# Electron Acceleration & movement between a uniform magentic field

#### HarleyM

1. Homework Statement
An electron is fired at 4.0x106 m/s horizontally between the parallel plates as shown, (see diagram) starting at the negative plate. The electron deflects downwards and strikes the bottom plate. The magnitude of the electric field between the plates is 4.0 x102 N/C. The seperation of the plates is 2.0 cm

A) find the acceleration of the electron
b)find the horizontal distance traveled by the electron when it hits the plate
c) find the velocity of the electron as it strikes the plate

2. Homework Equations

a)ma=εq
b) Δdy=V1yΔT + 1/2 aΔT2
ΔDx= V1xΔT
qelectron= -1.6x10-19
Melectron= 9.11x10^ -31 kg
3. The Attempt at a Solution

A) ma=εq
a= (4.0x10^2)(1.6x10^-19)/(9.11x10^-31)
a= 7.03 x10^13 m/s^2 [down]

B)Δdy=V1yΔT + 1/2 aΔT2
0.02=0ΔT +1/2 (7.03x1013ΔT2 (the 0Δ T cancels?)
ΔT= √(0.02)(2)/(7.03x1013
ΔT= 2.3x10-8s

ΔDx= V1xΔT
ΔDx= (4.0x10^6)*(2.3x10^-8)
=0.092 m
=9.2 cm

C) V2x=√V1x + 2aΔdy
V2= √(4.06x10^6)+2(7.03x10^13)(0.02) <--(should I use horizontal or vertical distance here)
V2= 4.34x10^6 m/s ( it gets faster which is a good sign lol)

My biggest confusion is when to use vertical distance 2 cm or horizontal distance 9.2 cm .

I took a lot of time to write this up so any help is appreciated immensely. I feel like I did everything right I just need re assurance. Im a perfectionist and I want to get full marks.

Sorry forgot to attach diagram, its crude but its exactly what is happening Last edited:
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#### technician

You know the electric field strength = 400N/C so you can find the force on an electron.
If you know the Force and the mass of an electron then you can calculate the acceleration.
You know the horizontal velocity of the electron so the question is a sort of projectile question..... does this help??

#### HarleyM

Uhh to be honest, It has only made me second guess myself lol :P

#### technician

OK... I will go through your numbers
I used F = E x q to get Foce on electron = 6.4 x 10 ^-17N
Then used F= ma to get acceleration a = 7.05 x 10^13 .... same as you
I got time to move between the plates (use 2cm as S in equation S = 0.5at^2) to be
2.39 x 10^-6 s (you got x 10^-8).... I will check mine again
(I think you have made a mistake where you have written 'the ΔT cancels'....it does not !!!)

I will be back soon with my complete calculation....

I made a mistake... also got the time to be 2.38 x 10^-8 s....
back soon

I got the same as you for the velocity..... the length of the plates does not come into the calculation.... they tell you the electron hits the plates and that confirms it....

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#### technician

You got it right..... do you want a challenge?????
At what angle does the electron strike the plate ????
I am surprised they did not ask you that #### HarleyM

Thank you very much for the help, as for the challenge I have no real idea how to even begin to find the angle..

I appreciate the input however, you seem to be helping me a lot lately !

Good vibes :)

#### Vanishd

You got it right..... do you want a challenge?????
At what angle does the electron strike the plate ????
I am surprised they did not ask you that The question asked for Velocity. Not speed. Velocity has direction.
in this case, the direction would be the angle at which it strikes the plate.
evidently they did ask you for that :P

Tan(theta) = V_f_y / V_f_x
(theta) = 23degrees

"Therefore the final velocity of the electron was 4.3 x 10^6 m/s [ E 26(degrees) S ]"

P.S. Also remember to show in your diagram which direction is North. (You can use up,down,left,right, or [Towards negative plate] if you wish

Last edited:

#### HarleyM

The question asked for Velocity. Not speed. Velocity has direction.
in this case, the direction would be the angle at which it strikes the plate.
evidently they did ask you for that :P

Tan(theta) = V_f_y / V_f_x
(theta) = 23degrees

"Therefore the final velocity of the electron was 4.3 x 10^6 m/s [ E 26(degrees) S ]"

P.S. Also remember to show in your diagram which direction is North. (You can use up,down,left,right, or [Towards negative plate] if you wish
How are you getting Vfx I used your answer of 23 degrees and solved backwards to get a vfx that should equal about 1.08x10-5 but I can't seem to get that answer as my vfx for the life of me.

Thanks!

#### HarleyM

Bump!

I hope im allowed to bump threads i need an answer !

#### technician

I said I would post my calculations and forgot !!!!! Here are the vital steps.
Force on electron = 6.4 x 10^-17 N
Gives vertical acceleration = 7 x 10^13 m/s^2
using separationof plates as distance travelled this gives vertical velocity = 1.67 x 10^6 m/s
Horizontal velocity is constant therefore velocity of electron is given by:
v^2 = (4 x 10^6)^2 + (1.67 x 10^6)^2
gives velocity = 4.34 x 10^6 m/s
Just as Vanished has pointed out
The angle of the velocity is given by Tanθ = vertical velocity/horizontal velocity
Tanθ = 1.67 x 10^6/4 x 10^6 = 0.4175 gives θ = 23 degrees to horizontal
Hope this is what you wanted.... ask again if you are not sure of anything

#### HarleyM

I said I would post my calculations and forgot !!!!! Here are the vital steps.
Force on electron = 6.4 x 10^-17 N
Gives vertical acceleration = 7 x 10^13 m/s^2
using separationof plates as distance travelled this gives vertical velocity = 1.67 x 10^6 m/s
Horizontal velocity is constant therefore velocity of electron is given by:
v^2 = (4 x 10^6)^2 + (1.67 x 10^6)^2
gives velocity = 4.34 x 10^6 m/s
Just as Vanished has pointed out
The angle of the velocity is given by Tanθ = vertical velocity/horizontal velocity
Tanθ = 1.67 x 10^6/4 x 10^6 = 0.4175 gives θ = 23 degrees to horizontal
Hope this is what you wanted.... ask again if you are not sure of anything

Uh, how is vertical velocity found, its driving me insane. I know you sort of explained it but Ive tried to find it, and tried to get the same number as you but I can't.

If you can explain it a little further that would be great !

thanks

#### Delphi51

Homework Helper
Vertically you have constant force so constant accelerated motion, so use Vf = Vi + a*t.
Vi is zero vertically. Never mind signs; it accelerates down and you know the velocity is downward.

I think you will be disappointed with those answers because they are not very accurate after part (a). You have 2.3 x 10^-8 for the time. I have 2.386. If you want 2 digit accuracy in final answers after this point you must start with 3 digit accuracy for the time. If you want 3 in final answers, keep 4 along the way.

#### technician

Do you see the vertical acceleration is 7.0 x 10^13 m/s^2 ?
I then used the distance between th 2 plates (2 x 10^-2 m) to find the vertical component of velocity using
V^2 = u^2 + 2aS and got v =1.7 x 10^6 m/s (1.68 to 3 figures)

As far as significant figures are concerned you are not given any numbers to more than 2 significant figures so it is not legitimate to give an answer to more than 2 figures
I used 3 figures in my calculations
by definition any more than 2 figures is insignificant

Hope you see the physics behind the calculation