A screen for detecting electrons

[C6ZR1]

Homework Statement

In Fig. 22-56 an electron is shot at an initial speed of v0 = 4.14 × 106 m/s, at angle θ0 = 43.6 ˚ from an x axis. It moves through a uniform electric field = (5.34 N/C). A screen for detecting electrons is positioned parallel to the y axis, at distance x = 1.64 m. What is the y component of the electron's velocity (sign included) when the electron hits the screen?

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c22/q50.jpg


Given:
θ=43.6˚
Vi=4.14 E 6 m/s
E=5.34 N/C
x=1.64 m

Homework Equations

E=(KQ)r^2

The Attempt at a Solution

What I first decided to do was try breaking V into x and y components..
x direction: 4.14 E6 (cos 43.6) ≈2.998 E 6
y direction: 4.14 E6 (sin 43.6) ≈ 2.855 E 6

After I did that I started to get confused on what other step I should take. Would I add the vector E to Y component then use basic trig to find the height at that x distance?

Help, please and thanks :)

 

[SammyS]

Homework Statement

In Fig. 22-56 an electron is shot at an initial speed of v0 = 4.14 × 106 m/s, at angle θ0 = 43.6 ˚ from an x axis. It moves through a uniform electric field = (5.34 N/C). A screen for detecting electrons is positioned parallel to the y axis, at distance x = 1.64 m. What is the y component of the electron's velocity (sign included) when the electron hits the screen?

http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c22/q50.jpg


Given:
θ=43.6˚
Vi=4.14 E 6 m/s
E=5.34 N/C
x=1.64 m

Homework Equations

E=(KQ)r^2

The Attempt at a Solution

What I first decided to do was try breaking V into x and y components..
x direction: 4.14 E6 (cos 43.6) ≈2.998 E 6
y direction: 4.14 E6 (sin 43.6) ≈ 2.855 E 6

After I did that I started to get confused on what other step I should take. Would I add the vector E to Y component then use basic trig to find the height at that x distance?

Help, please and thanks :)

If the x component of the velocity is 2.998×10⁶m/s, how long does it take an electron to reach the screen?

 

[C6ZR1]

well V=d/t so

t= 1.64/2.998 E 6

t≈5.47 E -7 s

 

[SammyS]

well V=d/t so

t= 1.64/2.998 E 6

t≈5.47 E -7 s

Good !

Now, what is the acceleration of the electron? In what direction does the electron accelerate?

 

[C6ZR1]

d=Vi*t+1/2 at^2

d-Vi*t=1/2at^2

1.64-(4.14 E 6 * 5.47 E -7 )= 1/2*a*(5.47 E -7 )^2

-0.625=1.496 E -13*a

a= -4.175 E 12m/s^2I'd assume that it would accelerate in the positive Y direction since the number is negative and the Electric field is in that direction.

 

[SammyS]

d=Vi*t+1/2 at^2

d-Vi*t=1/2at^2

1.64-(4.14 E 6 * 5.47 E -7 )= 1/2*a*(5.47 E -7 )^2

-0.625=1.496 E -13*a

a= -4.175 E 12m/s^2

This is completely wrong.

You used a value for d that is purely in the x direction.

The acceleration, a, has a component of zero in the x direction.

The initial velocity has both x and y components (neither of which is zero) but you used the magnitude of the initial velocity.

I'd assume that it would accelerate in the positive Y direction since the number is negative and the Electric field is in that direction.

The Electric field is in the +y direction, but the charge of the electron is negative.

How is the force on a charged particle related to the electric field ?

 

[C6ZR1]

F=QE E=KQ/r^2

F=KQ^2/r^2

 

[SammyS]

F=QE E=KQ/r^2

F=KQ^2/r^2

It's the F=QE that you need for this. You have E, and Q . Then if you have the force, can you find acceleration from that ?

 

[C6ZR1]

F=QE where E= 5.34 N/C Q=1.602 E -19 C so F=8.555 E -19 N

F=ma 8.555 E -19 = 9.11 E -31 *a a=9.39E 11 m/s^2

is that correct? My found acceleration seems to be really high.

 

[SammyS]

F=QE where E= 5.34 N/C Q=1.602 E -19 C so F=8.555 E -19 N

F=ma 8.555 E -19 = 9.11 E -31 *a a=9.39E 11 m/s^2

is that correct? My found acceleration seems to be really high.

Yes, that seems high. ...

However, the time interval is quite small, and the initial velocity is quite large.

See how it all works out,

 

[C6ZR1]

Wait, I'm still some what confused. so the acceleration is the y component they are asking for?

 

[SammyS]

Wait, I'm still some what confused. so the acceleration is the y component they are asking for?

