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Charges on a axis

  1. Feb 3, 2012 #1
    HELP I don't understand at all

    1. The problem statement, all variables and given/known data
    Two positive point charges on the y-axis, each of which has a charge of 5.0 10-9 C, are located at y = +0.60 m and y = -0.60 m. Find the magnitude and direction of the resultant electric force acting on a charge of 7.6 10-9 C located at x = 0.80 m on the x-axis.


    2. Relevant equations

    dont know

    3. The attempt at a solution

    I have tired to many ways with to many numbers.
    My teacher hasn't taught these!
     
  2. jcsd
  3. Feb 3, 2012 #2

    SammyS

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    Re: HELP I don't understand at all

    Some teachers are like that, however, a few of my students said the same thing about me.

    Find the force on the 7.6 nC charge due to each of the other charges, individually.
     
  4. Feb 3, 2012 #3
    Re: HELP I don't understand at all

    so I do 7.6e-9*9e9/70^2
     
  5. Feb 3, 2012 #4
    1. The problem statement, all variables and given/known data
    Two positive point charges on the y-axis, each of which has a charge of 5.0 10-9 C, are located at y = +0.60 m and y = -0.60 m. Find the magnitude and direction of the resultant electric force acting on a charge of 7.6 10-9 C located at x = 0.80 m on the x-axis.


    2. Relevant equations
    F=kq1q2/r^2


    3. The attempt at a solution
    I don't even know where to start my teacher hasn't taught us this. I have tried to do a vector diagram but it didn't work.
     
  6. Feb 3, 2012 #5

    berkeman

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    Can you show us your diagram? You will get two force vectors acting on the 3rd charge, and the net force is just the componen-wise sum of those forces...
     
  7. Feb 3, 2012 #6
    I attached my picture down below from their I do not know where to go
     

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  8. Feb 3, 2012 #7

    berkeman

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    Good start. Now draw the two force vectors -- one each from the charges on the y-axis. You draw the force vector arrows at the test charge, pointed along the line between the charges. So for the force on the test charge from the top y-axis charge, you draw the force vector at the test charge along the extended line between those two charges. Whether you draw the force vector arrow pointing back up at the top charge, or pointing away down and right, depends on whether the force is attractive or repulsive. Opposite charges attract, right?

    Then once you have the two force vector arrows at the right test charge, divide them up into their x & y components, add components, and then combine those two final components to get the total force vector.
     
  9. Feb 3, 2012 #8
    what are X Y components?
     
  10. Feb 3, 2012 #9

    berkeman

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  11. Feb 3, 2012 #10
    is the x y (40,70)

    is this how you find the magnitude
     
  12. Feb 3, 2012 #11

    berkeman

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    Actually I now notice that your diagram does not match the problem definition. The problem definition mentions y= +/- 0.60m and x = 0.80m. Where did your 40 and 70 numbers come from in your diagram?

    Once you get the dimensions right on the figure, draw your two force vector arrows, and use basic trig to get the x and y components of each vector. Then add the components to get the components of the total resultant force.
     
  13. Feb 3, 2012 #12
    is the hypotenuse 100
     
  14. Feb 3, 2012 #13

    SammyS

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    Re: HELP I don't understand at all

    Where does the 70 come from?

    What is Coulomb's Law for electric force?
     
  15. Feb 3, 2012 #14
    Re: HELP I don't understand at all

    Do I do √0.60^2+0.80^2
     
  16. Feb 3, 2012 #15
    does anyone know how to do this
     
  17. Feb 3, 2012 #16
    Re: HELP I don't understand at all

    Im just trying to get the most help I can
     
  18. Feb 4, 2012 #17

    SammyS

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    Re: HELP I don't understand at all

    Certainly this quantity is involved in this problem.
    [itex]\displaystyle\sqrt{0.60^2+0.80^2}[/itex] is the distance (in meters) from the 7.6 nC charge to either of the 5 nC charges.​
    What is the force on the 7.6 nC charge due to the charge at y = +0.60 m ?
     
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