Calculate the velocity of electron flying through electric field

In summary, the conversation discusses the calculation of the angle and full energy of an electron traveling through an electric field with a given speed and field strength. The discussion involves using equations such as ##E=(E,0,0)## and ##F=(eE,0,0)## to calculate the velocity and energy of the electron, as well as considering factors such as time dilation. Ultimately, it is determined that the total kinetic energy of the electron is the sum of its initial kinetic energy and the energy gained from the electric field, and the resulting angle and speed can be calculated using this information.
  • #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, let's 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## let's 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?
 
Physics news on Phys.org
  • #2
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 traveling 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
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 let's 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
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
UltrafastPED said:
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
skrat said:
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
Ok I understand! Thanks for your help! =)
 
  • #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°##
 

FAQ: Calculate the velocity of electron flying through electric field

What is the equation for calculating the velocity of an electron in an electric field?

The equation is v = (E/m)^1/2, where v is the velocity, E is the electric field strength, and m is the mass of the electron.

What units are used for the electric field strength and electron mass in the velocity equation?

The electric field strength is typically measured in volts per meter (V/m) and the electron mass is measured in kilograms (kg).

How do you determine the direction of the electron's velocity in an electric field?

The direction of the electron's velocity is determined by the direction of the electric field. The electron will accelerate in the direction of the electric field if it is negatively charged, and in the opposite direction if it is positively charged.

Can the velocity of an electron in an electric field be greater than the speed of light?

No, according to the theory of relativity, the speed of light is the maximum speed at which any object can travel. Therefore, the velocity of an electron in an electric field cannot exceed the speed of light.

How does the velocity of an electron change as it travels through different electric fields?

The velocity of an electron will change as it travels through different electric fields. It will accelerate in a stronger electric field and decelerate in a weaker electric field. However, the direction of its velocity will always be determined by the direction of the electric field.

Similar threads

Back
Top