Electron accelerated between plates

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SUMMARY

An electron is accelerated horizontally from rest in a television picture tube by a potential difference of 6380 V, achieving a horizontal speed of 4.73E7 m/s. After passing through two horizontal plates with a potential difference of 118 V, the vertical acceleration calculated is 3.79E14 m/s², resulting in a final vertical speed of 2.04E5 m/s. The angle θ is determined using the final horizontal and vertical velocities, leading to the correct calculation of θ = 0.124 radians. The initial approach using distances was incorrect; the final velocities must be used to find the angle.

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  • Understanding of kinematics equations
  • Familiarity with electric potential and force equations (E=F/q, V=PE/q)
  • Knowledge of conservation of energy principles
  • Basic trigonometry for angle calculations
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  • Learn about the relationship between potential difference and kinetic energy
  • Explore advanced kinematics problems involving multiple dimensions
  • Investigate the behavior of electrons in different potential configurations
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Homework Statement



An electron is accelerated horizontally from rest in a television picture tube by a potential difference of 6380 V. It then passes between two horizontal plates 2.55 cm long and 5.47 cm apart that have a potential difference of 118 V. At what angle θ will the electron be traveling after it passes between the plates?

Homework Equations



Kinematics equations, E=F/q, V=PE/q, V= -Ed

The Attempt at a Solution



I first used conservation of energy to solve for the horizontal speed, Vq = (1/2)mv^2, to get 4.73E7 m/s. I then used that to find the time in between the plates, v = x/t getting t = 5.39E-10 s. Then I found the force due to the plates, F=Eq=-Vq/d= 3.45E-16 N = ma. Using this with the mass of the electron I solved for a, the vertical acceleration, to get a = 3.79E14 m/s^2 = (final vertical speed - initial vertical speed)/t. Since initial vertical speed = 0, a = v/t and I got the final vertical speed as 2.04E5 m/s. I then used (final speed)^2 = (initial speed)^2 +2ay to solve for y as the vertical distance travelled, which I got as 5.50E-5 m. I then set up a triangle so that tanθ = y/x = (5.50E-5)/(0.0255), and I finished with θ=0.124. This is the wrong answer, though and I'm not sure where I'm going wrong. Any elucidation is much appreciated!
 
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Just kidding, I figured it out. I need to use the final horizontal and vertical velocities, not distances, to find the angle.
 

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