It's a lot like the usual projectile problem.

The motion in the x-direction is independent of the motion in the y-direction.

Remember, in this case the electric force, and acceleration are in the negative y-direction because the charge of the electron is negative. Also, in this problem they just ask for the "y component of the electron's velocity (sign included) when the electron hits the screen".

 

[C6ZR1]

oh ok, so then if the acceleration is negative then would I just subtract it from its initial velocity?

y direction: 4.14 E6 (sin 43.6) ≈ 2.855 E 6 - a=9.39E 11 m/s^2 ?

 

[SammyS]

How is uniform acceleration related to change in velocity and elapsed time?

 

[C6ZR1]

Is it the rate of change of velocity/ time?

 

[SammyS]

Is it the rate of change of velocity/ time?

Yes. And since the acceleration is only in the -y direction, only the y component of the velocity changes.

 

[C6ZR1]

I understand that, but with acceleration found how would we find the y component at that distance x? Add it to the y component of the initial velocity?

 

[SammyS]

I understand that, but with acceleration found how would we find the y component at that distance x? Add it to the y component of the initial velocity?

According to what's in your original post, "What is the y component of the electron's velocity (sign included) when the electron hits the screen?"

It only wants the y component of the velocity just before the electron hits the screen.

But, finding the y position on thee screen isn't all that difficult.

y = y_i + (v_i)_yt + (1/2) a_y t²

 

[C6ZR1]

oh, so for this problem would it be?

y=0+2.855 E6*5.47 E-7+1/2-9.39 E11*(5.47E-2)^2

 

[SammyS]

oh, so for this problem would it be?

y=0+2.855 E6*5.47 E-7+1/2(-9.39) E11*(5.47E-2)^2

That looks good, if you insist on finding this quantity. (But the power of ten with the time looks wrong. Isn't the time of flight t≈5.47×10^-7 s ?)

This should give the same result as

(1.64 m)tan(43.6°)-(1/2)9.39×10¹¹(5.47×10^-7)²​

 

[C6ZR1]

ok so my result from first equation is 1.42121 and from second one which you provided is 1.42127. Now regarding the units I keep coming up with m^2, is that right?

 

[SammyS]

ok so my result from first equation is 1.42121 and from second one which you provided is 1.42127. Now regarding the units I keep coming up with m^2, is that right?

How do you get m² ?

(m/s)(s) → ?

m/s²)(s²) → ?

 

[C6ZR1]

I think I just randomly wrote it on my scratch paper.

I tried entering in the answer of 1.42 with m as units but its still wrong. :/ should I try going out a few more decimal places and putting the units as no units?

 

[SammyS]

I think I just randomly wrote it on my scratch paper.

I tried entering in the answer of 1.42 with m as units but its still wrong. :/ should I try going out a few more decimal places and putting the units as no units?

What is the wording of the question you're answering?

Are they asking for the position on the screen at which the electron makes impact,

... or are they asking for the y-component of the electron's velocity at impact?

 

[C6ZR1]

"What is the y component of the electron's velocity (sign included) when the electron hits the screen?"

Sounds to me like its velocity

 

[C6ZR1]

So in that case would it be 2.855 E6?

 

[SammyS]

So in that case would it be 2.855 E6?

No.

How much does the (negative) acceleration decrease the vertical component of the velocity in a time of 5.47×10^-7 seconds?

 

[C6ZR1]

well if acceleration is the change in velocity/ time then would it be -9.39E11= 2.855 E6- vy/ 5.47×10-7 ?

 

[SammyS]

well if acceleration is the change in velocity/ time then would it be -9.39E11= (2.855 E6- vy)/ 5.47×10^-7 ?

If you use parentheses correctly, it's almost OK.

The change in velocity (y-component) is (v_f)_y - 2.855×10⁶ .

You have these reversed.

 

[C6ZR1]

ohhhh gotcha, and I guess I should start using those parentheses more often hehe.

So solving for Vfy I get that it is about 2.34137 E6

 

[SammyS]

I think that's about right.

 

[C6ZR1]

ok the numerical value is correct but the units are still wrong! lol

 

[SammyS]

ok the numerical value is correct but the units are still wrong! lol

You're now answering a question regarding velocity, not position so it has different units than the last time you asked about units.

 

[C6ZR1]

* face palm * I need to stop doing homework so late, I got it now. Thank you soooo much for your help and time. I really appreciate it! This will certainly be a question I'll review for an upcoming exam

 

[SammyS]

* face palm * I need to stop doing homework so late, I got it now. Thank you soooo much for your help and time. I really appreciate it! This will certainly be a question I'll review for an upcoming exam

You're welcome.

Time for me to get some sleep too.