Positrons magnitue and direction with velocities

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SUMMARY

The discussion focuses on calculating the required electric field to alter the trajectory of a positron fired in the +y direction at a velocity of v0=4.00E6 m/s, aiming for a final x-direction velocity of vx=1.50E6 m/s. Two methods were proposed: the first involves using trigonometric relationships to find the angle theta, resulting in a calculated angle of 68 degrees, while the second method applies kinematic equations to derive the acceleration and subsequently the electric field strength. The second method yields an electric field magnitude of E=6.56E30 N/C in the positive X direction, which is confirmed as the correct solution.

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  • Understanding of kinematics, specifically the equations of motion.
  • Familiarity with electric fields and forces, particularly F=qE.
  • Knowledge of vector components and trigonometry.
  • Basic principles of particle physics, specifically regarding positrons.
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  • Study the derivation of electric fields from forces using F=qE.
  • Learn about kinematic equations in two dimensions.
  • Explore the concept of vector addition and resolution of forces.
  • Investigate the behavior of charged particles in electric fields.
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Physics students, educators, and anyone interested in particle dynamics and electric field interactions.

mudrashka
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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

Homework Equations



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

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:
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I'm not sure which of these solutions is correct and I don't know how to check if it's right. Could someone please help me out?
 

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