Calculating Vertical Deflection of Electron in Parallel Plate Setup

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SUMMARY

The discussion focuses on calculating the vertical deflection of an electron moving through a parallel plate setup with a horizontal velocity of 3.00 x 106 m/s and an electric field strength of 120 N/C. The key steps involve determining the force acting on the electron using the formula F = qE, where q is the charge of the electron and E is the electric field. The resulting acceleration can be calculated using Newton's second law, and the time spent between the plates is derived from the distance and horizontal velocity. The vertical deflection can then be computed using kinematic equations.

PREREQUISITES
  • Understanding of electric fields and forces (F = qE)
  • Basic knowledge of kinematics (displacement, velocity, acceleration)
  • Familiarity with the properties of electrons (charge and mass)
  • Ability to apply Newton's laws of motion
NEXT STEPS
  • Calculate the force on the electron using F = qE
  • Determine the acceleration of the electron using a = F/m
  • Calculate the time the electron spends between the plates using t = d/v
  • Use kinematic equations to find the vertical deflection of the electron
USEFUL FOR

Physics students, educators, and anyone interested in electromagnetism and particle motion in electric fields.

XJellieBX
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Question:
An electron with a velocity of 3.00x10^6m/s[horizontally] passes through two horizontal parallel plates, as in the attached diagram. The magnitude of the electric field between the plates is 120N/C. The plates are 4.0cm across. Edge effects are negligible.
a) Calculate the vertical deflection of the electron.

I'm not sure what I should do here and I would appreciate any help. I think it might be [tex]\epsilon[/tex]=[tex]\frac{kq}{r^2}[/tex], but I'm not exactly sure. Thanks.
 
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What's the force on the electron while it's between the plates? The resulting acceleration? How long does it spend between the plates?
XJellieBX said:
I think it might be [tex]\epsilon[/tex]=[tex]\frac{kq}{r^2}[/tex], but I'm not exactly sure.
That's an equation for the field from a point charge, which is not relevant here. You are given the field, so find the force.
 

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