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Electric Field (velocity of a particle)

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Homework Statement


At some instant in the velocity components of an electron moving between two charged parallel plates are [tex]v_x=1.5x10^5m/s[/tex] and [tex]v_y=3.0x10^3m/s[/tex]. Suppose that the electric field between the plates is given by [tex]\vec{E}=(120N/C)j[/tex].

a)what is the acceleration of the electron?

b)what will be the velocity of the electron after its x coordinate has changed by 2.0cm?

Homework Equations


[tex]F=ma[/tex]
[tex]\frac{F}{m}=a_y[/tex]
[tex]a_y= \frac{q\vec{E}}{m}[/tex]?

The Attempt at a Solution


Do I just have to sub in the values for part a?
 
Last edited:

Answers and Replies

  • #2
Dick
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Yes. I think you have everything you need. Now get started!
 
  • #3
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so [tex]a_y = \frac{(1.6x10^{-19}C)((120N/C)j)}{9.1x10^{-31}Kg}[/tex]
[tex]a_y=(2.108x10^13m/s^2)j[/tex]?
 
  • #4
Dick
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Careful. What sign is the charge on an electron?
 
  • #5
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righht, negative, so does that mean that it is accelerating downwards?
 
  • #6
Dick
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righht, negative, so does that mean that it is accelerating downwards?
Don't you believe your equations?
 
  • #7
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I suppose...is there a horizontal component of the acceleration?
 
  • #8
Dick
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I suppose...is there a horizontal component of the acceleration?
Is there a horizontal component of the field?
 
  • #9
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thats what I figured, just making sure.
 
  • #10
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and for part b, would I use :[tex]\vec{E} = k Q / r2 [/tex]?
 
  • #11
Dick
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and for part b, would I use :[tex]\vec{E} = k Q / r2 [/tex]?
No, now that you have the accelerations just use kinematics. How long does it take the electron to go 2cm horizontally? Change in velocity=acceleration*time, etc, etc.
 
  • #12
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pardon my stupidness, but Ive always been bad at this kinematic stuff.
Once I find the change in velocity what do I do?
 
  • #13
Dick
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This is gonna make you really feel dumb, but you asked for it. Add the change in the velocity to the initial velocity to get the final velocity????
 
  • #14
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hahaha, its just one of those days....
 

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