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# Electron moving across a parallel plate capacitor in a B field

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1. Electron moving across a parallel plate capacitor in a B field
I'm having trouble formulating an answer for this question.

There are two parallel plates acting as a capacitor, arranged in the y-z plane, one at 0 volts and one at +V volts. A uniform magnetic field of flux density B field acts parallel to these plates in the z direction.

An electron is emitted from the surface of the 0 volt plate, with negligible velocity. Find the minimum value of +V required for the electron to be able to reach the opposite plate.

Obviously, initially with velocity effectively 0, there is no force from the magnetic field, only from the electric field (which I believe should be providing an acceleration of e(sigma)/2(epsilon) in the x direction, where sigma is charge density of the plates and epsilon is the vacuum permittivity).

But as soon as the electron starts to move in the x direction it will start to experience an increasing force from the magnetic field (equal to ev cross B).

The problem I have is trying to combine these two effects. The normal result of motion in a B field would be circular or helical rotation, but here there is also a constant acceleration in the x direction - which will feed back into the acceleration the B field causes, by increasing v over time (then decreasing it when the electron is headed in the other direction along the x axis).

So, how can I take these effects into account?

Also, am I correct in thinking that if the electron does not reach the far plate before it is turned through 90 degrees, that it will take up a stable loop and will never reach the far plate?
2. Re: Electron moving across a parallel plate capacitor in a B field
At a guess I think you have to take the limiting case: which is where an electron is travelling parallel to the positive plate of the capacitor (at the maximum speed it can be accelerated to). The electron will remain in this position if the force due to the magnetic field is equal to the force form the E field, but will move into the positive plate if the attractive force is larger.

Spoiler:
Show
(Sigma)/(2epsilon) = VB, where v is the maximum velocity and is a function of the potential difference.
.
3. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by 99wattr89)
I'm having trouble formulating an answer for this question.

There are two parallel plates acting as a capacitor, arranged in the y-z plane, one at 0 volts and one at +V volts. A uniform magnetic field of flux density B field acts parallel to these plates in the z direction.

An electron is emitted from the surface of the 0 volt plate, with negligible velocity. Find the minimum value of +V required for the electron to be able to reach the opposite plate.

Obviously, initially with velocity effectively 0, there is no force from the magnetic field, only from the electric field (which I believe should be providing an acceleration of e(sigma)/2(epsilon) in the x direction, where sigma is charge density of the plates and epsilon is the vacuum permittivity).

But as soon as the electron starts to move in the x direction it will start to experience an increasing force from the magnetic field (equal to ev cross B).

The problem I have is trying to combine these two effects. The normal result of motion in a B field would be circular or helical rotation, but here there is also a constant acceleration in the x direction - which will feed back into the acceleration the B field causes, by increasing v over time (then decreasing it when the electron is headed in the other direction along the x axis).

So, how can I take these effects into account?

Also, am I correct in thinking that if the electron does not reach the far plate before it is turned through 90 degrees, that it will take up a stable loop and will never reach the far plate?
When the electron has turned through 90 degs exactly, it is travelling parallel to the plates. At this moment the B and E forces must be equal.
(And won't it just continue to move parallel to the plates with constant velocity in the absence of a resultant force?)
Surely the limiting case is the one where the electron has turned through 90 degs just as it reaches the positive plate.
You can work out the speed of the electron from purely energy considerations in the electric field. The magnetic field imparts no kinetic energy to the electron.
This means you can equate the two forces because you know the speed from the separation and pd between the plates and hence the value of the magnetic force. This gives a formula relating V to the other quantities.
Last edited by Stonebridge; 13-03-2012 at 21:55.
4. Re: Electron moving across a parallel plate capacitor in a B field
An alternative approach is to use the Lorentz force

You know what E and B are:

So just solve the resulting ODE in the x position. It's sinusoidal so if the maximum value of x is less than d the particle won't reach the other plate.

edit: I'm taking d to be the separation between the plates.
Last edited by alice_in_video; 14-03-2012 at 12:49.
5. Re: Electron moving across a parallel plate capacitor in a B field

After reading your very helpful replies I thought I knew what I needed to do with this question, but on taking another look at the answer and thinking some more I'm not so sure I do after all, I think I'm a bit confused. The answer is that the minimum value of V has to be [e(dB)^2]/2m, where m is the mass of the electron.

I initially thought after reading some of the responses that what I need to do is show that for the force from the E field to be greater than (or is it greater than or equal to?) the force from the B field requires V to be [e(dB)^2]/2m, but I don't see how to do that - I thought I'd be able to cancel out terms, but what I actually get is V/d for the E field and vB for the B field. These are equal and opposite, so V/d - vB = positive if V is high enough, and negative if V is not high enough. But where does [e(dB)^2]/2m come into that?

But after reading the replies more carefully I’m now wondering if perhaps the electron can reach the far plate even if the force from the B field would be greater were it at 90 degrees – because if the plates are close enough and the difference in the E and B forces relatively small then perhaps it would just effectively run out of time to turn before it hit the far plate? Since the B force only partially opposes the E force at first, meaning that there would still be a net acceleration towards the plate until the electron had gone far enough to turn to close to 90 degrees. This does at least mean that d has to be involved in the answer, as required.

(Original post by XiaoXiao1)
At a guess I think you have to take the limiting case: which is where an electron is travelling parallel to the positive plate of the capacitor (at the maximum speed it can be accelerated to). The electron will remain in this position if the force due to the magnetic field is equal to the force form the E field, but will move into the positive plate if the attractive force is larger.

