Calculate the deflection sensitivity in mm per volt

[moenste]

Where did B come from? What's it's value?

Maybe then v_y = d / t?

 

[gneill]

Maybe then v_y = d / t?

You're just guessing. What force is acting on the electron? What's the acceleration? How long does the acceleration last? What's the resulting speed?

You need to be able to identify the salient details of a scenario, identify the forces acting, and write equations for them. Throwing random equations at the problem won't yield the desired results.

 

[moenste]

You're just guessing. What force is acting on the electron? What's the acceleration? How long does the acceleration last? What's the resulting speed?

You need to be able to identify the salient details of a scenario, identify the forces acting, and write equations for them. Throwing random equations at the problem won't yield the desired results.

I can only think of electric force E.

v = u + a t
a = F / m = e E / m, u = 0
v = (e / m) E t.

But that's wrong, as you earlier said.

So I've no idea what else is applicable here.

 

[gneill]

I can only think of electric force E.

v = u + a t
a = F / m = e E / m, u = 0
v = (e / m) E t.

But that's wrong, as you earlier said.

So I've no idea what else is applicable here.

It was wrong before because you were using the wrong potential difference to set the electric field. Remember, you plugged in 5 kV for the potential difference on the plates when that potential difference isn't associated with the plates at all. It just happened to be a handy voltage that was mentioned in the problem...

This time you've agreed that the plate potential difference is the "test value " V_p. That is a correct approach. Write E in your equations above using V_p and the plate geometry and you're on the right track.

 

[moenste]

It was wrong before because you were using the wrong potential difference to set the electric field. Remember, you plugged in 5 kV for the potential difference on the plates when that potential difference isn't associated with the plates at all. It just happened to be a handy voltage that was mentioned in the problem...

This time you've agreed that the plate potential difference is the "test value " V_p. That is a correct approach. Write E in your equations above using V_p and the plate geometry and you're on the right track.

Alright, so now we have:
v_x = √ 2 e V / m.

v_y = e V_P t / m d.

Then we find angle: tan θ = v_y / v_x = [e V_P t / m d] / [√ 2 e V / m].

 

[gneill]

Sure. But you can replace the time t, also, since you know v_x and the horizontal distance traveled within the plates, right? Just be sure to distinguish between the plate separation d which you've already used and the length of the plates. Maybe call them d_y and d_x?

I'd suggest picking a "test" value for V_p and calculating v_x and v_y separately for now.

 

[moenste]

Sure. But you can replace the time t, also, since you know v_x and the horizontal distance traveled within the plates, right? Just be sure to distinguish between the plate separation d which you've already used and the length of the plates. Maybe call them d_y and d_x?

I'd suggest picking a "test" value for V_p and calculating v_x and v_y separately for now.

v_x = √ 2 * 1.6 * 10^-19 * 5000 / 9.1 * 10^-31 = 41 931 393.47 m / s.

V_P = 100 V
v_y = e V_P * (4 d / √ 2 e V / m) / m d = 1.6 * 10^-19 * 100 * (4 * 5 * 10^-3 / √ 2 * 1.6 * 10^-19 * 5000 / 9.1 * 10^-31) / 9.1 * 10^-31 * 5 * 10^-3 = 1 677 256 m / s.

tan θ = 1 677 256 / 41 931 393.47
θ = 2.29 °

 

[gneill]

Okay. Now how will you find the deflection on the screen?

 

[moenste]

Okay. Now how will you find the deflection on the screen?

d_y = 0.16 tan 2.29 ° = 6.4 * 10^-3 m.

 

[gneill]

The 0.16 came out of nowhere, but I'm guessing that you used the distance from the front edge of the plates to the screen in meters?

That is not too bad an approximation, but it underestimates the curvature of the trajectory while the electron is between the plates. The electron is accelerating and covers more distance in the latter half of the plates than the front half. A better approximation can be made by taking the center of the plates as the vertex of your triangle. Triangle abc in the image:

You should also note that the velocity triangle depicted is similar to triangle abc. You didn't need to find the angle, you could simply have set up a ratio.

 

[moenste]

The 0.16 came out of nowhere, but I'm guessing that you used the distance from the front edge of the plates to the screen in meters?

That is not too bad an approximation, but it underestimates the curvature of the trajectory while the electron is between the plates. The electron is accelerating and covers more distance in the latter half of the plates than the front half. A better approximation can be made by taking the center of the plates as the vertex of your triangle. Triangle abc in the image:
View attachment 108077

You should also note that the velocity triangle depicted is similar to triangle abc. You didn't need to find the angle, you could simply have set up a ratio.

Yes, I used 15 cm + 2 cm / 2 = 16 cm and 0.16 in m.

I don't understand, what do you mean by a ratio?

 

[gneill]

I don't understand, what do you mean by a ratio?

Did you study similar triangles in geometry?

 

[moenste]

Did you study similar triangles in geometry?

0.15 / 41 931 393.47 = d_y / 1 677 256
d_y = 6 * 10^-3 m.

 

[gneill]

0.15 / 41 931 393.47 = d_y / 1 677 256
d_y = 6 * 10^-3 m.

Yes.

That's the deflection for your choice of deflection potential $V_p$. Use it to write the sensitivity (mm/V).

 

[moenste]

Yes.

That's the deflection for your choice of deflection potential $V_p$. Use it to write the sensitivity (mm/V).

Hm, so we change m into mm 6 * 10^-3 m → 6 mm and then divide by V_P, which I took as 100 V: 6 mm / 100 V = 0.06 mm V^-1. That's indeed the correct answer!

Thank you!