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Quick vector question (involving coulomb's law)

  1. Apr 6, 2008 #1
    1. The problem statement, all variables and given/known data
    [​IMG]


    2. Relevant equations

    It's given in the answer box.

    3. The attempt at a solution

    Basically, I found that equation for the magnitude of the force experienced from particle 0 by particle 3. I know it to be correct. However, vectors have always confused me, could anybody bump me in the right direction on how to label the equation appropriately with vectors? (using y hat and z hat). The distance between particle 0 and 2 is d2, and between particle 2 and 3 is d3. So it's a 45* angle at the hypotenuse.

    Would it be F = (kq0q3)y/(sqrt(2d2^2)z)? Thanks!
     
  2. jcsd
  3. Apr 6, 2008 #2

    G01

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    Mastering Physics, my favorite!:rolleyes:

    You have the magnitude of the force on charge zero. You now need to find the vector describing the force, which will have the form:

    [tex]\vec{F}=F_y \hat{y} + F_z \hat{z}[/tex]

    where F_y and F_z are the y and z components of the force, respectively.

    Does this help?
     
  4. Apr 6, 2008 #3

    Kurdt

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    One has to split the force due to each pair of particles into their separate components then apply vector addition to find the total force.
     
  5. Apr 6, 2008 #4
    Thanks for the replies guys, and I know that's exactly what I have to do, I'm just unsure of how the equation would split up. I'm confused because the force is the magnitude of the force along the hypotenuse, how can that be split up into two vectors?

    PS. I'm not a huge fan of mastering physics, or online homework in general. :)
     
    Last edited: Apr 6, 2008
  6. Apr 6, 2008 #5

    Kurdt

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    How would you find the length of the sides of a right-angled triangle given the hypotenuse and some other information (such as angle or one of the other sides)?
     
  7. Apr 6, 2008 #6
    The pythagorean theorem, which is what I used to distance for the d^2 in the equation for the magnitude I found. Both the base and height of the triangle are d2. So the hypotenuse is [tex]\sqrt{2d\stackrel{2}{2}}[/tex]. But how can that factor into the coulomb equation I found?

    EDIT: Would I just use the equation I found, solve for [tex]\sqrt{2d\stackrel{2}{2}}[/tex], and plug that into [tex]\sqrt{2d\stackrel{2}{2}}[/tex] for each side, and label it appropriately?
     
    Last edited: Apr 6, 2008
  8. Apr 6, 2008 #7

    Kurdt

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    Well like you say the magnitude you've worked out is like the force along the hypotenuse of a triangle. So you just need to take the components of that vector. See the following web page for more details about that.

    http://hyperphysics.phy-astr.gsu.edu/hbase/vect.html
     
  9. Apr 6, 2008 #8
    Okay, so I'll call the equation I found A. So it would simply be

    F= Acos(45)y + Asin(45)z. But sin(45) is undefined, which just brings up another question, haha.

    EDIT: STupid mistake, sorry. arcsin(45) is undefined.
     
    Last edited: Apr 6, 2008
  10. Apr 6, 2008 #9
    Okay, well I tried this answer, which makes sense, but mastering physics says it's wrong.

    [​IMG]

    What looks wrong?
     
  11. Apr 6, 2008 #10
    Bump! Still seeking some assistance with this :). Anybody have any idea what's wrong with my answer above? Thanks!!
     
  12. Apr 6, 2008 #11

    Kurdt

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    Ok, what is the charge of q0? What direction will the overall force be acting in. that should help you with the components.
     
  13. Apr 6, 2008 #12
    The charges of q0 and q3 are positive, therefore it's a repulsion affect. However, I'd think the magnitude of the overall force would depends on how big the charge of q3 is.

    I have a feeling I'm overlooking something blatantly obvious. Thanks again for helpin out.
     
  14. Apr 7, 2008 #13

    Kurdt

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    So whats the force on particle 0 from particle 3 then? That is what direction will it be traveling if it is being repulsed?
     
  15. Apr 7, 2008 #14
    It will be traveling down and to the left, the particle will go southwest of its current position if it's allowed to move.
     
    Last edited: Apr 7, 2008
  16. Apr 7, 2008 #15

    Kurdt

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    ..and your vector is heading north east.
     
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