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Components of electric force

  1. Sep 22, 2008 #1
    1. The problem statement, all variables and given/known data


    Particle B, with positive charge q_a, fixed in the yz-plane at (0,d_2,d_2). What is the net force F on particle A (positive) due to this charge?
    Express your answer (a vector) using k, q_0, q_3, d_2, x hat, y hat, and z hat. Include only the force caused by particle B.

    2. Relevant equations

    electric force F = k(q_a)(q_b)/(d_2)^2
    k = 9*10^9
    q_a,q_b = point charges
    d_2 = distance between charges

    3. The attempt at a solution

    i really not sure how to go about determining the components of the force due to particle B
    could someone help me with some guidelines on how to do so pls or a link to somewhere, couldn't find too much when i searched

    do i use the z and x distances to form a triangle and then use the side lengths to determine the angle theta?

    thanks so much
     

    Attached Files:

  2. jcsd
  3. Sep 22, 2008 #2

    LowlyPion

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    Coulomb's law is based on the distance between the charges.

    What is the distance if the charge is located d2 y and d2 z away?
     
  4. Sep 22, 2008 #3
    it would just be the hypotenuse using d2_y and d2_z as the sides, so distance = sqrt(d2_y^2 + d2_z^2), so is the triangle formed a 45-45-90?
     
  5. Sep 22, 2008 #4

    LowlyPion

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    Yes, making the distance effectively then (2)1/2 * d2
     
  6. Sep 22, 2008 #5
    so using the constants stated in problem the force would be:

    F = (k*q_0*q_3)/ (d_2(2^1/2))^2, how would i state the unit vectors the problem involves components, i know it involves x hat and z hat?
     
    Last edited: Sep 22, 2008
  7. Sep 22, 2008 #6

    LowlyPion

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    [tex] \vec{F} = 0 \hat{x} + \frac{F}{\sqrt{2}}\hat{y} + \frac{F}{\sqrt{2}}\hat{z}[/tex]

    Where

    [tex] F = \frac{k*q_0*q_3}{(d_2* \sqrt{2})^2} [/tex]
     
    Last edited: Sep 22, 2008
  8. Sep 22, 2008 #7
    ohhh, but why is F divided by 2? i must've missed something
     
  9. Sep 22, 2008 #8

    LowlyPion

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    Because those are the components of the force in each direction.

    But oops that should be over the sqrt of 2. Sorry.

    The Root Sum of the Squares = the F

    Edit: I just changed it.
     
  10. Sep 22, 2008 #9

    LowlyPion

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    In actual practice that would be the Cos or Sin of the angle contribution of the vector components.

    It so happens in this case that the angles are Sin45 and Cos45 and = 1/2*sqrt(2) or 1/sqrt(2)
     
  11. Sep 22, 2008 #10
    thanks so much, actually the force components were supposed to be negative since the two point charges were of the same charge and thus repelled eachother, so in other words replace the + with -
     
  12. Sep 22, 2008 #11

    LowlyPion

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    Yes, that's correct. The question was with respect to forces at A.
     
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