1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted