# Positrons magnitue and direction with velocities

1. Sep 7, 2010

### mudrashka

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)
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
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