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Vector Velocity and Coulomb's Law

  1. May 9, 2010 #1
    1. The problem statement, all variables and given/known data
    Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 3.5 cm. Two of the particles have a negative charge: q_1 = -8.0 nC and q_2 = -16.0 nC. The remaining particle has a positive charge, q_3 = 8.0 nC. Assume that particle 3 is no longer fixed to a corner of the triangle and is now allowed to move. In what direction would particle 3 move the instant after being released???

    **How do I draw the velocity vector for particle 3

    2. Relevant equations
    Coulomb's Law: F = [k*q1*q2]/r2

    3. The attempt at a solution
    F13 (on q3 due to q1) = [k *q3*q1]/r2

    = [(9*109)(8*10-9)(8*10-9)]/(.035)2

    = 4.702*10-4 N

    F23 (on q3 due to q2) = [k * q3*q2]/ r2

    = [(9*109)(8*10-9)(16*10-9)]/(.035)2

    = 9.404*10-4 N

    F13x = F13 * cos(60) = + 2.351*10^-4 N
    F13y = F13 * sin(60) = + 4.072*10^-4 N
    F23x = F23 * cos(0) = + 9.404 *10^-4 N
    F23y = 0

    Fx = 2.351*10^-4 + 9.404*10^-4 = 1.176*10^-3 N
    Fy = 4.072*10^-4 N

    ΣF = 10^-3√(1.176)2 + (0.4072)2 = 1.2445*10^-3 N
    θ = arctan (0.4072/1.176) = 19.1

    The velocity of the body will be directed with the force as the body was initially at rest
    F gives the direction of acceleration a
    a = (change in velocity ) /time
    a=(v-0)/t
    a is directed towards F but a, v, and t are unknowns

    How do I draw the velocity vector?? Please help.
     
  2. jcsd
  3. May 9, 2010 #2

    diazona

    User Avatar
    Homework Helper

    Yes. Now do you know the direction of the force? (Hint: yes you do)
     
  4. May 9, 2010 #3
    so it would go in the (+x axis, +y axis) with θ = 19?
     
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