Calculate the velocity of electron flying through electric field

  • Thread starter skrat
  • Start date
  • #1
skrat
748
8

Homework Statement


An electron with speed ##v_0=3/5c## flies 1m through electric field with ##E=1MV/m##. Calculate the angle of the flight when the electron exits the field and it's full energy.



Homework Equations





The Attempt at a Solution


The velocity changes only in one direction, lets say that this is direction of axis x, than ##E=(E,0,0)## and ##F=(eE,0,0)## while ##v_1=(0,3/5c,0)## and ##v_2=(v_x,3/5c,0)##
##eE=m\gamma \frac{dv}{dt}##
##\frac{eE}{m}t=\gamma v_x## lets say ##A=\frac{eE}{m}t##
##A=\frac{v_x}{1-\frac{v_x^2}{c^2}}## or do I have to use ##\frac{1}{1-\frac{v_x^2-v_y^2}{c^2}}## for ##\gamma ## ??
so ##v_x=\frac{cA}{\sqrt{A^2+c^2}}=286 779 393 m/s##

and therefore ##tan\alpha =\frac{v_y}{v_x}## so ##\alpha =32°##.

Is this ok?

Now I'm really having some troubles with calculating the full energy...
For example: ##w=m\gamma c^2=m\frac{1}{1-\frac{v_x^2}{c^2}-\frac{v_y^2}{c^2}}c^2## How does this sound?

One more thing, I know I am not allowed to calculate the speed when exiting as ##v_2=\sqrt{v_{x}^{2}+v_{y}^{2}}## but how can I?
 

Answers and Replies

  • #2
UltrafastPED
Science Advisor
Gold Member
1,914
216
So it will add 1 MeV of energy to the motion in the direction of the field; what is this as a fraction of c?

Previously it had enough energy to be travelling at 0.6c in it's original direction. How much energy does this correspond to?

Having both energies, what should the total energy be? For that energy what should its net speed be?
 
  • #3
skrat
748
8
One more thing I have to say before I try to answer your questions... The electron enters perpendicular to the electric field.

So, if I understand you correctly you are trying to say that the total energy of the system remains the same.

So, at the beginning, when the electron enters the field it's full energy is ##\sqrt{p^2c^2+m^2c^4}+1MeV## where ##p=\gamma mv_y##.
And at the exit ##\sqrt{p_0^{2}2c^2+m^2c^4}##

This should me than equal (if ##1MeV=W_e##):
##\sqrt{p^2c^2+m^2c^4}+W_e=\sqrt{p_0^{2}2c^2+m^2c^4}##
##p^2c^2+m^2c^4+2W_e\sqrt{p^2c^2+m^2c^4}+W_e^{2}=p_0^{2}2c^2+m^2c^4##
##p_0^{2}=p^2+2W_e\sqrt{p^2/c^2+m^2}+W_e^{2}/c^2##
Now lets say that ##A=p^2+2W_e\sqrt{p^2/c^2+m^2}+W_e^{2}/c^2##.
The equation is than ##p_0^{2}=\frac{m^2v^2}{1-v^2/c^2}=A##
so ##v=\sqrt{\frac{Ac^2}{A+m^2c^2}}##
where ##m^2c^2## is really small! so ##v=\sqrt{c}##

Or... ?
 
  • #4
UltrafastPED
Science Advisor
Gold Member
1,914
216
What I am saying is that the electron had some initial kinetic energy (which you can calculate from its speed), and it will acquire an additional 1 MeV of kinetic energy from the electric field (1 MV/m for 1 meter is a 1 MV potential difference over that 1 meter).

So if you start with a speed for one, and the additional energy for the other ... so find the energy for the first, and the speed for the second.

Your total kinetic energy is just the sum of the two kinetic energies.
 
  • #5
skrat
748
8
What I am saying is that the electron had some initial kinetic energy (which you can calculate from its speed), and it will acquire an additional 1 MeV of kinetic energy from the electric field (1 MV/m for 1 meter is a 1 MV potential difference over that 1 meter).

So if you start with a speed for one, and the additional energy for the other ... so find the energy for the first, and the speed for the second.

Your total kinetic energy is just the sum of the two kinetic energies.

Well, that sounds a bit too easy? What about time extension? Time is slower for the electron and therefore "feels" the force longer and changes direction more - meaning, the total travel distance in the field he makes is no longer 1 m but more, due to the electric force.
 
  • #6
UltrafastPED
Science Advisor
Gold Member
1,914
216
Well, that sounds a bit too easy? What about time extension? Time is slower for the electron and therefore "feels" the force longer and changes direction more - meaning, the total travel distance in the field he makes is no longer 1 m but more, due to the electric force.

I'm not doing it by force x distance; you can just take the change in electric potential from one side to the other ... here it is 1 MV, so the electron picks up 1 MeV. Yes, it is that simple!
 
  • #7
skrat
748
8
Ok I understand! Thanks for your help! =)
 
  • #8
skrat
748
8
If 1MeV is th kinetic energy than i can calculate one component of speed:

##T=mc^2\gamma =10^6eV##
##\gamma =\frac{10^6eV}{mc^2}##
##\frac{1}{\sqrt{1-v^2/c^2}}=\frac{10^6eV}{mc^2}##
##1-v^2/c^2=(\frac{mc^2}{10^6eV})^2##
so
##v=c\sqrt{1-(\frac{10^6eV}{mc^2})^2}\doteq c##

am... Is that really possible? If yes, than for ##v_y=3/5c## ---- ##tan\phi =\frac{v_x}{v_y}=\frac{5}{3}## so ##\phi =60°##
 

Suggested for: Calculate the velocity of electron flying through electric field

  • Last Post
Replies
5
Views
4K
  • Last Post
Replies
0
Views
7K
Replies
2
Views
1K
Replies
3
Views
6K
  • Last Post
Replies
8
Views
2K
Replies
8
Views
7K
Replies
4
Views
10K
Replies
6
Views
1K
Replies
1
Views
912
Top