What is the speed of the electron as it strikes the second plate?

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Homework Statement




A uniform electric field exists in a region between two oppositely charged plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of the opposite plate, 2.6 cm away, in a time 3.8 x 10-8 s. (a) What is the speed of the electron as it strikes the second plate? (b) What is the magnitude of the electric field?

Homework Equations



F = q E
E = F/ q = q/4pi[tex]\epsilon[/tex] r^2
V[tex]_{F}[/tex]^2 = V[tex]_{0}[/tex]^2 + 2a( x - x[tex]_{0}[/tex])
F = ma

The Attempt at a Solution



I thought finding the speed was obvious
i just converted the cm into meters and divided by the time
it came out to about 6.8e5
...which was wrong

I know how to relate the E field equations to speed and acceleration
and this was my final simplified eqation without pluging in the numbers

E = (V^2)m / (2 q x)

q was the charge of an electron
x was 0.026 meters
V was the 6.8e5
m was the mass of an electron

is the velocity the only thing holding me back from the right answer?
or is there something else that's wrong?
 

Answers and Replies

  • #2
collinsmark
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Homework Equations



F = q E
E = F/ q = q/4pi[tex]\epsilon[/tex] r^2
V[tex]_{F}[/tex]^2 = V[tex]_{0}[/tex]^2 + 2a( x - x[tex]_{0}[/tex])
F = ma

The Attempt at a Solution



I thought finding the speed was obvious
i just converted the cm into meters and divided by the time
it came out to about 6.8e5
...which was wrong
Yes, it is not correct. What you've done is assumed the electron moved from one plate to the other at a constant velocity.

But it doesn't do that. It experiences a constant force, meaning it travels from one plate to the other under constant acceleration. Use your kinematics equations for constant acceleration. (You have one already listed in your relevant equations, but you'll need to use at least one other one that incorporates time).
I know how to relate the E field equations to speed and acceleration
and this was my final simplified eqation without pluging in the numbers

E = (V^2)m / (2 q x)

q was the charge of an electron
x was 0.026 meters
V was the 6.8e5
m was the mass of an electron

is the velocity the only thing holding me back from the right answer?
or is there something else that's wrong?
Yeah, I think your simplified equation is correct for this particular problem. Just find the correct velocity. :smile:
 
Last edited:
  • #3
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yesss
thank you X)
 

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