Acceleration of an electron through a uniform electric field

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  • #1
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Question:
An electron is accelerated through a uniform electric field of magnitude 2.5x10^2 N/C with an initial speed of 1.2x10^6 m/s parallel to the electric field.

-----------<---------------
------------<--------------
e- ----------------------->
------------<--------------
-----------<---------------

a) Calculate the work done on the electron by the field when the electron has travelled 2.5cm in the field.
b) Calculate the speed of the electron after it has travelled 2.5 cm in the field.

My attempt:
For a), I used W=[tex]\epsilon[/tex]q[tex]\Delta[/tex]d
and I got 1.0x10^-18 J.

For b), I tried
W=[tex]\Delta[/tex][tex]E_{K}[/tex]
=0.5m[tex]v_{2}^{2}[/tex]-0.5m[tex]v_{1}^{2}[/tex]
After rearranging the equation to solve for [tex]v_{2}[/tex],
[tex]v_{2}[/tex]=[tex]\sqrt{2W/m}[/tex]+[tex]v_{1}[/tex]
=2.7x10^6m/s
for m, I plugged in the mass of an e-.

However, the answer is supposed to be 1.9x10^6 m/s.
Can someone please tell me what I'm doing wrong?
 

Answers and Replies

  • #2
Doc Al
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For b), I tried
W=[tex]\Delta[/tex][tex]E_{K}[/tex]
=0.5m[tex]v_{2}^{2}[/tex]-0.5m[tex]v_{1}^{2}[/tex]
This is OK.
After rearranging the equation to solve for [tex]v_{2}[/tex],
[tex]v_{2}[/tex]=[tex]\sqrt{2W/m}[/tex]+[tex]v_{1}[/tex]
But this is not OK.

Try this:

[tex]1/2 m v_f^2 = 1/2 m v_i^2 + W[/tex]
[tex]v_f^2 = (2/m)(1/2 m v_i^2 + W)[/tex]

...and so on.
 
  • #3
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Thank you =)
 

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