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Electric Field Magnitude

  • Thread starter Xels
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Homework Statement


The figure shows an electron entering a parallel-plate capacitor with a speed of 5.4×106m/s. The electric field of the capacitor has deflected the electron downward by a distance of 0.618cm at the point where the electron exits the capacitor.

Find the magnitude of the electric field in the capacitor.


Homework Equations


f=ma
f=k(q/r2)

The Attempt at a Solution



I'm trying to figure out how to gauge the force applied from the field, which I'm assuming can be taken from the displacement & speed traveled. I'm also assuming that gravity isn't a factor here, so if E=f/q0 where am I getting my f from?

Relevant data:
length of capacitor: 2.25E-2m
e-m: 9.11E-31kg
e-q: 1.60E-19C
e-v: 5.4E6m/s
[tex]\theta[/tex]:15.36[tex]\circ[/tex]
 
Last edited:

Answers and Replies

  • #2
cepheid
Staff Emeritus
Science Advisor
Gold Member
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You're getting your f from basic dynamics (Newton's laws of motion). In a parallel plate capacitor, the electric field is uniform (constant in magnitude and direction). So, throughout its journey across the capacitor plates, the electron has a constant vertical force applied to it. Therefore, it experiences a constant vertical acceleration (vertical meaning perpendicular to its direction of motion). You know how long it takes it to traverse the plates, because you know its horizontal speed (which won't change) and you know the length of the plates. Call this time interval "t". So, the question reduces to: a particle undergoes a displacement d in a time t when moving with constant acceleration a. What must "a" have been in order for it to have travelled that far in that amount of time? Once you know a, you know f.

This is no different from a projectile motion problem. Only the nature of the force (electrostatic vs. gravitational) has changed.
 
  • #3
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Great, thats exactly the kick I needed. Thank you.
 
  • #4
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so what is exactly the answer?
 

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