Electron accelerated by potential difference

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Homework Help Overview

The discussion revolves around the acceleration of an electron by a potential difference, specifically examining the relationship between voltage and the final speed of the electron. The subject area includes concepts of energy, kinetic energy, and the effects of varying potential differences on particle motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between potential difference and kinetic energy, questioning how changes in voltage affect the final speed of the electron. Some participants attempt to derive relationships using equations related to kinetic energy.

Discussion Status

The discussion includes various interpretations of how voltage influences speed, with some participants suggesting that increasing the voltage leads to a quadratic increase in speed, while others propose a linear relationship. There is no explicit consensus, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are working within the framework of basic physics principles, specifically relating to energy conservation and the definitions of kinetic energy. There may be assumptions regarding the mass of the electron and the nature of the potential difference that are not fully articulated.

gillgill
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An electron is accelerated by a potential difference of, say, 100V. How much greater would its final speed be if it were accelerated with four times as much voltage?
 
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The corresponding energy is eV where e is the elementary charge and v is the applied potential. Note that this energy is kinetic in nature and therefore equal to 1/2mv²

regards
marlon
 
Ep=Ek
eV=1/2mv²
so...four times as much voltage= 16 times greater speed?
 
So if the potential is 4 times bigger then the kinetic energy will get four times bigger. If A = mv²/2 or v = sqrt(2A/m) ; well then after the increase : 4A = mv²/2 or mv² =8A or v = sqrt(8a/m). Now if you divide sqrt(8A/m) by sqrt(2A/m) you acquire : sqrt(4) = 2. So the speed would be doubled...you see

regards
marlon
 

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