Calculating the magnitude of an electric field

In summary: Physics Forums ! :) In summary, an electron enters a parallel plate capacitor at the lower left side and exits at the upper right side, clearing the upper plate. The initial velocity of the electron is 7*10^6m/s parallel to the plates. With a capacitor length of 2cm and plate separation of 0.15m, assuming a uniform electric field between the plates, the magnitude of the electric field is found to be 2.09*10^5N/C. The path of the electron is parabolic, similar to that of a thrown stone in a uniform gravitational field.
  • #1
Haddon
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1

Homework Statement


An electron enters the lower left side of a parallel plate capacitor and exits precisely at the upper right side (just clearing the upper plate). The initial velocity of the electron is 7*10^6m/s parallel to the plates. The capacitor is 2cm long and its plates are separated by 0.15m. Assume that the electric field is uniform at every point between the plates and find its magnitude.
THE ANSWER IS = 2.09*10^5N/C
Don't mind my weak attempt at diagram.

Homework Equations


U(y-direction) = U(x-direction)*tanΦ
F=ma=qE
x=U(x-direction)t
y=U(y-direction)+½at2

The Attempt at a Solution


tanθ=0.15/0.02=7.5
U(x-direction)= 7*106m/s
U(y-direction)= U(x-direction)*tanΦ= 7.6*106*7.5=5.25*107m/s
x=2cm=0.02m, y=0.15m
x=U(x-direction)*t
0.02 = 7*106*t
t = 2.86*10-9s
y = U(y-direction)+½at
0.15 = 5.25*107*2.86*10-9 + ½(2.86*10-9)2a
0.15=0.15015 + 4.0898*10-18a
a= 3.668*1013ms-2
F = ma = qE
= 9.11*10-31 * 3.668*1013 = qE
⇒ 1.6*10-19 * E = 3.3415*10-17
E=208.85NC-1
 

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  • #2
I would guess you don't need any angle since the initial direction is parallel to plate and that component of velocity will not change. You just use it to get Δt which can then be used to determine acceleration perpendicular to plate.
 
  • #3
The problem statement says:
Haddon said:
The initial velocity of the electron is 7*10^6m/s parallel to the plates.
(my emphasis on the direction). Why have you assigned a y-component to the initial velocity?

Your diagram showing a diagonal trajectory does not make sense. The electron will be accelerating in the y-direction, remaining at constant velocity in the x-direction (no forces acting that way). So the path will be parabolic in form.
 
  • #4
gneill said:
The problem statement says:

(my emphasis on the direction). Why have you assigned a y-component to the initial velocity?

Your diagram showing a diagonal trajectory does not make sense. The electron will be accelerating in the y-direction, remaining at constant velocity in the x-direction (no forces acting that way). So the path will be parabolic in form.

But the electron entered at the lower left side and exited at the upper right side of the capacitor.. That's from the question. So, I thought it's path should be diagonal.
 
  • #5
neilparker62 said:
I would guess you don't need any angle since the initial direction is parallel to plate and that component of velocity will not change. You just use it to get Δt which can then be used to determine acceleration perpendicular to plate.
I think I did that.. from my solution but are you insinuating I use the same velocity?
 
  • #6
Haddon said:
But the electron entered at the lower left side and exited at the upper right side of the capacitor.. That's from the question. So, I thought it's path should be diagonal.

Its path is a curve, but INITIALLY the velocity is horizontal and the U in those equations is INITIAL velocity.

The idea is the same as throwing a ball horizontally off a cliff. Initial velocity is all horizontal. In a ball off a cliff, gravity acts on one component but not the other, so the equations you use for one component are accelerated and the equations for the other component use a = 0.
 
  • #7
Haddon said:
But the electron entered at the lower left side and exited at the upper right side of the capacitor.. That's from the question. So, I thought it's path should be diagonal.
No. It'll be parabolic, like a thrown stone in a uniform gravitational field.
upload_2018-12-9_10-27-33.png
 

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  • #8
gneill said:
No. It'll be parabolic, like a thrown stone in a uniform gravitational field.
View attachment 235479
TYSM
 
  • #9
gneill said:
No. It'll be parabolic, like a thrown stone in a uniform gravitational field.
View attachment 235479
I'm sincerely grateful. I get it now
 
  • #10
Great! Cheers!
 
  • #11
Haddon said:
I'm sincerely grateful. I get it now
@Haddon ,

By the way: . . . . :welcome: .
 
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Related to Calculating the magnitude of an electric field

1. What is the formula for calculating the magnitude of an electric field?

The formula for calculating the magnitude of an electric field is E = F/q, where E is the electric field strength in newtons per coulomb (N/C), F is the force in newtons (N), and q is the charge in coulombs (C).

2. How is the magnitude of an electric field affected by distance?

The magnitude of an electric field is inversely proportional to the distance from the source charge. This means that as the distance increases, the magnitude of the electric field decreases.

3. Can the magnitude of an electric field be negative?

Yes, the magnitude of an electric field can be negative. This indicates that the direction of the electric field is opposite to the direction of the force on a positive test charge placed in the field.

4. How does the magnitude of an electric field change with increasing charge?

The magnitude of an electric field is directly proportional to the amount of charge present. This means that as the charge increases, the magnitude of the electric field also increases.

5. What is the unit for measuring the magnitude of an electric field?

The unit for measuring the magnitude of an electric field is newtons per coulomb (N/C). This unit represents the force per unit charge experienced by a test charge in the electric field.

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