# Homework Help: Force acting on anode by electron current

1. Sep 1, 2016

### JulienB

1. The problem statement, all variables and given/known data

Hi everybody! I'm unsure how to tackle the following problem, hopefully some of you can give me some hints on how to do so. :)

How big is the force acting on an anode by an electron current of strength $I=0.3$A if the electrons are accelerated with a voltage $U=300$V?

2. Relevant equations

$\vec{F} = q \cdot \vec{E}$
$I = dQ/dt$

3. The attempt at a solution

I think I'm missing something here. I've drawn a picture of how I understand the problem, please let me know if something is wrong with my interpretation.

So the force should be $F = q \cdot E$ with $q = -e$ being the charge of an electron. My problem is the electric field, I cannot seem to find a way to calculate it! Other things I've tried/noticed:

$I = q/t$ since the current is constant, so I could solve for $t$ for an elementary charge. Does that make sense? I think I'd rather have $q = I \cdot t$, and then $F = I \cdot t \cdot E$. When I try to find $E$ using Gauss law, I get $E = \frac{q}{\epsilon_0 A}$ but that's a dead end since I don't have the area.

Surely I am missing an equation relating $U$ and $E$. Does anybody have a suggestion?

Julien.

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2. Sep 1, 2016

### Staff: Mentor

$F = q \cdot E$ tells you the force on charge q due to it being affected by an electric field E.

Think about the problem as a stream of small, massive particles striking the anode. An analogy might be a stream from a water hose striking a wall. Each particle carries a bit of momentum that must go somewhere when the anode captures it.

Can you think of a way to determine the momentum of individual electrons when they reach the plate?

3. Sep 1, 2016

### JulienB

Hi @gnell and thank you for your answer and for clarifying my misinterpretation of the force equation. What about using the electric potential energy then? When the electrons leave the cathode, they have a potential energy $U_e = F \cdot d = q \cdot E \cdot d = q \cdot U = I \cdot t \cdot U$. Now I would remove the $t$ by simply converting ampere into coulomb per second and assuming the time is one second. Does that work? I would then obtain:

$U_e = 0.3 \cdot 300 = 90 J$

Hopefully I did not commit the same mistake as before, but I think that using the force equation that way should give me the potential energy of an electron as it leaves the cathode.

Now mm... Can I assume that all of that potential energy was converted into kinetic energy as the electron reached the anode? If so:

$U_e = \frac{1}{2} m v^2 \implies v = \sqrt{\frac{2 U_e}{m}} \implies p = mv = m \sqrt{\frac{2 U_e}{m}} = \sqrt{2 m U_e}$

That all seems kind of strange and quite different from what I've done so far in EM... Please let me know if I am wrong before I go any further.

Thank you very much in advance.

Julien.

4. Sep 1, 2016

### Staff: Mentor

You're sort of on the right track. But recall that current times voltage yields power in watts (Joules per second). So 90 J of energy are delivered per second. The question is, what is the momentum associated with that energy? You'd need to look at what mass of electrons are delivered over that same second.

If I might suggest, you could determine the speed of an individual electron first. Then the momentum delivered per electron.
You can assume that all the KE is dumped into the anode, but what is more useful is the rate that momentum is transferred. Each electron gives up its momentum to the plate when it arrives and is captured. Recall what the definition of force is: $F = \frac{dp}{dt}$, the rate of change of momentum. So find the rate at which electrons arrive and thus the rate at which momentum arrives...

5. Sep 3, 2016

### JulienB

Hi @gneill and thank you for your answer. And sorry it took me a few days to answer again, I guess I am still somewhat confused.

How about that: the current $I = 0.3A = 0.3 C \cdot s^{-1}$ describes the flow of electric charge between the cathode and the anode. To me that means that $I/e$ would describe the number of electrons reaching the anode during a second, right? Does that answer your question about the "rate at which electrons arrive"?

Now the power (as you said) is known to be $P = I \cdot U = 90 W = 90 J \cdot s^{-1}$, that is how much work is done per second. If I would divide the power by the rate at which electrons arrive, wouldn't I get the work done by the electrons on the anode? That would be:

$W = \frac{P}{I/e} = P \cdot \frac{e}{I} = 90 J \cdot s^{-1} \cdot \frac{e}{0.3 C \cdot s^{-1}} = 300e J \cdot C^{-1}$

Does that make any kind of sense? The units work since $e$ can of course be expressed in Coulomb, but the work done seems to be very small. Could that be?

Please let me know if I am on the right track. I know I didn't quite follow all of your suggestions (like about the speed of an electron), this problem has confused me a lot.

Thank you very much in advance.

Julien.

Edit: Argh that can't be correct. I just realised that I wouldn't even need to know $I$ if that method was valid...

6. Sep 3, 2016

### JulienB

Okay new try. I kind of forgot that the problem was asking for the force and not for the work... I tried to follow your suggestions now:

$I = \frac{q}{t} \implies \frac{I}{e} = \mbox{rate at which electrons reach the anode, unit: } s^{-1}$
$\implies \frac{I}{m_e \cdot e} = \mbox{rate at which electrons reach the anode pro second pro electron mass, unit: } kg^{-1} \cdot s^{-1}$
$\implies P \cdot \frac{I}{e \cdot m_e} = I^2 \cdot U \cdot \frac{1}{e \cdot m_e} \mbox{ squared velocity of electrons, unit: } (m \cdot s^{-1})^2$
$\implies v = \sqrt{I^2 \cdot U \cdot \frac{1}{e \cdot m_e}} = \mbox{ velocity of electrons, unit: } m \cdot s^{-1}$
$\implies p = m_e \cdot v = m_e \cdot I \sqrt{U \cdot \frac{1}{e \cdot m_e}} = \mbox{ momentum of electrons, unit: } kg \cdot m \cdot s^{-1}$
$\implies F = \frac{dp}{dt} = I \sqrt{U \cdot \frac{m_e}{e}} = \mbox{ force applied by electrons on anode, with dt = 1s, } = 1.24 \cdot 10^{-5} N$

Well that is quite small but not crazy...The second and third steps are kind of clumsy, I tried to stick with the units as you can see. Could that be correct?

