Calculating Maximum Voltage of a Plate for electron movement

[QuickSkope]

Homework Statement

Drawing of the Problem

An electron rests on plate A and is accelerated by a 400V Plate (B). From there, Calculate the Maximum Voltage a 3cm plate (C) can be charged before the electron makes contact with the plate. Directly opposite Plate (C), is a +100V Charged Plate D.

Homework Equations

Electrostatic forumulas

The Attempt at a Solution

I assume the first step would be to calculate the velocity at which the electron leaves Plate B.

Voltage is 400V, and Ep can be calculated using Ep = q * V

Ep = (1.60 * 10^-19) * 400
= 6.4 * 10^-17
Ek = Ep
Ek = 6.4 * 10^-17

1/2mV^2 = 6.4 * 10^-17
V = 1.2 * 10^7

Is that correct for the velocity when the electron leaves Plate B?

After this, I have to calculate the maximum Voltage. That really isn't a massive problem as I can do that part by myself. The part that confuses me is that now I have a 400V current in the plate A pulling my electron back. Can someone show me how id do that?

 

[Simon Bridge]

The speed you got is 0.04c. i.e. 4% of the speed of light.
That seems fast doesn't it - and there's all these nasty numbers: it would be easy to miss out a power of ten somewhere wouldn't it?

Note: when it reaches plate B, it has a kinetic energy of 400eV ;)
(one electron, 400 volts: 400 electron volts ... easy aye?)

In the same units, the electron mass is 511keV/c² - so you can get it's speed without having to deal with nasty numbers.

If the electron is not relativistic:
$E_K=\frac{1}{2}mv^2 = \frac{1}{2}(mc^2)(v/c)^2$Plate B is 400V with respect to plate A.
Usually plate B will be +200V and plate B will be -200V and they will be close enough together that the electric field on the axis, outside the plates, can be neglected.

But if you really want to take it literally, then you'll need to add the two fields together (or work the forces in components in a fbd).
Do you want to approximate plate B as infinite in extent? (Or how will you handle the edge effects?)

The whole setup reads funny - these plates must be insulators for eg.
Since the separation of C and D is big compared with their extent, there will be non-trivial differences from linear at the ends.

 

[QuickSkope]

Sorry, not followiing a ton of what youve stated there. So my velocity is incorrect? How would I go about gettig the right velocity?

Also, I believe plate B is just 400V, and Plate A carries no charge.

So you're saying that the 400V will have no affect on how far the electron will travel?

 

[Simon Bridge]

Sorry, not followiing a ton of what youve stated there.

Where did I lose you?

So my velocity is incorrect? How would I go about getting the right velocity?

I'm not saying your velocity is correct or not - I have, instead, provided a way you can check your own results.

If you change your units, to electron-volts, you get an easier calculation - one where you are less likely to mess up the entry into the calculator. 1eV is the amount of kinetic energy gained by an electron accelerated though a potential difference of 1V... so KE calculations are easy.

If I use $E_0=mc^2$, then the mass of the electron comes to 511000eV in energy.

To use these, I have to change the KE equation a bit ... just multiply by $c^2/c^2$ like this:

$$E_K=\frac{1}{2}mv^2 = \frac{c^2}{c^2}\left ( \frac{mv^2}{2}\right ) = \frac{1}{2}(mc^2)\left (\frac{v}{c}\right )^2 = \frac{1}{2}E_0(v/c)^2$$... solve for v/c.

Also, I believe plate B is just 400V, and Plate A carries no charge.

Draw a free body diagram for the electron and do $\sum \vec{F} =ma$.

The effect is of the ballistics setup - only the charge on plate C changes the direction of "gravity".

So you're saying that the 400V will have no affect on how far the electron will travel?

No - I am saying that the usual way this problem is set up - the accelerating potential does not have a significant effect on the horizontal direction once the electron has left it. In your case, it is not clear if this is intended.
It's a judgement call and you are the one on the spot.

What level are you doing this at?

 

[QuickSkope]

Where did I lose you?I'm not saying your velocity is correct or not - I have, instead, provided a way you can check your own results.

