Calculate the deflection sensitivity in mm per volt

[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!