# Calculate the deflection sensitivity in mm per volt

• moenste
In summary, the charge/mass ratio of the electron has been determined by calculating the deflection sensitivity of a cathode ray tube using various data such as the potential difference, length and separation of deflector plates, and distance from the screen. The energy of the electrons can be found by using the formula E = V/d, and their initial velocity can be calculated using the formula v = √(2eV/m).
Sure. But you can replace the time t, also, since you know vx 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 dy and dx?

I'd suggest picking a "test" value for Vp and calculating vx and vy separately for now.

moenste
gneill said:
Sure. But you can replace the time t, also, since you know vx 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 dy and dx?

I'd suggest picking a "test" value for Vp and calculating vx and vy separately for now.
vx = √ 2 * 1.6 * 10-19 * 5000 / 9.1 * 10-31 = 41 931 393.47 m / s.

VP = 100 V
vy = e VP * (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 °

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

moenste
gneill said:
Okay. Now how will you find the deflection on the screen?
dy = 0.16 tan 2.29 ° = 6.4 * 10-3 m.

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
gneill said:
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?

moenste said:
I don't understand, what do you mean by a ratio?
Did you study similar triangles in geometry?

moenste
gneill said:
Did you study similar triangles in geometry?
0.15 / 41 931 393.47 = dy / 1 677 256
dy = 6 * 10-3 m.

moenste said:
0.15 / 41 931 393.47 = dy / 1 677 256
dy = 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
gneill said:
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 VP, which I took as 100 V: 6 mm / 100 V = 0.06 mm V-1. That's indeed the correct answer!

Thank you!

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