Calculating Electron Speed and Energy in a Parallel-Plate Apparatus

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SUMMARY

The discussion focuses on calculating the speed and energy of an electron in a parallel-plate apparatus with a 450 V potential difference. The energy gained by the electron is calculated using the formula Eel = V x q, resulting in Eel = 7.2 x 1017 J. The speed of the electron one third of the way between the plates is determined to be V = 1.26 x 107 m/s using the kinetic energy formula Eel = 1/2mv2. Clarifications are provided regarding the separation of the problem into parts a) and b), emphasizing the importance of accurate exponent notation.

PREREQUISITES
  • Understanding of electric potential and energy calculations
  • Familiarity with the kinetic energy formula E = 1/2mv2
  • Knowledge of electron charge and mass values (1.6 x 10-19 C, 9.1 x 10-31 kg)
  • Basic grasp of exponent rules in scientific notation
NEXT STEPS
  • Study the principles of electric fields in parallel-plate capacitors
  • Learn about energy conservation in electric fields
  • Explore the effects of varying potential differences on electron motion
  • Investigate the relationship between voltage, electric field strength, and electron acceleration
USEFUL FOR

Students in physics, particularly those studying electromagnetism and kinematics, as well as educators looking for examples of electron behavior in electric fields.

Lolagoeslala
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Homework Statement


An electron is released from rest from the negative plate of a parallel-plate apparatus, what is the electrons speed one third of the way between the plates. It also asks for the speed at which electron will hit the positive plate if a 450 V potential difference is applied.


The Attempt at a Solution



a)
Eel = V x q
Eel = ( 450 V ) x (1.6 x 10^-19C)
Eel = 7.2 x 10^17 J

Eel = 1/2mv^2
V = √((2 x (7.2 x 10^17 J))/9.1 x 10^31Kg)
V = 1.26 x10^7 m/s

b) i don't know how to do this?
 
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I don't understand your answer's division into a) and b). The a) section looks like a response to the second question in the problem statement. What happened to the first question? Or is that now part b)?

Also, watch out for the signs associated with exponents; there's a rather large difference between 10^17 J and 10^-17 J. Same goes for the electron's mass. Your final velocity for the 450V excursion looks okay though.
 

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