An electron is released from rest in a uniform electric field

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SUMMARY

An electron released from rest in a uniform electric field accelerates vertically upward, covering a distance of 4.50 meters in 3.00 microseconds. Using the kinematic equation s = a t²/2, the acceleration (a) can be calculated as 3.00 m/μs². Subsequently, the magnitude of the electric field (E) can be determined using the relationship E = ma, where m is the mass of the electron (9.11 x 10^-31 kg) and a is the calculated acceleration. This results in an electric field magnitude of approximately 1.74 x 10^5 N/C.

PREREQUISITES
  • Understanding of kinematic equations, specifically s = a t²/2
  • Knowledge of electric fields and forces
  • Familiarity with the mass of an electron (9.11 x 10^-31 kg)
  • Basic principles of acceleration and its relation to electric fields
NEXT STEPS
  • Study the derivation of kinematic equations in physics
  • Learn about the relationship between electric fields and forces on charged particles
  • Explore advanced concepts in electromagnetism, including field strength calculations
  • Investigate the effects of electric fields on particle motion in different mediums
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of charged particles in electric fields.

hitek131
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I need help with this question if anyone can give me an idea of what to do. Any help would be greatly appreciated.

An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3.00 micro seconds after it is released.
What is the magnitude of the electric field?
 
Last edited:
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HINT: [itex]s = a t^2/2[/itex]
 

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