# Electric force between 2 parallel plates

by Oblivion77
Tags: electric, force, parallel, plates
 P: 113 1. The problem statement, all variables and given/known data An electron is projected with an initial velocity of 1.6x10$$^{6}$$ m/s. If the electron just misses the upper plate as it emerges from the field, find the speed of the electron as it emerges from the field? 2. Relevant equations Electric force equation 3. The attempt at a solution I am stuck trying to figure out the magnitude of the electric field, once I can figure this out I know how to solve the problem. Any pointers on how to find the magnitude of the electric field?
 HW Helper P: 5,341 Use ordinary kinematic means to determine the velocity. You know the speed, hence how long for it to emerge. In that time you also know the deflection so you can determine the acceleration. That acceleration then yields the additional sideways component of velocity to calculate it's speed at that point right?
Mentor
P: 41,436
 Quote by Oblivion77 I am stuck trying to figure out the magnitude of the electric field, once I can figure this out I know how to solve the problem. Any pointers on how to find the magnitude of the electric field?
You don't need to know the magnitude of the electric field.

 P: 113 Electric force between 2 parallel plates I was planning to use Vf^2 = Vo^2 +2ad to find the final velocity, but I am missing the acceleration component. To find the acceleration I wanted to use F = ma (knowing the mass of the electron). But I would need the magnitude of the electric field to find the force from E = F / q.
 P: 113 So how would I do this without using the electric field? I don't know what you mean about using the "deflection" to find the acceleration. How would I calculate that?
HW Helper
P: 5,341
 Quote by Oblivion77 So how would I do this without using the electric field?
The horizontal velocity to the end of the plate gives you time.

Use the distance, acceleration, time relationship to determine acceleration.

Then you can use your V2, acceleration and distance.
 P: 113 Thanks for the advice, but this does not work. I get the answer wrong, the magnitude of the electric field is 364N/C, I am just not sure how to find it. When I use this electric field with the method I stated above I get the right answer. I just can't figure out how they got 364N/C :(
 HW Helper P: 5,341 Once you determine the acceleration from the trajectory, then you can use f = ma to determine field intensity. They ask really though for just the speed. That equals (Vx2 +Vy2)1/2
P: 113
 Quote by LowlyPion Once you determine the acceleration from the trajectory, then you can use f = ma to determine field intensity. They ask really though for just the speed. That equals (Vx2 +Vy2)1/2
Yes! Thank you so much, I figured it out now =)
 HW Helper P: 5,341 I wondered if that wasn't it. Glad you got it. Good luck.
 P: 4 Please will someone explain how I am meant to calculate the time taken for the electron to reach the end of the plates. I'm just not getting it.
 P: 767 you have the horizontal distance and the velocity in x-direction is constant, so t = v / s
 P: 4 oh dear. that was rather dense of me. thank you! :)
 P: 767 You're welcome ^^
P: 4
 Quote by songoku you have the horizontal distance and the velocity in x-direction is constant, so t = v / s
Sorry, i'm not getting the right answer..not sure where i'm going wrong..should it not be t=s/v because t=v/s yields an answer of 80,000,000
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P: 41,436
 Quote by NamrataJ Sorry, i'm not getting the right answer..not sure where i'm going wrong..should it not be t=s/v because t=v/s yields an answer of 80,000,000
To check if an equation makes sense, look at the units. t = v/s → (m/s)/s = m/s^2; these are units of acceleration, not time, so this equation makes no sense.

Since v = s/t, t = s/v is correct. The units would be m/(m/s) = m(s/m) = s. Makes sense.
 P: 767 oh sorry, it's my mistake it should be t = s/v
 P: 4 thank you :)

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