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Homework Help: Positrons magnitue and direction with velocities

  1. Sep 7, 2010 #1
    You fire a positron in the +y direction with a velocity v0=4.00E6 m/s. They pass through a region where a horizontal electric field exists. This region extends .06m in the y-direction. When they leave this region you wish for their velocity in the x-direction to be vx=1.50E6 m/s. What must be the magnitude and direction of the electric field to do this?
    q=1.6E-19C (positron)
    m=9.10E-31kg (positron)
    y=0.06m

    2. Relevant equations

    vx=v0cos(theta)
    vy=v0sin(theta)
    vfi2-vi2+2ad
    F=ma
    F/q=E

    3. The attempt at a solution
    There are 2 ways I have of possibly solving this problem. Hopefully one is right.
    1st: Using the:
    vx=v0cos(theta) to solve to (theta)= 68O
    Then using the 68o to solve for vy=vosin(theta)
    Then added the x- and y- components together and square rooted them like the a2 +b2=c2 method to get the magnitude of the resultant vector = 4.00E6m/s in positive X direction.

    2nd: Using the
    vfi2-vi2+2ad
    F=ma
    F/q=E
    1.50E62=4.00E62+2a(0.06m)
    a=-1.15E-14m/s2
    F=ma=(9.10E-3)(-1.15E14)=1.05E12
    F/q=E
    1.05E12/1.6E-19=E
    6.56E30 N/C (positive X direction)
     
    Last edited: Sep 7, 2010
  2. jcsd
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