Calculate the deflection sensitivity in mm per volt

[moenste]

Homework Statement

Describe a method by which the charge / mass ratio e / m of the electron has been determined.

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.

 

[mfb]

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?

E = V / d = 5000 / 0.005 = 1 000 000 V

The units don't match.

 

[moenste]

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.

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).

 

[gneill]

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.

 

[moenste]

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.

So, I should probably start by finding out the horizontal speed created by the potential difference?

And

What is the energy of the electrons once they reach the cylindrical shell?

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?

 

[gneill]

So, I should probably start by finding out the horizontal speed created by the potential difference?

Yes, that would be a good start.

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?

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?

 

[moenste]

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?

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.

 

[mfb]

But we don't know e and m.

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.

 

[gneill]

KE = e V → 0.5 m v² = e V → v = √ 2 e V / m?

Good.

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.

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.

 

[moenste]

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.

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.

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 m_Electron = 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.

 

[gneill]

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.

What do you think should be your next step?

 

[moenste]

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.

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.

View attachment 108020

What do you think should be your next step?

Find time, find electric field E, find PD and then divide 15 cm in mm by the PD?

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

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.

 

[gneill]

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.

Yes, that's valid.

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.

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.

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.

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 d_y 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.

 

[moenste]

I'd suggest that you find the vertical velocity of the electron as it leaves the plates.

v = v_initial - g t?

v = 41 931 393.47 - 10 * 4.77 * 10^-10 = 41 931 393.47 m s^-1.

 

[mfb]

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.

 

[gneill]

v = v_initial - g t?

v = 41 931 393.47 - 10 * 4.77 * 10^-10 = 41 931 393.47 m s^-1.

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.

 

[moenste]

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.

Electrons are accelerated by a PD of 5000 V. Then they just horizontally head towards the wall.

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.

But I don't have any acceleration.

 

[gneill]

Electrons are accelerated by a PD of 5000 V. Then they just vertically head towards the wall.

They are accelerated horizontally by the 5 kV PD, then they pass through the plates.

But I don't have any acceleration.

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.

 

[moenste]

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.

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?

 

[mfb]

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?

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.

 

[moenste]

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.

We then have a voltage V_P between the plates, like this?

 

[mfb]

Yes.

 

[moenste]

Yes.

I can only think of finding E. E = V_P / d.

 

[mfb]

Okay.

The electrons fly horizontally into the gap between the plates. There they are in an electric field in vertical direction. What happens (first in words, then in formulas)?

 

[moenste]

Okay.

The electrons fly horizontally into the gap between the plates. There they are in an electric field in vertical direction. What happens (first in words, then in formulas)?

The electric field between the two plates is going in some vertical direction and the electrons deviate from their horizontal movement towards some plate. I don't know why. Maybe because the electric field creates a force which affects the electrons and so they deviate.

 

[gneill]

The electric field between the two plates is going in some vertical direction and the electrons deviate from their horizontal movement towards some plate. I don't know why. Maybe because the electric field creates a force which affects the electrons and so they deviate.

moenste, you've done at least one other problem where an electric field between two parallel plates deflected the charged particle moving between them. This shouldn't be a new concept for you. Perhaps you need to review Coulomb's Law and electric fields?

 

[moenste]

moenste, you've done at least one other problem where an electric field between two parallel plates deflected the charged particle moving between them. This shouldn't be a new concept for you. Perhaps you need to review Coulomb's Law and electric fields?

We have PD V_P between the plates. We have distance d. We find the electric field E = V_P / d.

We have original velocity v = √ 2 e V / m and we have time to pass the two plates t = 4 d / v.

What do we need to find? The angle of deflection? Why is the angle in mm V^-1 and not degrees? What mm do represent here? 20 mm, 5 mm or 1500 mm? Voltage is V_P?

 

[gneill]

We have PD V_P between the plates. We have distance d. We find the electric field E = V_P / d.

We have original velocity v = √ 2 e V / m and we have time to pass the two plates t = 4 d / v.

What do we need to find? The angle of deflection? Why is the angle in mm V^-1 and not degrees? What mm do represent here? 20 mm, 5 mm or 1500 mm? Voltage is V_P?

The sensitivity of the device has been defined for you to be the linear deflection in mm per volt of PD applied on the deflection plates. They could have defined it to be "degrees per volt" but they didn't. The deflection in mm is more practical for such a device, because it can be measured on the screen (think of a CRT screen).

You want to determine how much the electron is deflected (in mm) for every volt applied as V_p. That's why V_p was not specified as a particular value. The deflection in mm is measured on the target screen, and it's the distance from the horizontal (undeflected) path to the impact point when the potential V_p is applied.

If you can find the deflection angle then you can work out the deflection by geometry. I'd suggest using the horizontal and vertical speeds of the electron as they leave the plates (i.e. the "launch velocity components") as the basis for determining the deflection angle, and using similar triangles rather than trigonometry to find the deflection distance on the screen.

Here's a diagram depicting the part of the problem you're working on now:

 

[moenste]

The sensitivity of the device has been defined for you to be the linear deflection in mm per volt of PD applied on the deflection plates. They could have defined it to be "degrees per volt" but they didn't. The deflection in mm is more practical for such a device, because it can be measured on the screen (think of a CRT screen).

You want to determine how much the electron is deflected (in mm) for every volt applied as V_p. That's why V_p was not specified as a particular value. The deflection in mm is measured on the target screen, and it's the distance from the horizontal (undeflected) path to the impact point when the potential V_p is applied.

If you can find the deflection angle then you can work out the deflection by geometry. I'd suggest using the horizontal and vertical speeds of the electron as they leave the plates (i.e. the "launch velocity components") as the basis for determining the deflection angle, and using similar triangles rather than trigonometry to find the deflection distance on the screen.

Here's a diagram depicting the part of the problem you're working on now:

View attachment 108072

We have v_x = √ 2 e V / m.

v_y: e (V_P / d) = B e v_y → v_y = V_P / B d.

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

 

[gneill]

We have v_x = √ 2 e V / m.

v_y: e (V_P / d) = B e v_y → v_y = V_P / B d.

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

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