If you change your units, to electron-volts, you get an easier calculation - one where you are less likely to mess up the entry into the calculator. 1eV is the amount of kinetic energy gained by an electron accelerated though a potential difference of 1V... so KE calculations are easy.

If I use $E_0=mc^2$, then the mass of the electron comes to 511000eV in energy.

To use these, I have to change the KE equation a bit ... just multiply by $c^2/c^2$ like this:

$$E_K=\frac{1}{2}mv^2 = \frac{c^2}{c^2}\left ( \frac{mv^2}{2}\right ) = \frac{1}{2}(mc^2)\left (\frac{v}{c}\right )^2 = \frac{1}{2}E_0(v/c)^2$$... solve for v/c.

Draw a free body diagram for the electron and do $\sum \vec{F} =ma$.

The effect is of the ballistics setup - only the charge on plate C changes the direction of "gravity".

No - I am saying that the usual way this problem is set up - the accelerating potential does not have a significant effect on the horizontal direction once the electron has left it. In your case, it is not clear if this is intended.
It's a judgement call and you are the one on the spot.

What level are you doing this at?

Im doing this at a Grade 12 level. I have yet to learn the ev measurements, as as such have no idea how to convert or manipulate them. Ill try some calculations

Also, Eo means what exactly?

Sorry, I'm not following this electron voltage stuff. Is there not a way to do it using Ek = Ep like I did above?

Giving it a shot here:

Ek = 1/2 (mc^2) (v/c)^2

400 / (1/2) = (511000) * ((3 * 10^8)^2) * (v/c)^2

800 / (4.6 * 10^22) = (v/c)^2

( v/c ) = 1.32 * 10^-10
v = (1.32 * 10^-10) * (3 * 10^8)
v= 0.04

Does that prove my answer, as this equation outputs in therms of % of speed of light?

 

[Simon Bridge]

Im doing this at a Grade 12 level. I have yet to learn the ev measurements, as as such have no idea how to convert or manipulate them. Ill try some calculations

Also, Eo means what exactly?

$E_0=mc^2$ it's the most famous equation in the world.

Sorry, I'm not following this electron voltage stuff. Is there not a way to do it using Ek = Ep like I did above?

Giving it a shot here:

Ek = 1/2 (mc^2) (v/c)^2

400 / (1/2) = (9.11 * 10^-31) * ((3 * 10^8)^2) * (v/c)^2

Not bad - but $mc^2=511000\text{eV}$ so you don't need to multiply out with $c^2$.
The trick to manipulating equations is to do the algebra first and put the numbers in last.

$$E_K=\frac{1}{2}E_0(v/c)^2 \Rightarrow \frac{v}{c}=\sqrt{\frac{2E_K}{E_0}}$$

$E_0=511000\text{eV}\\ E_K=400\text{eV}$

Leave your answer as a fraction of the speed of light.

One of the reasons you do this is to see if the speed is relativistic.
If v/c > 1 then there's something wrong.
You want v/c << 1.

eV are just units for energy - like Joules. You can look up the conversion.
Use them the same way. It's just that they are very convenient for things like electrons.

I had hoped it was just a quick check for you - but it seems that non-SI units is still a bit in the future for you ;) Still - may be a bit of fun to see what's coming up.

 

[Simon Bridge]

The next step is more important - it looks a bit on the tough side for 12th grade but that's your problem right?

Do you know how to do a free body diagram?

From that you can get the accelerations in the x and y directions (assuming simple fields).
You know the initial velocities.
You know the suvat equations.

 

[QuickSkope]

$E_0=mc^2$ it's the most famous equation in the world.

Not bad - but $mc^2=511000\text{eV}$ so you don't need to multiply out with $c^2$.
The trick to manipulating equations is to do the algebra first and put the numbers in last.

$$E_K=\frac{1}{2}E_0(v/c)^2 \Rightarrow \frac{v}{c}=\sqrt{\frac{2E_K}{E_0}}$$

$E_0=511000\text{eV}\\ E_K=400\text{eV}$

Leave your answer as a fraction of the speed of light.

One of the reasons you do this is to see if the speed is relativistic.
If v/c > 1 then there's something wrong.
You want v/c << 1.

eV are just units for energy - like Joules. You can look up the conversion.
Use them the same way. It's just that they are very convenient for things like electrons.

