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## 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?