Velocity of Electron in an electric field

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Homework Help Overview

The discussion revolves around the calculation of an electron's acceleration and velocity while moving in an electric field between charged parallel plates. The problem involves understanding the relationship between electric fields, forces, and motion in the context of particle dynamics.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of force using the electric field and the charge of the electron, followed by finding acceleration through division by mass. There are questions about the correctness of the calculations and the inclusion of direction in the results.

Discussion Status

Some participants have provided calculations and expressed confusion over discrepancies in their results. Others have confirmed the correctness of certain calculations while questioning the inclusion of directional components. The discussion is ongoing with attempts to clarify misunderstandings.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific formatting or answer requirements that are contributing to confusion. There is mention of a web-based assignment system that is marking answers as incorrect, prompting further inquiry into possible errors.

evanist00
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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?
 
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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?
 
physicsbhelp said:
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.
 
evanist00 said:
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?
 
That makes a lot of sense, if you mean m/sec^2. Put units on numbers please.
 
It'd doesn't work though for some reason. I have no idea what I'm doing wrong
 
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.
 
im having the same trouble with the acceleration i did the same process he did and my webassign says its wrong..
 
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
nvm it was just the negative sign i was missing n.n
 

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