I had hoped it was just a quick check for you - but it seems that non-SI units is still a bit in the future for you ;) Still - may be a bit of fun to see what's coming up.

Well, Its cool to see :D. And now I know. Maybe it will help me on my test tomorrow, who knows :P. Anyway, so by those calculations, that velocity is correct.

So the next step is seeing what will change the electrons movement. So if we assume the Perpedicular Plate has no effect on the electron, that let's use just use the Point potential between plates C and D, right?

The next step is more important - it looks a bit on the tough side for 12th grade but that's your problem right?

Do you know how to do a free body diagram?

From that you can get the accelerations in the x and y directions (assuming simple fields).
You know the initial velocities.
You know the suvat equations.

Yea, its supposed to be tough :P. Yes, I know how to do a free body diagram, just not sure how to do one in this situation. Would it look something like this?

http://draw.to/D1N5MAW

Those are the only 2 forces, correct? If the Perpendicular plate is negligible, and there's no force applied in the X for the electron.

If I break it down into X and Y components:

X:

V = 1.2 * 10^7
d = 0.03m
t = 2.53 * 10^-9 s (0.03/(1.2 * 10^7))

Y:
Vo = 0
Vf = X
d = 0.01m
a = ?
t = 2.53 * 10^-9s

d = vot+1/2at^2
a = 3.122 * 10^15 m/s^2

ma = F
(9.11 * 10^-31) * (3.122 * 10^15) = 2.84 * 10^-15 N

F /q = E

(2.84 * 10^-15) / (1.6* 10^-19) = 17777.78 N/C

E = V/d

17777.78 * 0.01 = 177.78

177.78 + 100V = 277.78 or 278V

Not sure if that's correct, just kinda went at it.

But if I ended up with a Voltage like this, you'd really think the perpendicular Plate would have an affect on the electron.

 

[gneill]

Well, Its cool to see :D. And now I know. Maybe it will help me on my test tomorrow, who knows :P. Anyway, so by those calculations, that velocity is correct.

So the next step is seeing what will change the electrons movement. So if we assume the Perpedicular Plate has no effect on the electron, that let's use just use the Point potential between plates C and D, right?



Yea, its supposed to be tough :P. Yes, I know how to do a free body diagram, just not sure how to do one in this situation. Would it look something like this?

http://draw.to/D1N5MAW

Those are the only 2 forces, correct? If the Perpendicular plate is negligible, and there's no force applied in the X for the electron.

If I break it down into X and Y components:

X:

V = 1.2 * 10^7
d = 0.03m
t = 2.53 * 10^-9 s (0.03/(1.2 * 10^7))

Y:
Vo = 0
Vf = X
d = 0.01m
a = ?
t = 2.53 * 10^-9s

d = vot+1/2at^2
a = 3.122 * 10^15 m/s^2

ma = F
(9.11 * 10^-31) * (3.122 * 10^15) = 2.84 * 10^-15 N

F /q = E

(2.84 * 10^-15) / (1.6* 10^-19) = 17777.78 N/C

E = V/d

17777.78 * 0.01 = 177.78

Ooooh. Doing so well up to this point! But the distance between the horizontal plates is 2cm, not 1 cm.

177.78 + 100V = 277.78 or 278V

Not sure if that's correct, just kinda went at it.

But if I ended up with a Voltage like this, you'd really think the perpendicular Plate would have an affect on the electron.

Ignoring "end effects", it's usually assumed that for charged plates E field is uniform and entirely confined to the space between them.

 

[QuickSkope]

Alright, so then 17777.78 * 0.02 = 355.56V

355.56 + 100V = 456V.

Sounds a little more reasonable. Is that the correct answer?

 

[gneill]

Alright, so then 17777.78 * 0.02 = 355.56V

355.56 + 100V = 456V.

Sounds a little more reasonable. Is that the correct answer?

It looks good to me

 

[QuickSkope]

Sweet Deal, thank you guys :)

 

[QuickSkope]

Sweet deal, thanks guys :)

 

[Simon Bridge]

Well done :)