Electric Field & Capcitor Problem HELP

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SUMMARY

The discussion focuses on calculating the electric field magnitude in a parallel plate capacitor where an electron enters with an initial speed of 4.00 x 106 m/s. The capacitor dimensions are 2.00 cm in length and 0.150 cm in plate separation. The force on the electron is determined using the formula F=ma=qE, where q is the charge of the electron, specifically -1.6 x 10-19 coulombs. The participant clarifies that the charge does not need to be computed but can be referenced from known values.

PREREQUISITES
  • Understanding of electric fields and forces
  • Familiarity with kinematics and acceleration calculations
  • Knowledge of the properties of electrons, including mass and charge
  • Basic grasp of parallel plate capacitor configurations
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  • Study the relationship between electric field strength and force on charged particles
  • Explore kinematic equations for charged particle motion in electric fields
  • Learn about the properties and behavior of parallel plate capacitors
  • Investigate the effects of varying electric field strengths on electron trajectories
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Students studying electromagnetism, physics educators, and anyone seeking to understand the dynamics of charged particles in electric fields.

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Electric Field & Capcitor Problem HELP!

Homework Statement



An electron enters the lower left side of a parallel plate capacitor & exits at the upper right side making an upward curve. The initial speed of the electron is 4.00 x 10^6 m/s. The capacitor is 2.00 cm long, and its plates are separated by 0.150 cm. Assume that the electric field between the plates is uniform everywhere and find its magnitude.

The Attempt at a Solution



So I understand that F=ma=qE. I also understand that I can use kinematics to find the acceleration and I can also use the mass of an electron to figure out the force. However, how do I find q? I'm so confused!:confused: Please help! Thanks!
 
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q is the charge of the electron. You don't compute it, you look it up. -1.6*10^(-19) coulombs.
 
Thank you so much! That's great...duh.
 

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