Motion of Charges in E and B Fields

AI Thread Summary
To determine the potential difference needed to accelerate an electron to a speed of 8.4x10^6 m/s, the relevant equation is 1/2mv^2 = eΔV. The mass of the electron is standard at 9.11×10^-31 kg, and ΔV represents the electric potential difference. Participants confirm the use of this equation and clarify that ΔV is not m/e, but rather the voltage difference. The discussion emphasizes the importance of correctly identifying variables in the equation. Understanding these concepts is crucial for solving the problem effectively.
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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