Electrostatics - Electron passing through charged plates

[fobbz]

Homework Statement

http://img716.imageshack.us/img716/7692/picture8sj.png

Homework Equations

ΔE_p=QV
E_k=0.5mv²

The Attempt at a Solution

I'm making sure I understand what's going on here. Please bare with me.
ΔE_p = E_k
.5mv² = Q(1200+V)

Solving for V, I get -642V.

In a nutshell, is this what is happening?:
So electron passes the first plate, and accelerates do to the potential of 1200V. It then becomes near the -642V plate, and stops. Why does it stop because of the plates negative voltage??

Also, since electric field runs from Postive to negative, essentially the field goes form left to right?

The answer by the way is A.

 

[Delphi51]

Looks good. I'm a little uncertain if it should be 1200+V or 1200-V.
I worked it out with just V and got 558 Volts for the potential difference between plates. The field has to be to the right to push left on a negative charge. So the 1200 plate has to have a higher voltage then the other one. It must be 558 V higher. That is, the other one is 1200 - 558.

 

[Simon Bridge]

Modelling the field outside the plates as negligible huh?
Note, electrons are negative - so the energy gain though potential difference ΔV is -eΔV.
The field has to point in the direction of travel (left to right) to slow the electron, so V < 1200V. (electrons like to roll uphill, so they slow-down going downhill)

Since this is multi-choice, that's all the info needed to answer the question:
there is only one option with the right field direction and with V < 1200.
You don't actually have to crunch the numbers!

Just for fun - trying a different way:
1eV is the energy lost by an electron when decelerated through a potential difference of 1V. So this seems like a convenient unit to use.

rest mass of an electron is 511keV/c²
(personally I find this easier to remember than whatever the kg amount is.)

v/c = 0.0467, so gamma (γ) is 1.0011 ergo: non-relativistic. Still...

K = (γ-1)E₀ = 0.0011x511x10³eV = 558.11eV

[Note, if you still wanted to use mv²/2 you just have to realize that m = E₀/c² so:
K ≈ 511x10³x(0.0467)²/2 = 557eV]

So it has to go through a potential drop of about 560V.

 

[fobbz]

Modelling the field outside the plates as negligible huh?
Note, electrons are negative - so the energy gain though potential difference ΔV is -eΔV.
The field has to point in the direction of travel (left to right) to slow the electron, so V < 1200V. (electrons like to roll uphill, so they slow-down going downhill)

Since this is multi-choice, that's all the info needed to answer the question:
there is only one option with the right field direction and with V < 1200.
You don't actually have to crunch the numbers!

Just for fun - trying a different way:
1eV is the energy lost by an electron when decelerated through a potential difference of 1V. So this seems like a convenient unit to use.

rest mass of an electron is 511keV/c²
(personally I find this easier to remember than whatever the kg amount is.)

v/c = 0.0467, so gamma (γ) is 1.0011 ergo: non-relativistic. Still...

K = (γ-1)E₀ = 0.0011x511x10³eV = 558.11eV

[Note, if you still wanted to use mv²/2 you just have to realize that m = E₀/c² so:
K ≈ 511x10³x(0.0467)²/2 = 557eV]

So it has to go through a potential drop of about 560V.

I'm sorry but I think your calculations are a bit advanced for me... I'm just a senior in high school and we've never seen those types of equations.. Regarding the movement of electrons however I believe you can help me with. I don't understand what's happening physically to the electrons through the electric field. Could you please explain why as you put it "electrons like to roll uphill, so they slow-down going downhill" as this is new to me. Basically I think what you're saying is electrons want to be charged more, so they go for the higher potential? As soon as you introduce a lower voltage though - what exactly happens?

Thanks for replying, I have a physics exam tomorrow and I'm pretty much screwed for electrostatics and magnetism, but this is helping!

 

[Simon Bridge]

I'm sorry but I think your calculations are a bit advanced for me... I'm just a senior in high school and we've never seen those types of equations.

You'll do relativity later this year then. I was using Einstein's E=mc² and changed units to makes the math easy. gamma is \gamma = 1/\sqrt{1-(v/c)^2} and is pretty much the only remotely hard equation at this level.

That "rolls uphill" thing is a phrase which makes a fundamental relationship easier to remember for some people. What is does is relate electric potential to gravitational potential ... so you think of a high voltage being physically higher-up than a low voltage. Positive charges will go from a high voltage to a low one ... down-hill see? But electrons are negatively charged, so they go the other way: uphill.

Linking to something you have better intuition for helps you get a feel for how things will behave. There are differences in the details later but it serves well and for lots of situations it will get you in the right ball-park.

Note: electrons are charge. They don't gain charge by attaining higher potentials. What happens is; they lose energy by rolling up hills. A high potential for them is a large negative potential energy. Potential energy, remember, is qV.

If an electron is already in an area with a high potential and there's a lower one someplace else, the electron stays where it is. But if you introduce a higher one it will want to go there by the most direct path.

Similarly, a positive charge wants to stay in the holes and hollows.

Some people think of the electric potential as being like those roads the light-cycles run on in Tron: where you can ride either side of them. So negative charges right the underside and feel gravity as upwards, while positive charges ride on top feeling gravity downwards.

Pick the one that's easy to remember and good luck on the exam.
Your basic grasp seems to be OK, you'll be fine.