Calculating the magnitude of an electric field

[Haddon]

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

 

[neilparker62]

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.

 

[gneill]

The problem statement says:

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.

 

[Haddon]

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.

 

[Haddon]

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?

 

[RPinPA]

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.

 

[gneill]

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.

 

[Haddon]

No. It'll be parabolic, like a thrown stone in a uniform gravitational field.
View attachment 235479

TYSM

 

[Haddon]

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

 

[gneill]

Great! Cheers!

 

[SammyS]

I'm sincerely grateful. I get it now

@Haddon ,

By the way: . . . . .