Effects of an electric field on a motion of an electron

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SUMMARY

The discussion focuses on calculating the time it takes for an electron, projected at an angle of 31.7° with an initial speed of 8.40×105 m/s, to return to its initial height in an electric field of E = 383j N/C. The participant correctly applies the formula F=qE to find the electric force and subsequently calculates the acceleration of the electron using a=(q/m)E. The initial attempt yields a time of 1.31x10-11 seconds, which is later corrected due to a miscalculation of the electron charge exponent, confirming the importance of precise values in physics calculations.

PREREQUISITES
  • Understanding of electric fields and forces (F=qE)
  • Knowledge of kinematic equations (vf=vi + at)
  • Familiarity with projectile motion concepts
  • Basic knowledge of electron properties (charge and mass)
NEXT STEPS
  • Review the principles of electric fields and their effects on charged particles
  • Study kinematic equations in the context of non-gravitational forces
  • Explore the relationship between electric force and acceleration in charged particle motion
  • Investigate the significance of precision in scientific calculations, particularly in physics
USEFUL FOR

Students and educators in physics, particularly those focusing on electromagnetism and kinematics, as well as anyone interested in the motion of charged particles in electric fields.

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Homework Statement



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.

Homework 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)


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! :)
 
Physics news on Phys.org
Uh, the exponent of the electron charge is -19, not -16. But there is nothing wrong with your method.
 
omg... wow... I can't that was all it was!

THANK YOUUU DICK! :D
 

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