Motion of Charges in E and B Fields

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Homework Help Overview

The discussion revolves around the motion of an electron being ejected from a heated filament and the potential difference required to accelerate it to a specific speed. The subject area includes concepts from electromagnetism and kinematics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevant equation for relating kinetic energy to electric potential energy, questioning the variables involved, particularly the mass of the electron and the meaning of the potential difference.

Discussion Status

Some participants have confirmed the equation being used and clarified the meaning of the variables. There is ongoing exploration regarding the mass of the electron and its implications for the problem.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available or the methods they can use to solve the problem.

PatrickGeddes
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Homework Statement


Consider and electron being ejected from the surface of a heated filament at nearly zero speed. Through what potential difference must it be accelerated to achieve a speed of 8.4x10^6 m/s?

Homework Equations


I have found that the equation I should be using is 1/2mv^2=eΔV (where I believe ΔV is m/e?)


The Attempt at a Solution


No attempt finding the equation is the work I have done.
 
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Hi Patrick! :smile:

(try using the X2 button just above the Reply box :wink:)
PatrickGeddes said:
I have found that the equation I should be using is 1/2mv^2=eΔV

yes :smile:

(but do you mean that you've worked out that this is the correct equation, or that you found it somewhere and it looks right? :confused:)
(where I believe ΔV is m/e?)

no, ∆V is the voltage difference (electric potential difference)

(V means voltage, ∆ means difference in)
 
I have a question on this one...what would the mass be? You have the V and e correct but you don't have m??
 
OR would the mass be just the standard electron mass at 9.11×10^-31 kg?
 
yup! :smile:
 

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