
#1
Jul707, 10:26 AM

P: 15

1. The problem statement, all variables and given/known data
An electron is projected at an angle of 31.7° above the horizontal at a speed of 8.40×10^5 m/s in a region where the electric field is E = 383j N/C. Neglecting the effects of gravity, calculate the time it takes the electron to return to its initial height. 2. Relevant equations I first found the electric force that the E field exerts on the electron by using F=qE. Then I found the acceleration of the electron with a=(q/m)E Thought perhaps electron would undergo motion similar to a projectile, so then first found the y component of the velocity sin 31.7=opp/8.40x10^5s. Then applied vf=vi + at solving for t (with vf=0 because it will be when the electron has all potential energy at its highest point) 3. The attempt at a solution F=383x(1.6x10^16C) =6.13x10^14N a=6.13x10^14N/9.11x10^31kg (mass of electron) =6.73x10^16m/s^2 sin 31.7=opp/8.40x10^5m/s opp= 4.41x10^5m/s vf=vi +at 0=4.41 + 6.73x10^16m/s^2t t=6.55x10^12s (this would be the time to reach max height, so the amount of time to get back down to its initial height would be double this) so t= 1.31x10^11s But this answer is not right... and I'm not show how else to relate t to an electric field. Can anyone point out what i'm doing wrong here?? Thanks SOOO much!! :) 



#2
Jul707, 10:43 AM

Sci Advisor
HW Helper
Thanks
P: 25,175

Uh, the exponent of the electron charge is 19, not 16. But there is nothing wrong with your method.




#3
Jul707, 10:57 AM

P: 15

omg... wow... I can't that was all it was!!
THANK YOUUU DICK!!! :D 


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