# Cathode Ray Tube Questions

## Homework Statement

Have a few questions on CRT's. If we are given the accelerating voltage (250V) how would I calculate the kinetic energy of the electron as it leaves the positive end of the charge area? What is the formula for the speed of an electron along the tube (vz) as it leaves the positive charge area, in terms of the accelerating potential Va and the mass and charge of an electron.

F = qE=ma
KE= (1/2) mv^2

## The Attempt at a Solution

Having only been given the accelerating voltage (250V) I'm not sure where to begin. We did all kinds of stuff with kinetic energy in Newtonian physics but I'm not sure how to apply it with voltage.

I think at some point I arrived at

qV=(1/2)mv^2

(-1.6x10^-19)(250) = (1/2)(9.11x10^-31)v^2

v = +/- 9.371x10^6 i

KE = (1/2)(9.11x10^-31)(9.37x10^6) = 4.268035x10^-24

If a=qE/m and we know the q charge value for an electron, we know the mass of an electron, how do I find the E value produced by 250V?

Last edited:

Delphi51
Homework Helper
I didn't check the math, but I think you have it. By definition, electric potential is "energy per charge" so V = E/q, E = qV is the kinetic energy given to charge q as it passes between the plates.

That's the short way. The long way is to put together
V causes E causes F causes a causes Δv
E = V/d, F = qE, F = ma, Vf² = Vi² + 2ad

The second part, about "leaving the positive charge area" is interesting. Usually we take the E field to be isolated between the plates and say no further acceleration takes place after the electrons pass through the positive plate. But is that really true? Can we tell by imagining a Gaussian surface through the cathode-anode system?

berkeman
Mentor

## Homework Statement

Have a few questions on CRT's. If we are given the accelerating voltage (250V) how would I calculate the kinetic energy of the electron as it leaves the positive end of the charge area? What is the formula for the speed of an electron along the tube (vz) as it leaves the positive charge area, in terms of the accelerating potential Va and the mass and charge of an electron.

F = qE=ma
KE= (1/2) mv^2

## The Attempt at a Solution

Having only been given the accelerating voltage (250V) I'm not sure where to begin. We did all kinds of stuff with kinetic energy in Newtonian physics but I'm not sure how to apply it with voltage.

I think at some point I arrived at

qV=(1/2)mv^2

(-1.6x10^-19)(250) = (1/2)(9.11x10^-31)v^2

v = +/- 9.371x10^6 i

KE = (1/2)(9.11x10^-31)(9.37x10^6) = 4.268035x10^-24

If a=qE/m and we know the q charge value for an electron, we know the mass of an electron, how do I find the E value produced by 250V?
The handy thing to remember is that eV is a unit of energy. So if you accelerate an electron through a voltage difference of 1kV, it has 1keV of kinetic energy.

In a CRT, the end of the electron gun is usually at the anode voltage, so the electrons do not accelerate the rest of the way to the anode and phosphur. They get deflected (by plates or coils) after the gun, but they move at a constant speed toward the anode/faceplate after leaving the electron gun.

BTW, I think you forgot to square the velocity in your equation here:

KE = (1/2)(9.11x10^-31)(9.37x10^6) = 4.268035x10^-24

Thanks for the help, I'm definitely further along now. I have just one more question.

I've found that the speed of the electron as it leaves the anode can be given as:

Vz = ((2)(e)(Va)/(m))^(1/2) (e=electron charge, Va=accel voltage, m=electron mass)

Once the electron enters the vertical deflection plates the velocity of the x component can be shown as (y isn't affected of course):

Vx = ((e)(Vd)(l)) / ((m)(d)(Vz)) (Vd=deflection voltage, l=plate length, d=distance between plates)

Now if the distance from the end of the deflecting plates to the phosphor screen is L, at what distance D from the center will the electron beam impact? (Hint given: the two velocity components of an electron and the two displacement components L and D are sides in similar triangles). I should end up with D = ((l)(L)/(2)(d)) ((Vd)/(Va))

berkeman
Mentor
Thanks for the help, I'm definitely further along now. I have just one more question.

I've found that the speed of the electron as it leaves the anode can be given as:

Vz = ((2)(e)(Va)/(m))^(1/2) (e=electron charge, Va=accel voltage, m=electron mass)

Once the electron enters the vertical deflection plates the velocity of the x component can be shown as (y isn't affected of course):

Vx = ((e)(Vd)(l)) / ((m)(d)(Vz)) (Vd=deflection voltage, l=plate length, d=distance between plates)

Now if the distance from the end of the deflecting plates to the phosphor screen is L, at what distance D from the center will the electron beam impact? (Hint given: the two velocity components of an electron and the two displacement components L and D are sides in similar triangles). I should end up with D = ((l)(L)/(2)(d)) ((Vd)/(Va))
Sorry, I don't understand your x, y, z coordinates. And can you show your work for the last part please?

Sorry, I don't understand your x, y, z coordinates. And can you show your work for the last part please?
Sorry, I'm using the perspectives that my lab book uses so I'll correct them for conventional usage.

I've found that the speed of the electron as it leaves the anode can be given as:

Vx = ((2)(e)(Va)/(m))^(1/2) (e=electron charge, Va=accel voltage, m=electron mass)

Once the electron enters the vertical deflection plates the velocity of the x component can be shown as (y isn't affected of course):

Vy = ((e)(Vd)(l)) / ((m)(d)(Vx)) (Vd=deflection voltage, l=plate length, d=distance between plates)

The last equation that I entered: D = [(l)*(L)/(2)*(d)]*[(Vd)/(Va)] is given in the lab. I have to use what I've put together so far to derive it.

*EDIT* Ok, I've pretty much got it figured out. Just enter the equations into the proportion and solve for D. That provides the correct equation.

Last edited:
Delphi51
Homework Helper
D = [(l)*(L)/(2)*(d)]*[(Vd)/(Va)] is given in the lab
This is puzzling. For instance, suppose L = 0 (the phosphor screen is right at the end of the deflecting plates). This formula gives D = 0. But the electrons are still deflected while between the plates and D is NOT zero. Are the units of the formula even correct? Looks like distance = distance cubed!!

I don't see the idea of the similar triangles. The path of the electron is parabolic while between the plates; no triangle can represent that. The similar triangle may make some sense for the last part of the trajectory - the distance L from the end of the plates to the screen where the trajectory is a straight line.

I see the problem as
D = vertical distance travelled while accelerating vertically between plates
+ distance traveled while going the horizontal distance L after plates
= ½a⋅t² + Vy*T
t is the time to pass through the plates, T the time to cover distance L.
These times are easy to find since the horizontal motion is uniform.
t = l/Vx = l*sqrt(m/(2eVa) and T = L*sqrt(m/(2eVa)
It all works out to
D = Vd/Va*(l²/2 + l*L)/(2d)
which at least is dimensionally correct, is zero when l = 0 (no deflecting plates) and still has a deflection when L = 0.