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Coulomb's Law

  1. Jul 7, 2006 #1
    I have a question that I've worked out and am having a bit of difficulty with. The problem is:
    http://synthdriven.com/images/deletable/help3.jpg

    This is what I've worked out:
    (sorry, it's a bit faint.)
    http://synthdriven.com/images/deletable/help2.jpg


    What I did. I drew a diagram of the electrical forces acting on the point charge in question. I used Coloumb's law to find F for each. I then broke F1 and F2 down into vector components. It's obvious from the diagram that the x-components are going to cancel each other out... So that should leave a resultant force going in the -y direction. Can someone look at this and tell me what I've done wrong? According to this I should be getting a 0 resultant force. Which is stupid.

    I've been long out of practice when it comes to vectors, but I think what I've done wrong lies more around the F1 and F2 magnitude... Is it the signs??


    Thanks!
     
  2. jcsd
  3. Jul 7, 2006 #2

    nazzard

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    Hello verd,

    if you apply the scalar form of Coulomb's law, and that's what I think you were trying to do, you shouldn't get a (-) sign for any of the forces.

    [tex]|F|=k\frac{|q_1||q_2|}{r^2}[/tex]

    In this case it is ok to calculate the norm of F1 and F2 and the norm of their components, but always make it clear in which directions the forces are acting in your sketch (as you did).

    As a formal sidenote: if you insert explicit values into your equations you'd have to write the correct unit everytime, not only in your endresult.

    You could also use the vector form of Coulomb's law right from the start.

    [tex]\vec F=k\frac{q_1q_2}{r^2}\vec e_r[/tex]

    [tex]\vec e_r=\frac{\vec r}{r}[/tex]

    It's the [itex]\vec e_r[/itex] that differs when calculating the forces. That's where you get the correct signs for their components.

    Regards,

    nazzard
     
    Last edited: Jul 7, 2006
  4. Jul 7, 2006 #3
    ...Alright, so... In using the correct version of Coulomb's law, the one with the absolute values around the charge, and applying signs to direction, I get:

    F1x=-0.00084
    F1y=-0.00048

    F2x=0.00084
    F2y=-0.00048

    Which would cancel the x-component and leave -0.00048-0.00048 for the y, which would equal -0.00096. Is this correct? Do I need to leave to leave the sign on that one? Or is it 0.00096?


    Thanks for your help
     
  5. Jul 7, 2006 #4

    nazzard

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    Hello vred,

    Let's call the resulting force F.

    [tex]\vec F=\vec F_1 + \vec F_2[/tex]

    The y-component would be:

    [tex]F_y=\vec F \vec e_y=-0.00096\,N[/tex]

    It's correct to have the (-) sign in your final answer.

    Regards,

    nazzard
     
    Last edited: Jul 7, 2006
  6. Jul 7, 2006 #5
    ...So the resultant force, F is going to be -0.00096... I tried that and it doesn't seem to be correct.

    I mean, the resultant x-component is 0 and the resultant y-component is -0.00096. Right? So the resultant force F is going to be -0.00096 in the y-direction.

    Am I doing this wrong?
     
  7. Jul 7, 2006 #6

    nrqed

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    Your mistake is that you must use the distances in meters, not in centimeters.
     
  8. Jul 7, 2006 #7

    nazzard

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    Ah, thank you nrqed. I just had a look at the sketch again and noticed that there's something wrong :rolleyes:

    So verd, do you get 9,6 N as the resulting force pointing at the negative y-direction?
     
    Last edited: Jul 7, 2006
  9. Jul 7, 2006 #8

    nrqed

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    So a few comments. As you know by now, the *magnitudes* of the forces must always be positive. That was a mistake in your written notes.

    About the sign of the components, it is possible to always use [itex] F_x = +F cos \theta [/itex] and [itex] Fy = +F sin \theta [/itex] (where F is the magnitude so it's positive), BUT if you use that, you must always use an angle measured from the positive x direction with the convention that counterclockwise is positive.

    However, if you use different definitions of the angle (as you did here), there you have to include by hand a sign + or - in the equation, depending on where your angle is.

    EDIT: Not only the sign must be put by hand but of course whether one uses a sin or a cos depends on where the angle is measured.
    Patrick
     
    Last edited: Jul 7, 2006
  10. Jul 7, 2006 #9
    Awesome. Yeah. Thank you. I miss details like that. I'm sorry.


    Thanks again guys.
     
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