Spoiler:
Show
(Sigma)/(2epsilon) = VB, where v is the maximum velocity and is a function of the potential difference.
.
Also, I get that (Sigma)/(2epsilon) = vB not VB though, just to make sure that's what you meant.

(Original post by Stonebridge)
When the electron has turned through 90 degs exactly, it is travelling parallel to the plates. At this moment the B and E forces must be equal.
(And won't it just continue to move parallel to the plates with constant velocity in the absence of a resultant force?)
Surely the limiting case is the one where the electron has turned through 90 degs just as it reaches the positive plate.
You can work out the speed of the electron from purely energy considerations in the electric field. The magnetic field imparts no kinetic energy to the electron.
This means you can equate the two forces because you know the speed from the separation and pd between the plates and hence the value of the magnetic force. This gives a formula relating V to the other quantities.
I don’t think the B and E forces would necessarily be equal at 90 degrees though. See the first part of this post for my confusion about this.

When you say I can work out the speed from purely energy considerations I’m not quite sure what you mean. Are you saying that its kinetic energy will be equal to the loss of potential due to position within the B field? Because I thought that some of it would be wasted overcoming the B field, since the B force acts partially or wholly against the E force.

Also, if the forces would be equal and opposite at 90 degrees then would it not have taken infinite time for the electron to turn through 90 degrees? Also, also, if the B force were greater the electron would keep turning and take up a loop, but would that loop be stable? As it came back around to complete the first full circuit, it would now have no velocity in the x direction as before, but a high velocity in the y direction, so it would have both the B and F forces accelerating it in the x direction initially, so it would go much faster in the x direction. But maybe the increased deflective effect of B matches that increase and negates the effect?

(Original post by alice_in_video)
An alternative approach is to use the Lorentz force

You know what E and B are:

So just solve the resulting ODE in the x position. It's sinusoidal so if the maximum value of x is less than d the particle won't reach the other plate.

edit: I'm taking d to be the separation between the plates.
That seems to effectively be what I tried today – it gives F = V/d – vB in the x direction and no other forces (when the particle is at 90 degrees), but that doesn’t seem to give the solution required.
6. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by 99wattr89)

After reading your very helpful replies I thought I knew what I needed to do with this question, but on taking another look at the answer and thinking some more I'm not so sure I do after all, I think I'm a bit confused. The answer is that the minimum value of V has to be [e(dB)^2]/2m, where m is the mass of the electron.
Have you covered the situation of charged particles in crossed fields? It's a pretty standard lecture derivation. The solution for the motion is circular motion about a guiding centre. Once you work through this you need to set the diameter of the circle equal to the plate separation. This does give the answer you were given.

Spoiler:
Show
You know vz is zero so you end up with two coupled differential equations in vx and vy
7. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by suneilr)
Have you covered the situation of charged particles in crossed fields? It's a pretty standard lecture derivation. The solution for the motion is circular motion about a guiding centre. Once you work through this you need to set the diameter of the circle equal to the plate separation. This does give the answer you were given.

Spoiler:
Show
You know vz is zero so you end up with two coupled differential equations in vx and vy
Huh, that's strange, there was never anything like that in any of the lectures.

In mechanics lectures (but not the electromagnetism lectures) we did look at a slightly different setup, and the textbooks I have also address that other setup, which is the velocity filter. In the velocity filter the E field is in the y direction though, not the x direction.

But in any case, I'm closer now to the answer, so thank you for your post!

I now have:

From the two equations, one for x and one for y. For the limiting case the maximum value of x must be d, and at t=0, x and x dot are both zero.

But I can't solve this equation - the forms of solution I try aren't working. Can anyone help me get started?
8. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by 99wattr89)
Huh, that's strange, there was never anything like that in any of the lectures.

In mechanics lectures (but not the electromagnetism lectures) we did look at a slightly different setup, and the textbooks I have also address that other setup, which is the velocity filter. In the velocity filter the E field is in the y direction though, not the x direction.

But in any case, I'm closer now to the answer, so thank you for your post!

I now have:

From the two equations, one for x and one for y. For the limiting case the maximum value of x must be d, and at t=0, x and x dot are both zero.

But I can't solve this equation - the forms of solution I try aren't working. Can anyone help me get started?
I think the easiest way is to keep it in terms of vx and vy, and differentiate the equation for vx and then sub in the the equation for vy to get the solutions for vx and vy which you can then integrate to get the x and y solutions.

Edit: It is possible to solve the equation you have but you also need the y equation to determine some of the integration constants.
Last edited by suneilr; 15-03-2012 at 13:58.
9. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by suneilr)
I think the easiest way is to keep it in terms of vx and vy, and differentiate the equation for vx and then sub in the the equation for vy to get the solutions for vx and vy which you can then integrate to get the x and y solutions.
I'm not sure I understand. The equations I have are:

1.

and

2.

So differentiating 2, as it's the one with , would give

If I subbed that into 1, it would give the same sort of equation as I mentioned in my last post - although with a higher order differential. How does that help solve the question? Sorry if I'm being dense!
10. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by 99wattr89)
I'm not sure I understand. The equations I have are:

1.

and

2.

So differentiating 2, as it's the one with , would give

If I subbed that into 1, it would give the same sort of equation as I mentioned in my last post - although with a higher order differential. How does that help solve the question? Sorry if I'm being dense!
*
**

If you differentiate * you can sub in ** and you get an expression you can integrate directly, remembering the initial conditions to fix the constants.
11. Re: Electron moving across a parallel plate capacitor in a B field
(Original post by suneilr)
*
**

If you differentiate * you can sub in ** and you get an expression you can integrate directly, remembering the initial conditions to fix the constants.
I finally solved it. Turns out the trail solutions I tried were all wrong.

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