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Particle acceleration in electrical field

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A charge is placed on the x axis (q=+7.00 [tex]\mu[/tex]C, x=0.600 m), and another charge is placed on the y axis (q=+9.00 [tex]\mu[/tex]C, x=0.400 m). A third charge (q=-6.00 [tex]\mu[/tex]C, m= 5.00 x 10[tex]^{-8}[/tex] kg) is placed at the coordinate origin. If the charge at the origin were free to move, what would be (a) the magnitude of its acceleration and (b) the direction of its acceleration? Specify your answer in apart (b) as an angle relative to the +x axis.


    2. Relevant equations

    Okay, so I know that [tex]\sum[/tex]F = ma, so a= sum(F)/a

    Maybe this is relevant: F=k[tex]\frac{q1q2}{r^2}[/tex]


    3. The attempt at a solution

    First off, I just started the second semester of General Physics. I took the first semester of it 2 1/2 years ago.. So some of the real basic stuff is fuzzy. I know essentially what the question is asking and how to work it. My problem seems to be with the addition of the vectors.

    I have set it up as a right triangle (see attachment), so wouldn't I just use the Pythagorean theorem? And if so, do I use the calculated forces or the lengths of the sides? I calculated the F12 to be 3.03 N and F13 to be 1.05 N. After this, I am kind of lost.

    And just to be sure for part B, I did theta = arctan(.400/.600) = 33.7 degrees.

    Thanks!
     

    Attached Files:

  2. jcsd
  3. May 26, 2010 #2
    these forces act at 90 degree to each other take the vector product n u'll get total force
     
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