Help with electric fields, max height, and displacement?

  • Thread starter mohemoto
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  • #1
mohemoto
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Homework Statement


An electron is projected at an angle of 29.1° above the horizontal at a speed of 8.29×105 m/s in a region where the electric field is E = 378j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height, the maximum height it reaches, and the horizontal displacement when it reaches its maximum height.


Homework Equations


I solved for the time it takes for the electron to reach max height, and in doing so got the values for the following:
F = -6.048 x 10^-17 N
a = -6.639 x 10^13 m/s^2
vi = 4.03 x 10^5 m/s
t = 1.21 x 10^-8 s

To solve for max height, I tried using the equation ymax = y0 + vt(max) + 1/2at(max)^2, but it did not produce the correct answer, and I'm not sure why. I also have no idea where to begin to solve for the horizontal displacement. I'm aware that I need the maximum height before I can do so, but I don't know how to generate an equation for it.

The Attempt at a Solution


y(max) = y0 + vt(max) + 1/2at(max)^2
y(max) = (4.03 x 10^5 m/s)(1.21 x 10^-8 s) + (0.5)(6.639 x 10^13 m/s^2)(1.21x10^-8 s)^2
y(max) = 1.48 x 10^-2 m
^this answer is wrong, however.

Thanks for any help or tips!
 

Answers and Replies

  • #2
mohemoto
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Ahh, I see I was not using the correct time for t(max), I was using the time it takes for it to return to the initial position. So I solved the maximum height to be 1.22 x 10^-3 m, and that is correct. I now only need help on finding the displacement!
 
  • #3
ehild
Homework Helper
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The electron moves with constant horizontal velocity if the electric field is vertical.

ehild
 

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