# Help Conservation of Energy

1. Feb 20, 2007

### xxkbxx

Help!!! Conservation of Energy

The apparatus to be used is a classic one for measuring q/m for electrons. A photo of the cathode-ray tube together with a simplified graphic are shown below. A filament heated with a 6.3-V AC source "boils" off some electrons. These electrons have very little energy initially. However, some of them enter a uniform electric field between two plates maintained at a constant potential difference of V1. The electrons are accelerated in the field and leave the plates through an aperture at Q. The electrons then enter another electric field between two deflection plates a distance L long and separated by distance d. The plates are maintained at a constant potential difference of V2. Between the plates the electrons follow a parabolic path characteristic of charged particle motion in a uniform electric field.

The problem statement, all variables and given/known data

Begin this problem by determining an equation for the velocity, vo, of electrons as they leave the accelerating plates and enter the deflection plates. This is a conservation of energy problem. Clearly indicate your initial and final states and start with a correct statement for the law of conservation of energy as it applies to this problem. Assume that the electrons have negligible kinetic energy as they enter the region between the accelerating plates at point P. Give your result for vo in terms of V1 and q/m.

2. Relevant equations
Initial Energy = Final Energy
U = qV
a = qE/m

3. The attempt at a solution
I know that initial and final energy are equal so:
1/2mvi^2 + Ui = 1/2mvf^2 + Uf

I replace Ui and Uf with qVi and qVf
solving for initial velocity (v0 or vi) I get

v0 = sqrt (mvf^2 + 2(Ua + Ub) )

I need to get the equation down to terms of V1 and q/m, that's where I'm stumped

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2. Feb 20, 2007

### Dick

What are you thinking that vf is in this equation? Also looking at the previous equation I would expect a difference rather than a sum of the potentials in your equation for v0.

3. Feb 20, 2007

### Dick

You also haven't very clearly identified the U's in your solution with the V's in the problem.

4. Feb 20, 2007

### xxkbxx

vf is in the final equation because Ki + Ui = Kf + Uf, Kf is 1/2mvf^2, so it's part of the equation.

I did mess up on the potentials it should be:

v0 = sqrt (mvf^2 + 2(Ub - Ua) )

These would simplify further to:

v0 = sqrt (mvf^2 + 2q(V2 - V1) )

But I'm still stumped

5. Feb 20, 2007

### Dick

No, the question I mean to ask is where is vf in your diagram? What should it's value be in the equation? Also is V2 really needed in there to answer the question. Again where is it in your diagram? You're equations are evolving independently of your diagram.

6. Feb 20, 2007

### xxkbxx

Sorry, it's initially from point Q, as it leaves the accelerating plates, and it's final point is it entering the deflection plates. I'm not sure how to get vf though, but I have a feeling it's based on
a=qE/m

7. Feb 20, 2007

### Dick

The problem says: "Assume that the electrons have negligible kinetic energy as they enter the region between the accelerating plates at point P."

8. Feb 20, 2007

### Dick

You may be missing the fact that the actual question only refers to the first half of your diagram. I assume there are more question coming later...

9. Feb 20, 2007

### chaoseverlasting

OK. Since they are accelerated through a potential v1, their energy when they enter the second set of plates is ev2(in electron volts). You can find the ke of the electron here, but leave it in $$v^2$$ form.

At the point where the electron enters the second set of plates, the energy it gains will be ev2 which will be equal to its ke in the y direction. again, from here you can get the velocity in the y direction and leave it in the $$v^2$$ form.

Now, at this point, Im slightly confused. If the detector is beyond the second set of plates at a certain distance, then the tangent of the angle at which the electron leaves the plates will be $$tan\theta=\frac{v_y}{v_x}$$. Now, you would know where the electron strikes the detector and the distance at which the detector is kept to calculate $$tan\theta$$.

If, however, the electron does not escape, but just hits the upper plate of the second set, $$v_x$$ is a constant and you know the lenght L of the plate and hence the time taken by the electron to travel the distance. Since you know the distance travelled (d/2) and the time it took to travel that distance, you can again find the velocity required which is can be calculated by ev(in electron volts->convert to joules) =0.5mv^2, again from which you can find the e/v ratio.

10. Feb 20, 2007

### Dick

You are a bit confused. The electron doesn't fall through the full V2 potential. The purpose of the second potential is to apply a sidewise force to the electron via the electric field. You then compute the deflection just as you would for a projectile moving in a gravitational field.

11. Feb 20, 2007

### chaoseverlasting

Excuse me. Sidewise? Dont you mean upwards? And if youve been following my earlier post, its the same damn thing.

The ONLY point of confusion is whether the detector is imbedded into the plates (HIGHLY unlikely) or if the electron leaves the plate at a certain angle of deflection which can be calculated as stated above.

If you do not use an externat detector, I dont see how you can calculate the e/m ratio unless the electron just hits the upper end of the second set of plates.

How else would you propose to solve the problem? I.e, for the experiment to be successful, you need the e/m ratio. How would you get that.

You could have solved the question using the kinematical equations as well but here they have explicitly asked the use of conservation of energy.

I'd like to see you propose a solution. Now I'm really interested. Go ahead.

12. Feb 20, 2007

### Dick

As I pointed out to the poster. The ONLY question that is asked is what the velocity of the electron is after leaving the accelerating plates. This is the only place to explicitly apply conservation of energy. The deflection phase is kinematics.

13. Feb 20, 2007

### chaoseverlasting

Im sorry. My mistake. I thought we had to find the e/m ratio as well. The first part seemed too simple.

14. Feb 20, 2007

### Dick

No problem. The first part IS simple.