How Long Does It Take for an Electron to Return to Its Initial Height?

AI Thread Summary
An electron projected at a 30.8° angle with an initial speed of 8.20×10^5 m/s in an electric field of 388 N/C takes approximately 1.23e-08 seconds to return to its initial height. The calculations involve determining the components of velocity and the acceleration due to the electric field. The key insight is that the electron's velocity upon returning to its initial height is equal to its initial vertical velocity, leading to the equation V_y = -V_y + at. This results in a factor of two in the time calculation. The discussion emphasizes the parabolic trajectory of the electron and the importance of understanding its velocity at different heights.
Leeoku
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Homework Statement


An electron is projected at an angle of 30.8° above the horizontal at a speed of 8.20×105 m/s in a region where the electric field is E = 388j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height.

Answer: 1.23e-08 s

Homework Equations


v = v_0+at
F = qE = MA


The Attempt at a Solution


Components of Velocity
V_x = Vcos30.8
V_y = Vsin30.8

F = qE = ma
a = qE/m
= 1.6e-19*388/9.11e-31
= 6.81e13

V = v_0+at (set V = 0 because max height)
0 = v_y+at
t = 6.17e-9

Now i was playing with nums and found that if i multiplied V_y by 2 then divided it by a that is the right answer, why?
 
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I think you might have misinterpreted the question. The electron's path is parabolic. What is its velocity going to be when it returns to its initial height? Hint: non-zero.
 
so... if i assume it is a parabola... let's say it is initially at height H.

It starts going to a max with V initial at V_y. It goes to a max and starts coming down. Once it passes height H again.. will it have the same velocity again?

If so... does that mean for my equation v = v_0+at
-V_y = V_y+at
 
That's correct! Now solve that equation, and what do you get? :)
 
oh that's how the 2 comes about. thanks <3
 
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