Electric field and particles

1. Apr 22, 2008

~christina~

[SOLVED] Electric field and particles

1. The problem statement, all variables and given/known data
It is useful to separate fast-moving ions from slow-moving ones. To accomplish this, the ions enter a component of a device (modeled somewhat like a velocity selector) as a narrow horizontal beam that passes between two parallel plates that are 5.0 cm long and 4.0 cm wide. The separation between the plates is 3.0 cm. A high voltage applied to the plates accelerates the ions toward one of the plates and away from the other plate. This ion beam consists of a mixture of $$SO_2^{4+}$$ particles. These particles enter the space between the plates having initial velocities 5.60 × 04 m / s and 5.60 × 105 m / s in the positive x-direction. The mass of each ion is 1.06 x 10-25 kg and the applied potential is 150 V. The ions enter the space between the plates at its center. After leaving this space, the ions travel an additional 50 cm to the detector

(a) Give a brief description of the electric field between the plates if the ions' motion is in the positive xy-plane after leaving the gap.

(b) Calculate the magnitude of the acceleration that the particles have when they are between the plates.

(c) Find the angle(s) of the particles, relative to their initial motion, when they emerge from the plates.

(d) Will these ions strike the detector at the same spot? If yes, what is the y-coordinate of the ions; if no, what is the separation between the ions when they strike the detector? Ignore the gravitational force.

2. Relevant equations
$$F= qE$$

$$\Delta V= -Ed$$

$$a= \frac{qE} {m}$$

3. The attempt at a solution
I'm not sure what they mean by mix of particles of $$SO_2^{4+}$$ with different velocities since I thought that would be one molecule but I think that they mean that it is different charges thus different velocities so I would calculate them seperately.

$$m_{1&2}= 1.06 x 10-25 kg$$
$$v_1= 5.60 × 04 m/s$$
$$v_2= 5.60 × 105 m/s$$
$$\Delta V= 150V$$

a) Give a brief description of the electric field between the plates if the ions' motion is in the positive xy-plane after leaving the gap.

Not sure but would it be a good description if I said that the electric field goes from + to - and if the ion's molecules are going in the possitive xy direction, then the possitive plate must be on the bottom and the negative plate on the top.

(b) Calculate the magnitude of the acceleration that the particles have when they are between the plates.

$$\Delta V= -Ed$$

$$150V=E(0.03m)$$
$$E= 5e3$$V/m

$$a= \frac{qE} {m}$$

I have E but don't have q and I'm not sure how to find it either for both the particles (not sure if one is + and other - [very confusing]). (I don't think it would be the same as a electron)

(c) Find the angle(s) of the particles, relative to their initial motion, when they emerge from the plates.

not sure how to find this...

do I use the UAM equations after I find the accelerations? (not sure how to relate that to when they exit)

(d) Will these ions strike the detector at the same spot? If yes, what is the y-coordinate of the ions; if no, what is the separation between the ions when they strike the detector? Ignore the gravitational force.

not sure how to find this either.

Thanks

2. Apr 22, 2008

cristo

Staff Emeritus
You're told the charge on the ion (4+), thus the charge will be +4|charge on electron|. (i.e. 4 times the charge on a positron.)

Just consider one particle at a time. I'm not sure what you mean by "UAM," though, but if you mean the kinematic equations then, yes, use these. Note that the acceleration in the x direction is zero. All you need to find are the final components of the velocity in the x and y directions, then the result will follow.
Do part (c) first, then you should be able to tackle this.

Last edited: Apr 22, 2008
3. Apr 23, 2008

~christina~

so I assume I did find the electric field, correctly then? (I was thinking I had to use the area of the plates and then somehow find the electric field from that)

so if there is more then one particle the only difference is the speed right?

$$Q_{SO_2^{4+}} = 4(1.60e-19)= 6.4x10^{-19}C$$

aceleration would then be(assuming E is fine):

$$a_{SO_2^{4+}}= \frac{qE} {m}= [(6.4x10^{-19}C)(5x10^3 V/m)]/ (1.06 x 10^{-25} kg) = 3.0188 m/s$$

ok. (I meant uniform acelerated motion equations / kinematic equations)

I'm not quite sure but I think that they are launched straight but then turn so would the initial velocity be at an angle or only the final velocity? (I'm assuming that initially it is not at an angle)

time it takes to get to the end of the plates:
1st particle
$$x_f-x_i= vi_x(t)$$
(0.05m)= (5.60 ×10^4 m/s)t
$$t_1= 8.92x10^-7s$$

2nd particle
$$x_f-x_i= vi_x*t$$
(0.05m)= (5.60 ×10^5 m/s)t
$$t_2= 8.92x10^-8s$$

Okay, I found the time needed for the ion to reach the end of the plate but I can't figure out what equation (kinematic) to use.

