Finding the angle of an electron striking a plate

[chef99]

Homework Statement

An electron is fired at 4.0 x 106 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 102 N/C. The separation of the plates is 2.0 cm.

a) Find the acceleration of the electron between the platesa) Find the horizontal distance traveled by the electron when it hits the plate

Homework Equations

F_{net y} = F_E - F_g

ΔD_y =v_1y Δt + 1/2 a Δt²

ΔD_x = v_1x Δt

v_f = v_i + (a)(t)

The Attempt at a Solution

a)

Let right and down be positive.

F_{net y} = F_E - F_gma = qε - 0a = qε / ma =https://www.physicsforums.com/file://localhost/Users/jefferyhewitt/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image002.png (1.6 x10^-19C)(4.0 x10²N/C) / (9.11 x10^-31kg)a = 7.03 x10¹³m/s² [down]The acceleration of the electron is 7.0 x1013m/s2 [down].

b)

In order to calculate the distance traveled, the value of time (https://www.physicsforums.com/file://localhost/Users/jefferyhewitt/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image001.pngt) must be determined.https://www.physicsforums.com/file://localhost/Users/jefferyhewitt/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image002.pngΔD_y =v_1y Δt + 1/2 a Δt²0.02m = 0 Δt +https://www.physicsforums.com/file://localhost/Users/jefferyhewitt/Library/Group%20Containers/UBF8T346G9.Office/msoclip1/01/clip_image007.png1/2 (7.03 x1012) ΔtΔt = √(0.02m)(2) / (7.03 x 10¹³)Δt = 2.38 x10-8 sNow the value of time can be used to determine the horizontal distance traveled by the electron:

ΔD_x = v_1x Δt

ΔD_x = (4.0 x10⁶m/s)(2.38 x10^-8s)

ΔD_x = 9.5m

The horizontal distance traveled by the electron is 9.5m.c)

vf = vi + (a) (t)vf = (4.0 x106m/s)(7.03 x1013m/s)(2.38 x10-8s)vf = 6.69 x1012 m/sThe velocity of the electron as it strikes the plate is 6.7 x1012 m/s [ ].I think this is all right but, the one thing I don't know how to do but I think I have to determine the angle at which the electron strikes the plate. If anyone has suggestions I would greatly appreciate it.

 

[haruspex]

ΔD_x = (4.0 x10⁶m/s)(2.38 x10^-8s)

ΔD_x = 9.5m
...
vf = vi + (a) (t)
vf = (4.0 x106m/s)(7.03 x1013m/s)(2.38 x10-8s)

Check those two steps.

 

[chef99]

Check those two steps.

Thanks for pointing that out.

So
ΔDx = (4.0 x106m/s)(2.38 x10-8s)

ΔDx = 0.0952

ΔDx = 9.5 x10^-2m
and

v_f = v_i + (a) (t)

v_f = (4.0 x10⁶m/s)(7.03 x10¹³m/s)(2.38 x10^-8s)

v_f = 5673140

v_f = 5.7 x10⁶m/s

Also for c) do I have to determine the angle the electron strikes that plate or not?

 

[haruspex]

v_f = v_i + (a) (t)

v_f = (4.0 x10⁶m/s)(7.03 x10¹³m/s)(2.38 x10^-8s)

this is still not right. You need to distinguish between directions. The initial velocity is at right angles to the acceleration.
Also, you did not post the question for part c. If it asks for a velocity then the answer should consist of a speed and a direction.