Positrons magnitue and direction with velocities

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To achieve a final x-direction velocity of 1.50E6 m/s for a positron initially moving at 4.00E6 m/s in the +y direction, the required electric field magnitude is calculated. Two methods were attempted: the first involved using trigonometric functions to find the angle and resultant velocity, while the second applied kinematic equations to derive acceleration and force. The second method yielded an electric field of 6.56E30 N/C in the positive x-direction. The user seeks confirmation on the correctness of these calculations and methods. Clarification on the appropriate approach to solve the problem is requested.
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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