You fire a positron in the +y direction with a velocity v(adsbygoogle = window.adsbygoogle || []).push({}); _{0}=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 v_{x}=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

v_{x}=v_{0}cos(theta)

v_{y}=v_{0}sin(theta)

v_{fi}^{2}-v_{i}^{2}+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:

v_{x}=v_{0}cos(theta) to solve to (theta)= 68^{O}

Then using the 68^{o}to solve for v_{y}=v_{o}sin(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

v_{fi}^{2}-v_{i}^{2}+2ad

F=ma

F/q=E

1.50E6^{2}=4.00E6^{2}+2a(0.06m)

a=-1.15E-14m/s^{2}

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)

**Physics Forums | Science Articles, Homework Help, Discussion**

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

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**