# Calculate the velocity of electron flying through electric field

1. Oct 20, 2013

### skrat

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

2. Relevant equations

3. 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?

2. Oct 21, 2013

### UltrafastPED

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. Oct 21, 2013

### skrat

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. Oct 21, 2013

### UltrafastPED

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. Oct 22, 2013

### skrat

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. Oct 22, 2013

### UltrafastPED

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. Oct 22, 2013

### skrat

Ok I understand! Thanks for your help! =)

8. Oct 22, 2013

### skrat

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 $\phi =60°$