Question

In the figure, a proton, mass , is projected horizontally midway between two parallel plates that are separated by , with an electrical field with magnitude between the plates. If the plates are long, find the minimum speed of the proton that just misses the lower plate as it emerges from the field.

Answer 1

and this is what I did :
\[\Sigma F_y=m.a_y \rightarrow mg + Eq = m.a\]
\[a_y = 5.94*10^{13}\]
\[y=1/2 a_y t^2 \rightarrow t=1.3*10^{-8} \]
\[v_x = d/t \rightarrow v_x = 3.3*10^6\]
but this is not right for some reason and when I googled it they say you have to consider relativistic effects. but I have no idea how to do that please help.

Answer 2

At these speeds, you need to adjust the mass m by the relativistic adjustment factor. We normally write it as
\[ \gamma = \frac{1}{\sqrt{1-v^2/c^2}} \]
The adjusted (and correct!) mass is
\[ m' = \gamma m \]
Then in your first equation
\[ \gamma m a = \gamma m g + Ee \]
where \( e \) is the charge of a proton
(= with sign change, charge of electron = 1.6 x 10^(-19) C)

Now using this find your new \( a \) and then use standard kinematics to find the maximum length of the plates.

Answer 3

and how can I calculate the speed for γ is the speed the initial speed or final speed?

Answer 4

You know elementary ODEs?

Answer 5

yeah

Answer 6

Well, write out the ODE in v and dv/dt = a.

Answer 7

oh.. ok , I didn't think it's gonna involve difrentials ... thank you very much dude !

Answer 8

one more question ,
am I writing the equation right ?
\[m/\sqrt{1-v^2/c^2}(dv/dt-g)=Ee\]

Answer 9

Yes. This equation is separable, although messy.

Answer 10

If you have a tutor for this course, you might also run this by them. They may be expecting you to use some sort of approximation.