# Electron moving through uniform magnetic field

1. Mar 24, 2015

### MiniOreo1998

1. The problem statement, all variables and given/known data

An electron is fired at 4.0 x 10^6 m/s horizontally between the parallel plates, as shown, 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 x 10^2 N/C. The separation of the plates is 2.0 cm.

A) Find the acceleration of the electron between the plates.

B) Find the horizontal distance travelled by the electron when it hits the plate.

C) Find the velocity of the electron as it strikes the plate.

Starting to pick up steam and build some confidence. Still shaky on "proper notation" without flipping through pages, but I'm getting there (it's really just the beginning of each equation that makes me "iffy", I'm getting the hang of what order things should be in (which I'm starting to see will keep coming easier the more I work at it). Any suggestions are warmly invited!

2. Relevant equations

m a = ε q
Δdy = v1y Δt + 1/2 a Δt2
Δdx = v1x Δt
v2x = √ v1x + 2a Δdy

3. The attempt at a solution

A)

m a = ε q
a = ε q / m

a = (1.6 x 10-19) (4.0 x 102 / 9.11 x 10-31
a = 7.03 m/s2 [down]

B)

Δdy = v1y Δt + 1/2 a Δt2
0.2 = 0 Δt + 1/2 (7.03 x 1013)
Δt = (0.02) (2) / (7.03 x 1013)
Δt = 2.3 x 10-8

Δdx = v1 Δt
Δdx = (4.0 x 106) (2.3 x 10-8)
Δdx = 0.092 m
Δdx = 9.2 cm

C)

v2x = √ v1 + 2a Δdy
v2 = (4.0 x 106) + 2 (7.03 x 1013) (0.02)
v2 = 4.34 x 106 m/s2 [down]

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2. Mar 24, 2015

### ehild

There are some typing errors in your work, but the method and the results are correct. The vertical velocity component could have been easier to get from v=vo+aΔt. You knew the time and that vo=0.
And the field is electric, not magnetic as you wrote in the title.

Last edited: Mar 24, 2015
3. Mar 24, 2015

### BvU

You did quite well|: work with symbols, find right values, etc.
The working with symbols could be kept up a little longer: my Δdx = 9.5 cm because Δt = 2.38 10-8 s. You should continue to calculate with the correct value but for results you report a reasonably rounded value (2.4 in this case)

You do want to check more thoroughly afterwards: the 1013 you found and used in b is missing in a.
The 0.2 in b should be 0.02 (2 cm)
and the first term in the sqrt in c misses a square.
the last answer doesn't have the dimension of a velocity.
The numerical value of the answer is correct though, so you must have done the right thing.

One extra: velocity is a vector. You state 4.34 106 [down].
Bad-mood markers jump on that. Either 4.34 106 [$\theta = -.397$], or (vx, vy) = (4, -1.68) 106 m/s