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Calculate the deflection sensitivity in mm per volt

  1. Oct 24, 2016 #1
    1. The problem statement, all variables and given/known data
    Describe a method by which the charge / mass ratio e / m of the electron has been determined.

    d756ebb11e17.jpg

    Calculate the deflection sensitivity (deflection of spot in mm per volt potential difference) of a cathode ray tube from the following data: electrons are accelerated by a potential difference of 5.0 kV between cathode and annode; length of deflector plates = 2.0 cm; separation of deflector plates = 5.0 mm; distance of mid point of deflector plates from screen = 15.0 cm.

    Answer: 6 * 10-2 mm V-1.

    2. The attempt at a solution
    I found E = V / d = 5000 / 0.005 = 1 000 000 V and that's it. Not much clear or basic information on the deflection of spot in mm / V can be found online. And I don't think it's as easy as 5 mm / 5000 V.
     
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  3. Oct 24, 2016 #2

    mfb

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    The 5 kV are between the cylindrical shell and the electron source. There is no voltage given for the two plates to the right.

    What is the energy of the electrons once they reach the cylindrical shell? What is their speed? How long will they be between the two parallel plates?

    The units don't match.
     
  4. Oct 25, 2016 #3
    Yes, sorry, it's V m-1.

    I would say the energy is 1 000 000 V m-1, but you say the 5 kV have nothing to do with it then I don't quite understand what can we find having three numbers concerning the size (2 cm, 5 mm, 15 cm).
     
  5. Oct 25, 2016 #4

    gneill

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    Electrons are accelerated to some horizontal speed by the potential difference between the electron gun and the cylinder. They pass through and exit the cylinder with that speed.

    Next they enter the parallel plates where they are deflected by some potential difference, call it Vp, that's established between the plates. The problem is asking you to determine the amount of deflection seen on the target (in mm) for a given amount of that potential difference Vp.
     
  6. Oct 26, 2016 #5
    So, I should probably start by finding out the horizontal speed created by the potential difference?

    And
    Energy of the electrons is E, right? E = V / d, where d is the distance between plates. We can also see on the picture the shell is also 5 mm wide. Can't we say that energy is E = 5000 / 0.005 = 1 000 000 V m-1?
     
  7. Oct 26, 2016 #6

    gneill

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    Yes, that would be a good start.
    You're confusing energy (Joules) and electric field (V/m). An electric field is not the same as potential difference, although electric fields can be established by potential differences (and vice versa).

    There are no plates in the region where the electrons are picking up their initial velocity: They are being pulled from the electron gun by the potential difference between the gun and the cylinder. The plates don't become involved until the electrons have passed through the cylinder.

    You should be familiar with a few basic formulas when dealing with charges, electric fields, and potential differences:

    1. A potential difference V between plates separated by distance d creates electric field E = V/d between the plates.
    2. Charge q in electric field E experiences force F = qE.
    3. A charge q falling through a potential difference of V experiences a change in kinetic energy of ΔKE = qV

    You should memorize the units associated with these things:

    a. Potential difference is specified in volts (V). The volt is a compound unit: V = Joule/Coulomb, so energy per unit charge.
    b. Electric fields are specified in V/m or N/C (they work out to be the same in fundamental units).

    The electrons are given their initial velocity by having them "fall" through the potential difference between the electron gun and the cylinder. So what formula above will allow you to find the velocity?
     
  8. Oct 26, 2016 #7
    KE = e V → 0.5 m v2 = e V → v = √ 2 e V / m?

    But we don't know e and m. I know the elementary charge e = 1.6 * 10-19 C, but not sure whether this is applied here.
     
  9. Oct 26, 2016 #8

    mfb

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    You accelerate electrons, you can look up those values for electrons. Before doing this experiment people could not make the calculations you are asked to make, but that is not relevant for this question.

    There is a minus sign missing but working with absolute values should be fine here.
     
  10. Oct 26, 2016 #9

    gneill

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    Good.
    It's an electron. You should have a table of its properties in your text or notes (or just look them up online). For these problems some quantities, such as the mass and charge of common elementary particles, you'll be expected to have on hand.
     
  11. Oct 26, 2016 #10
    I was just looking at the electron mass which is 9.1 * 10-31 kg.

    So, we then have: v = √ 2 e V / m, where e (electron elementary charge) = 1.6 * 10-19 C, V = 5000 V and mElectron = 9.1 * 10-31 kg → v = √ 2 * 1.6 * 10-19 * 5000 / 9.1 * 10-31 = 41 931 393.47 m s-1 or 4.2 * 107 m s-1.

    Update
    Maybe we can use s = v t to find time, where v is the velocity of the electrons and s is the length of the deflector plates. So: t = s / v = 0.02 / 41 931 393.47 = 4.7696339 * 10-10 s.

