Calculating Electron Speed at 12kV Potential Difference on a Computer Screen

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Homework Help Overview

The problem involves calculating the speed of an electron accelerated by a potential difference of 12kV, starting from rest at the cathode of a computer screen. The subject area includes concepts from electromagnetism and energy conservation.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to apply energy conservation principles and questions the appropriate expressions for potential energy in this context. Some participants explore the relationship between electric potential and kinetic energy, while others suggest reconsidering the analogy with gravitational potential energy.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of energy conservation and the appropriate use of potential energy expressions. Some guidance has been offered regarding the simplification of the problem, but no consensus has been reached.

Contextual Notes

Participants note the potential difference and the initial conditions of the electron, while also questioning the relevance of gravitational forces in this scenario. The need for the distance between the cathode and the screen is mentioned, indicating a potential gap in the information provided.

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Homework Statement



The potential between the cathode and the screen of a computer is 12kV. What is the speed of an electron when it reaches the screen if it starts at rest at the cathode?

The Attempt at a Solution



First I found that the force that acts on the electron is 1.922E-15 N. I can then find the acceleration, but I need the distance between the cathode and the screen to find the speed.

Can I use energy conservation? Probably. Gravitational forces can be ignored. (1/2)mv^2 = Epot But what is the the expression for potential energy? Not mgh. Can I simply replace m with charge, g with E and h with the potential?

1/2)mv^2 = eEU

?
 
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Generally the electric and gravitational potentials will NOT take the same form - good try though. The answer is much simpler - it is contained in the problem statement (hint hint: what are the first 2 words?).
 
The potential. Probably eV then? But why?
 
The idea of replacing mgh with eEh is a bit careless but has some meaning if understood properly: Work done by constant forces is F*s, where F is -mg in the first case and eE in the second case. Since the change of energy equals -work, the analog expression of mgh is actually -eEh. However the electric field is not always uniform, so -Eh must generaly be replaced by Integral(-E*dh)=U
 
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