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Coulomb's law vector question.

  1. Feb 1, 2005 #1

    mad

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    Hello, I have a simple exercice I'm having trouble to solve.

    So we have a) and b)

    a) asked the force applied on a charge.... and the answer was [tex] \vec{F_{res}.}= Qkq ( -0,222 \vec{i} -0,250\vec{j}) N [/tex]
    that's not a problem here.. that was the good answer. Now in b), they're asking where do they have to put a +2,5Q charge to nullify the force exerced.. so I thought yeah really simple, the force just has to be the inverse:

    [tex] \vec{F_{12}}= Qkq ( 0,222 \vec{i} + 0,250\vec{j}) N [/tex]

    and since [tex] \vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}} [/tex]

    then I just have to equal this last equation to the 2nd to last one, and I equal the [tex] \vec{i} [/tex] and [tex] \vec{j} [/tex] vectors in both ones to find out the vector of where the charge should be placed to nullify the force applied on it, but I'm having trouble finding it. Believe me, I have tried more than 10 times, and I can't find the vector. Can someone please help me on this one?

    thanks a lot!
     
    Last edited: Feb 1, 2005
  2. jcsd
  3. Feb 1, 2005 #2

    quasar987

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    [tex]\vec{F_{12}}= \frac{2,5Qkq}{r^2_{21}} \vec{u_{21}} [/tex]

    can also be written as

    [tex]\vec{F_{12}}= \frac{2,5Qkq}{|\vec{r_{21}}|^3} \vec{r_{21}} [/tex]
     
  4. Feb 1, 2005 #3

    dextercioby

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    If you want help,then u have to help us too:the only way to do that is to post the ORIGINAL TEXT OF THE PROBLEM...

    Daniel.
     
  5. Feb 1, 2005 #4

    dextercioby

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    And BTW:if that N at the right of your first equation is what i think it is,then learn that VECTORS DO NOT HAVE "SI" UNITS...

    Besides,even if it were a scalar,you should have used the traditional notation
    [itex] [unit] [/itex]

    Daniel.
     
  6. Feb 1, 2005 #5
    Please give a description of the first part. Like:
    Is there another particle that causes this first force?
    Is q the charge on the main particle? Is it positive?
    If these are right then lets look at it as: (sorry im not goot at latex)
    charge 1 = main particle
    charge 2 = causes force in part a
    charge 3 is the 2.5Q charge

    Force on 1 due to 2:
    kqQ*(u12)/(r12)^2

    Force on 1 due to 3:
    kq(2.5Q)*(u13)/(r13)^2

    They must sum to zero
    cancel kqQ

    (u12)/(r12)^2 = -2.5*(u13)/(r13)^2

    since we know its on the other side, the unit vector would be the negative of the previous
    u12=-u13
    1/(r12)^2 = 2.5/(r13)^2

    so r13=sqrt(2.5)*r12

    The unit vector of the first is (-0.664i-0.748j) with the magnitude of 0.334 (u=r/|r|)

    so r13=sqrt(2.5)*(0.334)=0.528

    So the vector is magnitude*(0.664i+0.748j) = 0.351i+0.395j

    Thats what I get, but im not very confident in my skills.
     
    Last edited: Feb 1, 2005
  7. Feb 1, 2005 #6

    mad

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    A force has a unit, Newtons, that's why I posted the N.It has a force in a direction i and j.

    Healey01, thanks for the help but that's not the answer.

    The question was there's two charge that create a force (the first one I posted) on another charge. They then add a 2,5Q charge and ask where to put it in order to nullify the force applied on the charge..
    Thanks for everyone that helped, and I hope someone can figure this out...
     
  8. Feb 1, 2005 #7

    mad

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    Yes, that's how I started the problem, but it gives me a wrong answer..
     
  9. Feb 1, 2005 #8

    dextercioby

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    A SCALAR YES,but A VECTOR NO...
    [tex] F_{x}=... [N] [/tex]
    and similar for F_{y} woud have been the correct formulas...

    Daniel.
     
  10. Feb 1, 2005 #9

    mad

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    I have been using vectors with units for over a year in my physics courses. Every teacher in the physics department teaches this. A force has components in i, j and k.. how can it not have units? But I dont want to start an argument here about this, that's what we use here. Thanks for pointing that out, but we always put units, vectors or not, in my college. So please lets go back to the question.. do you have any clues/tips on how to do it.. I've been trying for a long time now.
     
  11. Feb 1, 2005 #10

    dextercioby

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    I'm asking you (for the second time) to post the original text of the problem...Else i cannot help,as i cannot lmake an idea about the charges' distribution and therefore i cannot compute any forces...

    Daniel.

    P.S.Ask a teacher at your school to measure (in Newtons) the vector [itex] 2\vec{i} + 3 \vec{j} [/itex]. :wink:

    P.P.S.Take the table of units (SI-MKgs) and search for FORCE.Tell me if they put the vector on the "oblect"...
     
  12. Feb 1, 2005 #11

    quasar987

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    Ok, my first post was a little reckless.

    Here's a way to do it...

    Draw a cartesian coordinate with the charge q at the center and draw also the vector force. Find the angle it makes with the x axis using the sine law or something.

    You know the charge 2,5Q will have to be put somewhere along the line along which the actual force act. So you can use the scalar version of coulombs law to find the distance r along this line where you must but the 2,5Q charge. And use the angle found previously to find the components. Sclalar version of coulomb's law:

    [tex]|\vec{F}| = \frac{k(2,5Q)q}{r^2}[/tex]

    I hope that's clear enough.
     
  13. Feb 1, 2005 #12

    mad

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    Sorry for the question, I posted the one they asked but forgot to add that they're ponctual charges. It's in french, that's why I translated it. But basically, the two charges create a resultant force of the one I wrote in my post. They then ask where to put a ponctual 2,5Q charge to nullfy the force applied on the charge. So that's why the force has to be the inverse of the other one. You can just take on from there, I just need to equal the 2 equations I posted to find the 2,5Q charge's position vector.

    I'll ask the teacher, that is if he doesnt kill me because he's nuts about units.

    Take a position vector R= (3i + 2j) m . It has to have units, or else what is 3 in i and 2 in j ?.. anyways, thanks
     
  14. Feb 1, 2005 #13
    sorry to interfere, but for

    [tex]|\vec{F}| = \frac{k(2,5Q)q}{r^2}[/tex]

    I was just wondering do the commas represent decimals?
    if not, then what?

    Thanks,
    MS
     
    Last edited: Feb 1, 2005
  15. Feb 1, 2005 #14

    mad

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    yes, decimals, as in 2 and a half Q
     
  16. Feb 1, 2005 #15

    dextercioby

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    Yes in his and (incidentally on mine) side of the world the comma is a decimal and the decimal point indicates 3 consecutive zero-s in a number which are at the left of the comma...
    [tex] 1000\frac{3}{10} =1,000.3=1.000,3 [/tex]

    Daniel.
     
  17. Feb 1, 2005 #16
    Thanks for the explanation. I should have figured it was just a crazy "Euro-thing" :biggrin:
     
  18. Feb 1, 2005 #17

    quasar987

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    mad,

    I'm guessing your textbook is by Harris Benson?
     
  19. Feb 1, 2005 #18

    mad

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    Exactly :)

    And I'm not in Europe ;) in Montreal , Canada
     
  20. Feb 1, 2005 #19

    quasar987

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    That's where I'm from too. Actually South shore of montreal. I used to go to Cegep Edouard-Montpetit; where do you go?
     
  21. Feb 1, 2005 #20
    My bad,,, I should have said crazy "wannabe Euro-thing" :smile:
    j/k
     
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