Electron moving between 2 charged plates

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SUMMARY

The discussion centers on the motion of an electron between two charged parallel plates, with specific velocity components of vx = 1.20 x 105 m/s and vy = 3.70 x 103 m/s, and an electric field strength of E = 120 N/C in the j direction. The user calculated the electron's acceleration using the formula A = (E*Q)/M, resulting in an acceleration of 2.11 x 1013 m/s2 in the negative j direction. The user also computed the final velocity after the x coordinate changed by 1.4 cm, yielding a final velocity of -2.46 x 106 m/s in the j direction. Ultimately, the user discovered that their calculations were correct, but the input format was not accepted by the system.

PREREQUISITES
  • Understanding of classical mechanics, specifically Newton's laws of motion.
  • Familiarity with electric fields and forces, particularly F = E*Q.
  • Knowledge of kinematic equations, including V_0 + A*T = V_1.
  • Basic proficiency in scientific notation and unit conversions.
NEXT STEPS
  • Review the application of Newton's second law in electric fields.
  • Explore the concept of electric field strength and its effects on charged particles.
  • Study kinematic equations in detail, focusing on their application in two-dimensional motion.
  • Investigate common formatting issues in physics problem-solving software.
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and motion of charged particles, as well as educators looking for examples of problem-solving in electric fields.

juggalomike
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Homework Statement


at some instant the velocity components of an electron moving between two charged parallel plates are vx = 1.20 105 m/s and vy = 3.70 103 m/s. Suppose that the electric field between the plates is given by Evec = (120 N/C) jhat.

(a) What is the electron's acceleration in the field?
-2.11*E13 m/s2 jhat Incorrect

(b) What is the electron's velocity when its x coordinate has changed by 1.4 cm?
1.20*E5 m/s ihat + -2.46*E6 m/s jhat Incorrect


Homework Equations


F=ma
F=E*Q
V*T=D
V_0+A*T=V_1

The Attempt at a Solution


Rearranging the above equations i get A=(E*Q)/M

so A=(120*1.6*10^-19)/(9.1*10^-31)=2.11*10^13 jhat in the negative direction

Because acceleration is in the J direction there is no Iforce, so I is constant. so for D=t*V or t=d/v t=.014m/1.2*10^5=1.17*10^-7

then V_F=V_I+A*T so V_F=3.7*10^3+-2.11*10^13*1.17^10^-7=-2.46*10^6

Yet all of my answers are incorrect, what am i doing wrong?
 
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Never mind, my answer was correct but the system wasn't accepting the format i was inputting it as(which is ironically the format they fed the information to me in)
 

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