomb
     
  11. Sep 28, 2008 #10
    yup, thats the right answer =)
     
  12. Sep 29, 2008 #11
    thanks it was right
     
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