# A heated cathode inducing thermionic emission of electrons.

1. Mar 28, 2015

### 3OPAH

Hello all. I have been looking at this problem:

I wrote three equations there, one for each part, which I think will help me solve each part; is my approach to the problem using those equations that you see there correct? I am just looking for some advice on where and how to start this problem. It looks like to me that I cannot calculate the velocity of the electrons until I get the volume charge density and current density between the two electrodes? To help you help me, here are some notes pertaining to this problem, but I have no similar examples to go off of.

Any and all help is greatly appreciated.

Last edited by a moderator: Mar 28, 2015
2. Mar 28, 2015

### uart

It looks like part "c" is the easy one. You don't want $J=\sigma E$ though, that would be more appropriate in a solid conductor/resistor. In this case I believe the curernt density would just be $J=q_e N$.

Parts "a" and "b" are the interesting ones. I agree with the equation $u = J / \rho_v$ and I think that is key. So clearly if we know the velocity "u" then we can find the volume charge density, and vice versa. But which one to find first and how?

Given the relatively simple nature of the problems in the notes you linked, my guess is that you are meant to make some simplifying assumptions about the spatial distribution of the electric field, and then to find the electron velocity from the electric field.

I think that you are meant to just take the electric field as uniform (even though it's not really the case, due to the presence of space charge and $\nabla \cdot E = \rho_v / \epsilon_0$).

To be honest this is not exactly my area, but a quick "back of the envelope" calculation using what I believe would be typical values for current density and electric field in such a tube, tells me that after the electrons have been accelerated through just a couple of volts that the space charge density is low enough that the change in electric field is very low compared to it's average value. So hopefully the constant field assumption is indeed an appropriate one.

Last edited: Mar 28, 2015
3. Mar 29, 2015

### 3OPAH

Ok, so current density between the electrodes is very simple as you showed, but I am having trouble with the volume charge density calculation. I just need to figure out a way to calculate this volume charge density and then the velocity of the electrons is simply $u = J / \rho_v$ . Any ideas, uart?

4. Mar 29, 2015

### uart

Yes, as I already stated above, if you assume that the electric field is constant then you can very easily deduce the velocity as a function of position.

Specifically, if you assume that the E field is constant then the voltage is a linear function of position. The velocity is then calculated by simply equating the changes in the electrostatic potential energy and the kinetic energy of the electrons.

5. Mar 29, 2015

### 3OPAH

I understand that the voltage vs position is linear in a constant E-field, but why wouldn't the next step after calculating the current density, part c, be to calculate volume charge density? Then simply plug those two values into the equation for velocity of the electrons, $u = J / \rho_v$ ?

6. Mar 29, 2015

### uart

That would be just fine, if you can find a way to calculate the volume charge density independently of the velocity. Can you please explain how do you propose to do that? And if as I suspect you do not know how to do that, then why reject my method which is simple and which works.

Look I'm just trying to understand your train of thought here. I've just outlined a very simple method to first find velocity and then find charge density, yet you think it would be easier to do it the other way around despite (so far) having no proposed way of doing so.

7. Mar 29, 2015

### 3OPAH

Ok, so E-field is constant which means that the plot for voltage vs position of electrons is linear inside the medium of the heated cathode. Then to calculate velocity of these electrons we simply equate the electrostatic potential energy and the kinetic energy of these electrons. Knowing that the potential energy for electrons is U=kQq / r, but to find the kinetic energy of a particle we would have to use some special relativity an Einstein's famous equation for energy E=mc^2. I am having trouble understanding how I can relate electrostatic potential energy and the kinetic energy of these electrons and go on to calculate the velocity from this.

8. Mar 29, 2015

### uart

No it's easier than that. The electrostatic PE is simply qV, so the change in PE is just the electron charge times the change in voltage.

Also I don't think you need to use relativistic KE unless the question specifically asked for it, or a very high voltage was specified. Under 300 volts the electron velocity is less than about 10^7 m/s, so relativistic effects are extremely small.

Last edited: Mar 29, 2015
9. Mar 29, 2015

### 3OPAH

So electrostatic potential energy is qV and KE=1/2 mv^2. So we have:

qV=1/2 mv^2

Solving for v, (velocity),

v=

Would that be the necessary calculation?

10. Mar 29, 2015

### 3OPAH

So here is my whole approach to the problem. What do you think, uart?

11. Mar 29, 2015

### uart

Watch your signs. The cathode is at z=0 and the anode at z=d (and I'm taking d to be positive), so the direction of the electrons is clearly the postive z direction.

Also you want the velocity and charge density as a function of position. So start with $V(z) = (V_A/d)z$, where $V_A$ is the anode voltage.

The KE of the electron is equal to the *decrease* in the electrostatic PE. So,
$$\frac{1}{2} m v^2 = q_e(0 - V(z))$$
where I'm using $q_e$ here to mean the *signed* charge of the electron. Of course you have a double negative there, so you can just use the elementary charge magnitude and drop the negative in front of V(z).

So use the above to put everything in terms of position "z", and only take the positive square root in your final expression for velocity.