Electric field decelerating electron

In summary, the conversation discusses the concept of work and how it applies to the deceleration of electrons by a voltage. The equation W=qU is used to calculate the work done, and the final velocity can be determined using the equation v_2=sqrt(2(W+KE_1)/m). The book mentioned uses a simplified version of this equation, but a more accurate calculation can be done using 1.3kV instead of 1.3V. The concept of work being negative and the relationship between voltage and charge is also discussed.
  • #1
Daltohn
30
0

Homework Statement


Electron are decelerated by the voltage 1,3V. Initial velocity is 22 Mm/s, what is the final velocity?

Homework Equations


W=qU, W=KE_2-KE_1

The Attempt at a Solution


From above equations I get v_2=sqrt(2(W+KE_1)/m). q is the elementary charge -e so the work is negative in the equation. The answer is 5,2 Mm/s though, I get 21,99 Mm/s so barely any work done. It feels simplified, that the voltage could be used like that.
 
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  • #2
1.3kV works better. :smile:
 
  • #3
NascentOxygen said:
1.3kV works better. :smile:
Aarghh, this book has so many errors... Thanks. :) While we're at it, our equation for work is charge*voltage, so here the voltage would be positive and the charge negative. Is that how you do it? I'm often confused whether work should be subtracted or added but have negative value. With vectors, when you solve the corresponding scalar equation the magnitude should come out positive, for example with forces even though a force is in the negative direction according to your diagram. But with work you should get a negative value if it is negative, since it is a scalar. Have I understood correctly?
 
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