    Then we use v = u + a t, where F = m a → a = F / m = e E / m → v = (e / m) * E * t → E = v / (e / m) t = 41 931 393.47 / (1.6 * 10-19 / 9.1 * 10-31) * 4.7696339 * 10-10 = 500 000 V m-1.

    Now we find PD, since we have the electric field E and distance between plates d: E = V / d → V = E d = 500 000 * 0.005 = 2500 V.

    And if we do (15 cm * 10) / 2500 = 0.06 mm V-1 -- this gives the right answer. Though this last step I don't understand.
     
    Last edited: Oct 26, 2016
  12. Oct 26, 2016 #11

    gneill

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    Yes that's fine. Although you really don't need to perform any calculations at this stage. What you need to do is establish the expressions for the various quantities. These expressions will be used in the derivation of the sensitivity. Many things will cancel along the way, so it's really a waste of effort to calculate everything at every stage.

    [Edit: Your "update" is not fine because you are not given any information about the field between the plates or the potential difference applied to the plates. It's a variable that you have to introduce to determine the sensitivity; you want to know what the deflection is per volt applied as the potential difference.]

    The important thing to take away from this part is that the horizontal speed of the electrons is given by

    ##v_x = \sqrt{\frac{2 e~V}{m}}##

    So you now have an expression for the horizontal speed of the electrons as they enter the deflection plates. You aren't given a particular value for the potential on the plates so you must use a symbol for it. You can choose a variable name for it (I chose U, but you can use what you want). The idea is to determine the distance that the electron is deflected from the centerline on the target screen for a given value U.

    upload_2016-10-26_12-31-55.png

    What do you think should be your next step?
     
    Last edited: Oct 26, 2016
  13. Oct 26, 2016 #12
    Find time, find electric field E, find PD and then divide 15 cm in mm by the PD?
    If we put all cm in mm and simplify what I have above I'll get 60 d / V = 60 * 5 / 5000 = 0.06 mm V-1.
     
    Last edited: Oct 26, 2016
  14. Oct 26, 2016 #13

    gneill

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    Yes, that's valid.
    That does not make sense since potential difference on the plates and the electric field between the plates has nothing to do with the horizontal velocity v, other than v determining the time spent between the plates.
    Nope. We have to supply the PD between the plates by way of introducing it as a variable. That last step is just numerology; by coincidence that combination of numbers gives you a number that happens to be the same as your desired result. The calculation itself has no physical meaning.

    We are not given the potential difference between the plates. Nor do we have the electric field there. That potential difference is some variable you must introduce in order to find the sensitivity. In other words, if you apply a potential difference U between the plates, what is the deflection on the screen? Or, turning it around, for some given deflection dy on the screen, what is the potential difference U of the plates. (This might be a profitable way to approach the problem. You could declare the deflection to be 1 mm and then find the value of U that produced it.)

    If you want to proceed dragging calculations along then I'd suggest that you set an arbitrary value for the plate potential difference U. Think of it as a "test value" that's applied and you want to see the resulting deflection. Pick a value that's easy to work with, something like 100 V.

    I'd suggest that you find the vertical velocity of the electron as it leaves the plates.
     
  15. Oct 27, 2016 #14
    v = vinitial - g t?

    v = 41 931 393.47 - 10 * 4.77 * 10-10 = 41 931 393.47 m s-1.
     
  16. Oct 27, 2016 #15

    mfb

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    You think gravity plays a role? Also, where does the initial vertical velocity come from?

    Please stop putting together random unrelated numbers. Start with thinking about the physical process that happens, then find suitable formulas based on this process.
     
  17. Oct 27, 2016 #16

    gneill

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    g? There's no gravity involved here. The time of flight is too short and the electrical forces too large by comparison for gravity to be anything but negligible in this problem. And if gravity was involved, it would work only in the vertical direction. The initial velocity you've used is the horizontal velocity. You need to think about the scenario before writing equations.

    But you do have an acceleration to work with: Between the plates is an electric field producing a force on the electron. Now you have to decide whether you're going to pick an arbitrary potential difference for the plates or just represent it by a variable and carry it through symbolically.
     
  18. Oct 27, 2016 #17
    Electrons are accelerated by a PD of 5000 V. Then they just horizontally head towards the wall.

    But I don't have any acceleration.
     
    Last edited: Oct 27, 2016
  19. Oct 27, 2016 #18

    gneill

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    They are accelerated horizontally by the 5 kV PD, then they pass through the plates.
    They are accelerated vertically by the electric field between the plates. Read what I said about that. You have to choose an arbitrary potential difference for the plates to define that electric field.
     
  20. Oct 27, 2016 #19
    Sorry, yes, horizontally.

    But nothing is said about the plates. There are just two metal plates d = 5 mm and length 2 cm. Why should the electrons deflect anywhere?
     
  21. Oct 27, 2016 #20

    mfb

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    There is a voltage between those two plates, that is part of the experiment. You don't know this voltage, you can introduce a variable for it, but if it helps you you can also consider an arbitrary value, like 100 V.
     
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