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Coulomb's law and electric fields

  1. Oct 18, 2011 #1
    I'm having a bunch of trouble with an electrostatics questions as well as an electric fields question. I'll start with the electrostatics problem.

    1. The problem statement, all variables and given/known data
    I am given the following equilateral triangle and asked to calculate the net electrostatic force on each charge.

    Q087F.png

    2. Relevant equations
    Fe=[kq1q2]/r2

    3. The attempt at a solution
    So since the charges are all equal and the distances between them are all equal, the force of the blue on red is equal to the force of red on green which is equal to the green on the blue. So I find:

    Fe=[kq1q2]/r2
    Fe=[(8.99x109)(2.50x10-6C)(2.50x10-6C)]/(0.150 m)2
    Fe=2.4972 N/C

    I don't really know how to proceed past this. I think I need to break things down into x and y components but I don't know how I would draw those here. I know that since all of the charges are negative they will be repelling each other but how do I figure out the magnitude and direction of that repulsion?
     
  2. jcsd
  3. Oct 18, 2011 #2

    vela

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    Consider just the red and blue charges. They sit in the xy-plane so the force on the blue charge will have no z-component. Make a drawing of just the two charges in the xy plane, and you should be able to calculate the components of the force. Do the same thing for the green and blue pair. This time they will lie in the yz plane, so you'll have no x-component and non-vanishing y and z components. The total force will be the sum of the two individual forces.

    You'll have to do some geometrical reasoning to figure out where in the axes each charge sits.
     
  4. Oct 18, 2011 #3
    Okay so here is what I have:

    FB on R=2.4972 N/C

    x1=sin30°(2.4972 N/C)
    x1=1.2486 N/C

    y1=cos30°(2.4972 N/C)
    y1=-2.162638638 N/C

    FB on G=2.4972 N/C

    x2=sin30°(2.4972 N/C)
    x2=-1.2486 N/C

    y2=cos30°(2.4972 N/C)
    y2=-2.162638638 N/C

    Is that correct so far? The next part confuses me where you say "This time they will lie in the yz plane, so you'll have no x-component and non-vanishing y and z components. The total force will be the sum of the two individual forces."

    I've never had to do any questions that used the z plane so I'm not sure what this means.
     
  5. Oct 18, 2011 #4

    vela

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    OK, I interpreted the dotted lines as the x, y, and z axes; otherwise, I'm not sure why they would be drawn in that way. Is that your drawing or the book's?

    If it's just a two-dimensional problem, then your calculations look okay. What you want to do, however, is find the two force acting on one charge and sum those. Right now, you're finding the components of the force acting on two different charges due to one charge.
     
  6. Oct 18, 2011 #5
    I pulled the picture straight from the assignment. There hasn't been any three dimensional examples in the text yet so I don't think they intended this one to be. So I've now done:

    x1+x2
    =1.2486 N/C + (-1.2486 N/C)
    =0

    y1+y2
    =-2.162638638 N/C + (-2.462638638 N/C)
    =-4.325277276 N/C

    So then the resultant force will be 4.33 N/C downward? But if they are all negative charges shouldn't they be repelling each other and the force would be upward? Or do I have the direction completely wrong?
     
  7. Oct 18, 2011 #6

    vela

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    Read the last paragraph of my previous response. I probably edited after you read it initially.
     
  8. Oct 18, 2011 #7
    Sorry, I'm not quite sure what you mean by
     
  9. Oct 18, 2011 #8

    vela

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    You're finding the components of the force the blue charge exerts on the red and green charges, i.e, "B on R" and "B on G". (If that's not what you intended, your math is wrong too.) Those forces act on different bodies, so it doesn't make sense to sum them.

    What you need is the force that R exerts on B and the force that G exerts on B. Those two forces both act on B, so you can sum them.
     
  10. Oct 18, 2011 #9
    Didn't I find that in my first post when I did:

    Fe=[kq1q2]/r2
    Fe=[(8.99x109)(2.50x10-6C)(2.50x10-6C)]/(0.150 m)2
    Fe=2.4972 N/C

    Since they all have the same charge and are the same distance from each other, aren't they all exerting a force of 2.4972 N/C on each other? And in my third post I found the components of blue on red and blue on green. Isn't this the same as finding the components for red on blue and green on blue since they have the same charge?
     
  11. Oct 18, 2011 #10

    vela

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    Sort of, but not for the reason you give, i.e., that they have the same charge. The forces a pair of charges exert on each other are an action-reaction pair, so by finding the components of one, you effectively find the components other other. This is true regardless of what the charges are. You have to remember that the action-reaction pair of forces are equal and opposite to get the directions right.
     
  12. Oct 18, 2011 #11
    I see. So then what exactly do I need to calculate here to find the total net electrostatic force on each charge? The x and y components of green on blue and the x and y components of red on blue, and then add those together and find the resultant?
     
  13. Oct 18, 2011 #12

    vela

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    Yup.
     
  14. Oct 18, 2011 #13
    I think I'm confused as to how to draw the vector diagram for this so I can find the components with proper directions. Here is a very quick MSPaint of how I've done it:

    http://i.imgur.com/sHSJt.png

    That isn't right though is it?
     
  15. Oct 18, 2011 #14

    vela

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    Looks fine so far. Why do you think there's a problem?
     
  16. Oct 18, 2011 #15
    Oh good. I wasn't sure because that is how I got the components in post #3 and I thought I was on the wrong track. So again:

    FR on B components

    x component
    x1=sin30°(2.4972 N/C)
    x1=1.2486 N/C

    y component
    y1=cos30°(2.4972 N/C)
    y1=-2.162638638 N/C

    FG on B components

    x component
    x2=sin30°(2.4972 N/C)
    x2=-1.2486 N/C

    y component
    y2=cos30°(2.4972 N/C)
    y2=-2.162638638 N/C

    Then I add the x components and the y components.
    x1 +x2=1.2486 N/C + (-1.2486 N/C)
    =0

    y1 +y2=-2.162638638 N/C + (-2.462638638 N/C)
    =-4.325277276 N/C

    I'm not sure about this next part.
    resultant=√x2+y2
    R=√(0)2+ (-4.325277276 N/C)
    R=4.325277276

    So the answer is 4.33 N/C ? In what direction?
     
  17. Oct 18, 2011 #16

    vela

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    All of the components you calculated have the wrong sign. The force of R on B points upward and to the left, so the y-component should be positive and the x-component, negative, etc.

    The resultant is the vector sum of the two forces. In this case, you'll get
    [tex]\vec{R} = 4.33\hat{y}~\mathrm{N/C}[/tex]and its magnitude, as you found, is [itex]\|\vec{R}\|=4.33~\mathrm{N/C}[/itex].
     
  18. Oct 19, 2011 #17
    Okay I see. And the force of green on blue will point upward and to the right. So the resultant will be in the positive y direction? Or could I call it 90°?
     
  19. Oct 19, 2011 #18

    vela

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    Yes, in the +y direction or 90 degrees. As long as you're clear about what you mean, it should be okay.
     
  20. Oct 19, 2011 #19
    Okay, thanks for all of your help. I think I understand it now.
     
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