Velocity of Electron in an electric field

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Homework Statement


At some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.6 x 105 m/s and vy = 3.8 x 103 m/s. Suppose the electric field between the plates is given by (120 N/C). In unit-vector notation, what are (a) the electron's acceleration in that field and (b) the electron's velocity when its x coordinate has changed by 1.4 cm?

Homework Equations



E = F / q

F = m*a

Vf = Vi + at

The Attempt at a Solution



First, I found the force by multiplying the electric field by the charge of the electron. I divided that force by the mass of the electron to find the acceleration and go it wrong. What should I do?
 

Answers and Replies

  • #2
Dick
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You should show us what you actually did. What numbers did you use and what did you get? It's hard to guess. You do know the force will be in the z direction, yes?
 
  • #3
Dick
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In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was 58 m/s (terminal speed), that his mass (including gear) was 85 kg, and that the force on him from the snow was at the survivable limit of 1.2 105 N.

(a) What is the minimum depth of snow that would have stopped him safely?

?m

(b) What is the magnitude of the impulse on him from the snow?


? kg·m/s

PLEASE HELP ON THIS PROBLEM
ANY HELP ON THIS PROBLEM WILL BE GREATLY APPRECIATED

Don't post your own problems on other peoples threads. It's rude.
 
  • #4
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First, I found the force by multiplying the electric field by the charge of the electron. I divided that force by the mass of the electron to find the acceleration and go it wrong. What should I do?

I start with F = E q to find the force. Then I divide the force by the mass because of a=F/m. So the whole thing is as follows:

= (120) (1.602 * 10^-19) / (9.109 * 10^-31) = 2.11044* 10^13

Does this part make sense?
 
  • #5
Dick
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That makes a lot of sense, if you mean m/sec^2. Put units on numbers please.
 
  • #6
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It'd doesn't work though for some reason. I have no idea what I'm doing wrong
 
  • #7
Dick
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What do you mean it doesn't work? Your answer is correct to like five decimal places. At least I get the same thing. Did you put in the direction in terms of the direction vector k? The question asks you to.
 
  • #8
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im having the same trouble with the acceleration i did the same proccess he did and my webassign says its wrong..
 
  • #9
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my numbers are Vx= 1.2 x 10 ^5 Vy=1.7x 10^3 and E= (105 N/C) j


first i tried to get the acceleration with the equation a = eE/m

1.602x10^-19 * 105 / 9.11x10^-31

and i get 1.844 x10^13 i think i did the math right but still it says its wrong...

anyy ideas what i might of done wrong?
 
  • #10
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nvm it was just the negative sign i was missing n.n
 
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