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Charged particles and vectors

  1. Sep 8, 2005 #1

    ranger

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    Gold Member

    Lets say that we have a four positive and equally charged particles. They are located on the corners of a square.
    Can someone show (by diagram) how the other three particles would affect any one particle. Please show how you reslove the vectors into its componets and how you get the resultant (formulas please :smile: ).

    --thanks
     
  2. jcsd
  3. Sep 8, 2005 #2

    sic

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    Projecting the forces onto axes:
    say we choose two axes parallel to the sides of the square, then if the square side length is a, each one of the two charges on the axes will apply force equal to q^2/a^2 on its axis and 0 on the second one, and the charge on the diagonal will apply sin(pi/4)*q^2/(sqrt(2)a)^2=cos(pi/4)*q^2/(sqrt(2)a)^2=q^2/(sqrt(2)a^2) on each one of the axes
     
  4. Sep 8, 2005 #3

    Astronuc

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    Staff: Mentor

    The force vectors on a point charge operate along the lines determined by the point charge and each of the other charges. If charges are opposite, the vector will point toward the other charge, and if the charge are the same sign, the vector will point away.
     
  5. Sep 8, 2005 #4

    ranger

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    Okay then, so since the charges are the same, they would repel. I've put together a little image based on that.
    http://img356.imageshack.us/img356/3092/vector1mb.png

    Now I want to calculate the final position and magnitude of the particle that is being acted upon. For that I need the resultant vector. But how do I get that? It looks like the brown charge is in the place of where the resultant sould be. I'm thinking tail-to-tip here. This is getting very confusing.
     
  6. Sep 9, 2005 #5

    Astronuc

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    Staff: Mentor

    The resultant force vector is the sum of each force vector contributing to the force.

    So if one has 3 vectors (i, j, k are orthogonal unit vectors).

    a1 i + a2 j + a3 k
    b1 i + b2 j + b3 k
    c1 i + c2 j + c3 k

    then the resultant vector is simply

    (a1 + b1 + c1) i +
    (a2 + b2 + c2) j +
    (a3 + b3 + c3) k

    Since the problem is 2D (coplanar), the coefficients of k are zero

    :smile:
     
    Last edited: Sep 9, 2005
  7. Sep 9, 2005 #6

    ranger

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    I'm still knida confused. I see that you have a1 + b1 + c1, what do thoes mean. Did you reslove them into conponenets? If you did then how come you have 3 components? Arent we just using x and y axis.

    thanks again
     
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