Julien.

Last edited: Sep 3, 2016
7. Sep 3, 2016

### Staff: Mentor

I think you're getting very close, but some items you've introduced confuse me a bit. For example, why would you want to know the rate at which electrons arrive per electron mass? Surely the simple rate at which they arrive (call it items per second) is sufficient to use to determine the momentum delivered if you know the "item" is momentum. You simply have to know the momentum of an individual item and multiply by the rate. So $r = \frac{I}{e}$ is all you need in terms of rate.

As for the velocity determination, I think you've really over complicated your approach. If you know the KE of one electron when it reaches the anode you can find its velocity by the usual approach:

$v = \sqrt{\frac{2 KE}{m_e}}$

and you've presented in post #3, for a charge e falling through potential U the change in KE is $KE = e \cdot U =$. Just combine the above to find the velocity of the electrons.

With the velocity you can find the momentum carried by the individual electrons. The rate tells you how many of those "momentum packets" arrive per second...

8. Sep 3, 2016

### rude man

What are the dimensions of your F expression?

Also: P (power) should not enter the computation at all, or at least it's distracting.
gneill in post 7 is giving you suggestions to be taken seriously ...

Last edited: Sep 3, 2016
9. Sep 3, 2016

### JulienB

@gneill @rude man Thx guys for your answers. I think I get it now, so I give it another go and try at the same time to give it some physical sense:

So the rate at which electrons reach the anode is $r = \frac{I}{e}$. That's how many electrons reach the anode in one second, and that's pretty clear to me.
Now the power $P = I \cdot U = 90 W$ is how much energy is delivered per second, or alternately the potential energy per second of a negative charge in the electric field created by the anode as it leaves the cathode (correct me if I am wrong here).

So as one second passes by, $\frac{I}{e}$ electrons reach the anode with an individual kinetic energy $KE = I \cdot U \cdot \frac{e}{I} = e \cdot U$. The velocity of the electrons can be derived using $KE = 1/2 \cdot m_e \cdot v^2$ which yields $v = \sqrt{\frac{2 \cdot KE}{m_e}}$. The momentum of each electron is then $p = m_e \cdot v = m_e \cdot \sqrt{\frac{2 \cdot KE}{m_e}} = \sqrt{2 \cdot e \cdot U \cdot m_e} = 9.36 \cdot 10^{-24} kg \cdot m \cdot s^{-1}$. Now I take $F = \frac{dp}{dt}$ with $dp$ being the amount of momentum of $\frac{I}{e}$ electrons in one second and $dt$ being that very same second, and the result is:

$F = 9.36 \cdot 10^{-24} N$.

Uf I feel like I have overcomplicated it one more time. I have attached another picture with my refreshed vision of what's going on in there. Please let me know if I misinterpreted something again.

Thanks a lot for your help, it's been very valuable. I hope I got it right now.

Julien.

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Last edited: Sep 3, 2016
10. Sep 3, 2016

### rude man

What happened to I/e? You're very close now ...

11. Sep 3, 2016

### JulienB

$dp = p \cdot \frac{I}{e}$
$F = \frac{dp}{dt} = 1.76 \cdot 10^{-5} N$

Was that what you meant?

Thanks a lot!

Julien.

12. Sep 3, 2016

### Staff: Mentor

That looks better! That last result looks good. But you need to forget the excursion into power and the involvement of rate when determining the value of the individual dp. Just concentrate on a single particle for that. Apply the rate after you have dp.

You've already presented suitable versions mixed in with all the rest over several posts. So picking out the relevant bits and discarding the rest it boils down to this:

For the KE of one electron falling through the given potential difference U, $KE = eU$. That gives us:

$v = \sqrt{\frac{2 eU}{m_e}}$

and so the momentum of a single electron arriving at the plate is:

$p_e = m_e \sqrt{\frac{2 eU}{m_e}} = \sqrt{2 eU m_e}$

Individually these bits of momentum are very tiny, and the current is made up of a great many of them. So we are justified in identifying $dp \approx p_e$. Thus we let:

$dp = \sqrt{2 eU m_e}$

Then apply your arrival rate to form $\frac{dp}{dt}$.

13. Sep 3, 2016

### JulienB

@gneill Thanks a lot for your help, that thread has been so helpful to me. With some practice it will surely go better and better.

Thank you again.

Julien.

14. Sep 3, 2016

### Staff: Mentor

You're very welcome. Good luck with your studies!

15. Sep 3, 2016

### rude man

Yay team!
But I wouldn't associate a numerical value to dp which is a differential quantity. Call it Δp instead, the finite change in momentum in 1 second.

16. Sep 4, 2016

### JulienB

@rude man Thx, I appreciate all your comments and it will definitely help me next time I face such a problem.

Julien.

17. Sep 5, 2016

### rude man

Fine! Try to appreciate the importance of dimensional analysis in all your work. There is nothing like it to find errors in your math! In fact it's the most important idea I can convey in physics, engineering, etc.
.