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Electrons moving in an electric field

  • #1
74
1

Homework Statement


Suppose electrons enter the electric field midway between two plates at an angle θ0 to the horizontal, as shown in the figure, where L = 5.1 cm and H = 1.1 cm. The path is symmetrical, so they leave at the same angle θ0 and just barely miss the top plate. What is θ0? Ignore fringing of the field.

21-66alt.gif


Homework Equations


F = qE

The Attempt at a Solution


Not sure where to start but I'm assuming I have to break things into components and that there's trig involved...
 

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
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Not sure where to start but I'm assuming I have to break things into components and that there's trig involved...
Very good - that is an excellent start.
You have actually done problems like this before.
Hint: the electron is a projectile fired at an angle to the horizontal in a uniform vertical force field
 
  • #3
Indeed you can solve it by using projectile motion. By getting the ratio of maximum heights (H) and the range (L), you can determine the angle.
 
  • #4
74
1
Got the answer! Thanks!
 
  • #5
gneill
Mentor
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2,730
Indeed you can solve it by using projectile motion. By getting the ratio of maximum heights (H) and the range (L), you can determine the angle.
Can you demonstrate that mathematically? Somehow I doubt that it is so o_O
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
17,841
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Got the answer! Thanks!
For the benefit of someone else with the same issue, please post what you did.
 
  • #7
Can you demonstrate that mathematically? Somehow I doubt that it is so o_O
The motion of electrons can be calculated by using the equations of projectile motion.
The electrons reach the maximum height at
H= v2*(sin(θ))2/2*g
and leave the plate at the range of
R=v2sin(2θ)/g
(Note: H is the distance between two plates, R is the length of plate, v is the initial velocity of electron, g is the acceleration acts on the electron )
by getting the ratio, we can eliminate the v and g by division thus leads to
H/R= sin2(θ)/2sin(2θ)
from the identity of trigonometry we know that 2sin(2θ)=2sinθcosθ
Therefore,
tanθ=2H/R. Put in the values of H and R we get
θ=tan-1(2H/R)
 
  • #8
gneill
Mentor
20,733
2,730
The motion of electrons can be calculated by using the equations of projectile motion.
The electrons reach the maximum height at
H= v2*(sin(θ))2/2*g
and leave the plate at the range of
R=v2sin(2θ)/g
(Note: H is the distance between two plates, R is the length of plate, v is the initial velocity of electron, g is the acceleration acts on the electron )
by getting the ratio, we can eliminate the v and g by division thus leads to
H/R= sin2(θ)/2sin(2θ)
from the identity of trigonometry we know that 2sin(2θ)=2sinθcosθ
Therefore,
tanθ=2H/R. Put in the values of H and R we get
θ=tan-1(2H/R)
Nice. I didn't see that as what you were implying from your hint.
 
  • #9
1
0
The motion of electrons can be calculated by using the equations of projectile motion.
The electrons reach the maximum height at
H= v2*(sin(θ))2/2*g
and leave the plate at the range of
R=v2sin(2θ)/g
(Note: H is the distance between two plates, R is the length of plate, v is the initial velocity of electron, g is the acceleration acts on the electron )
by getting the ratio, we can eliminate the v and g by division thus leads to
H/R= sin2(θ)/2sin(2θ)
from the identity of trigonometry we know that 2sin(2θ)=2sinθcosθ
Therefore,
tanθ=2H/R. Put in the values of H and R we get
θ=tan-1(2H/R)
2sin(2θ)=2sinθcosθ???
Shouldn't it be sin(2A)=2sinAcosA?
 

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