Velocity of Electron in an electric field

In summary, the conversation involves finding the acceleration and velocity of an electron moving between two charged parallel plates using unit-vector notation. The force is found by multiplying the electric field by the charge of the electron and dividing by the mass. Some individuals were having trouble with their calculations, but it was determined that the issue was with units and negative signs.
  • #1
evanist00
3
0

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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  • #2
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
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.
 
  • #4
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?
 
  • #5
That makes a lot of sense, if you mean m/sec^2. Put units on numbers please.
 
  • #6
It'd doesn't work though for some reason. I have no idea what I'm doing wrong
 
  • #7
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
im having the same trouble with the acceleration i did the same process he did and my webassign says its wrong..
 
  • #9
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
 

What is the velocity of an electron in an electric field?

The velocity of an electron in an electric field depends on the strength of the electric field and the mass of the electron. It can be calculated using the equation F = ma, where F is the force applied by the electric field, m is the mass of the electron, and a is the acceleration. The velocity can then be determined using the equation v = u + at, where u is the initial velocity and t is the time.

How does an electric field affect the velocity of an electron?

An electric field exerts a force on an electron, causing it to accelerate. The direction of the acceleration is determined by the direction of the electric field. As the electron accelerates, its velocity increases until it reaches a constant velocity, also known as the drift velocity. The strength of the electric field determines the magnitude of the force and therefore, the velocity of the electron.

What factors influence the velocity of an electron in an electric field?

The velocity of an electron in an electric field is influenced by the strength of the electric field, the mass of the electron, and any external forces acting on the electron. Additionally, factors such as the type of material the electron is moving through, temperature, and the presence of impurities can also impact the velocity of the electron.

How does the velocity of an electron in an electric field relate to current flow?

The velocity of an electron in an electric field is directly related to the current flow. When an electric field is applied, the electrons in a conductor accelerate and move in the direction of the electric field, creating a flow of charge or current. The greater the velocity of the electrons, the higher the current flow. This is why a stronger electric field can result in a higher current.

Can the velocity of an electron in an electric field be controlled?

Yes, the velocity of an electron in an electric field can be controlled by adjusting the strength of the electric field. By altering the strength of the electric field, the force applied to the electron can be changed, thus affecting its velocity. Additionally, the presence of external forces or factors, such as impurities, can also affect the velocity of the electron. Therefore, by controlling these factors, the velocity of the electron can be controlled to some extent.

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