I'm thinking of using this equation (Vf= Vi + a(t)) since I know the initial Vy is 0 and I found the acceleration of the particle in the electric field and I just found the time above.
so I think I could go and find Vf and then:$$Vfy= Vf(cos \theta)$$ and I think that's how I'd solve for the angle but maybe it's incorrect.

when I do find the velocity in the y direction when the particle leaves the plates and since the x velocity is the same at the begining of the plate and end of the plate, how would I find the point where the ion strikes the plate?
I do have the x distance (0.5m) and initial velocities (starting from end of plate to the detector). I can find the time it takes for the particle to strike the detector as well but I don't have an initial height or do I assume it is at 0 height initially and then travels up to the detector and calculate the height like that?.

Thanks for your help cristo

4. Apr 27, 2008

cristo

Staff Emeritus
Hmm... you might want to check this; I get the same number, but with extra "times ten to the power of .." Also, the units should be m/s^2, since it's an acceleration.

Ok, good.
Sounds good to me. If you find vfy, then together with vfx you can calculate the angle of v to the horizontal.
Yes, that's how I'd do it, since we are not asked for a definite height, but just whether they hit at the same point: so long as you take the same reference point, it doesn't matter which reference point this is.
You're welcome

5. Apr 28, 2008

~christina~

okay, fixing this I get 3.0188x10^10 m/s^2

hm trying this I get:

particle 1
$$Vfy= Viy + a(t)$$
$$Vfy= 0 + (3.0188x10^10 m/s^2)(8.92x10^{-7}s)$$
$$Vfy_1= 2.69x10^4 m/s$$

$$Vfx=Vix$$
$$Vfx= 5.60 × 10^4 m/s$$

$$tan \theta= (2.69x10^4 m/s)/(5.60 × 10^4 m/s)$$
$$tan^{-1}(0.048085)=2.752^o$$

particle 2
$$Vfy= Viy + a(t)$$
$$Vfy= 0 + (3.0188x10^10 m/s^2)(8.92x10^{-8}s)$$
$$Vfy_1= 2.69x10^3 m/s$$

$$Vfx=Vix$$
$$Vfx= 5.60 × 10^5 m/s$$

$$tan \theta= (2.69x10^3 m/s)/(5.60 ×10^5 m/s)$$
$$tan^{-1}(0.0048035)= 0.2755^o$$

somehow these angles look funny.
well assuming these angles are alright (I calculated it a few times) But I was wondering about the acceleration (would it be from the plates still?, but they are outside the plates when they hit the detector and the problem says to ignore the gravitational force so is there acceleration or not? I'm assuming it's still from the plates but it may be incorrect)

Particle 1
Xf-Xi=Vxi(t)
Particle 1[/tex]
$$0.5m= (5.60x10^4m/s)t$$
$$t= 8.92x10^{-6}s$$

$$Yfy= Yiy + Viy(t) + 0.5at^2$$
$$Yfy= 0 +(2.69x10^4 m/s)(8.92x10^{-6}s) + 0.5(3.0188x10^10m/s^2)(8.92x10^{-6}s)^2= 1.44m$$

Particle 2
Xf-Xi=Vxi(t)
$$0.5m= (5.60x10^5m/s)t$$
$$t= 8.928x10^{-7}s$$

$$Yfy= Yiy + Viy(t) + 0.5at^2$$
$$Yfy= 0 +(2.69x10^3 m/s)(8.92x10^{-7}s) + 0.5(3.0188x10^10m/s^2)(8.92x10^{-7}s)^2= 0.0144m$$

so they don't strike the same place if my numbers are correct I think.

Is this right?

Thanks again, cristo

6. Apr 28, 2008

cristo

Staff Emeritus
Good

I must admit I've not checked your calculations that throughly, but your method, and the numbers you used are correct so, since you say you've done it a few times, I'd say these were correct.

No: once the particles have left the electric field, since we are neglecting gravity, there is no force on the particles, and thus no acceleration. Work through the following without the y accelerations (i.e. Yf=Viy.t) and see what you get!

By the way, a little LaTeX tip: to make your x's look like multiplication signs, use the command \times.

7. Apr 28, 2008

~christina~

I was thinking that acceleration was zero but wasn't sure about that. Okay.

thanks for the tip.

Thanks for your all your help cristo.

8. Apr 28, 2008

cristo

Staff Emeritus
You're welcome!