Displacement of an electron between two plates

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SUMMARY

The discussion focuses on the calculation of the displacement of an electron between two deflecting plates in an oscilloscope. An electron is accelerated through a voltage of 1.14x103 V and then deflected by a voltage of 412 V across plates spaced 2.00 cm apart. The correct displacement toward the positive plate is determined to be 7.23x10-3 m, with key calculations involving the electron's velocity, time of travel, and electric field strength. The user initially miscalculated the time of travel along the plates, which was corrected to 2.0x10-9 s.

PREREQUISITES
  • Understanding of basic physics concepts such as electric fields and forces.
  • Familiarity with kinematic equations for motion in electric fields.
  • Knowledge of electron properties, including charge (1.6x10-19 C) and mass (9.11x10-31 kg).
  • Ability to perform calculations involving voltage, time, and displacement.
NEXT STEPS
  • Study the principles of electric fields and forces acting on charged particles.
  • Learn how to apply kinematic equations in the context of charged particle motion.
  • Explore the concept of time of flight in electric fields for charged particles.
  • Investigate the behavior of electrons in oscilloscopes and other electronic devices.
USEFUL FOR

Students studying physics, particularly those focusing on electromagnetism and particle motion, as well as educators and professionals working with oscilloscopes and electronic instrumentation.

oy_katet
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Homework Statement


In an oscilloscope, an electron is accelerated from rest through a voltage of 1.14x103 V. It then passes through deflecting plates 2.00 cm apart, across which there is a voltage of 412 V. Its velocity when it enters the plates is perpendicular to the electric field of the plates. When the electron leaves the plates, what distance will it have moved toward the positive plate? The length of the plates are 4.00 cm.

The answer is 7.23x10-3 m


Homework Equations



Givens:
vi= 0
Va= 1.14x103 V
Vd= 412 V
length of plate = 4 cm = 0.04 m
d= 2 cm= 0.02 m

Equations:
1/2mv2= Vaq
t=length of plate/v
E=Vd/d
F= Eq
a= F/m
x=1/2at2


The Attempt at a Solution



So, because it's an electron
q= 1.6x10-19
m= 9.11x10-31

Using the above equations and plugging in the givens, I've gotten:

v= 2.001x107 m/s

t=1.0x10-9 s

E= 20600 v/m

F= 3.296x10-15 N

a= 3.618x1015 m/s2

x= 1.809x10-3 m


which is not the right answer :( I'm not very good at physics so I'm having a hard time figuring out what I'm doing wrong.
 
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oy_katet said:
Using the above equations and plugging in the givens, I've gotten:

v= 2.001x107 m/s
Correct...

oy_katet said:
t=1.0x10-9 s

here is a mistake: The electron travels along the length of the plate with the v=2.00 m/s so the time is 2.0x10-9 s

ehild
 
ehild said:
Correct...



here is a mistake: The electron travels along the length of the plate with the v=2.00 m/s so the time is 2.0x10-9 s

ehild

Wellp, I'm just a big ol' dummy! I can't believe I made that oversight. Thanks so much!
 

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