Answer:
The speed is [tex]138852.4 \frac{m}{s} [/tex]
Explanation:
Electric field magnitude (E) and electric force magnitude (F) are related by the equation
[tex]F=Eq [/tex]
with q the charge of the proton ([tex]1.61\times10^{-19} C[/tex]). Because between parallel plates electric field is almost constant, electric force is constant too:
[tex]F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N [/tex]
because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:
[tex]a=\frac{F}{m} [/tex]
with m the mass of the proton ([tex]1.67\times10^{-27} kg[/tex]):
[tex]a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}[/tex]
Now with this constant acceleration we can use the kinematic equation
[tex]v^{2}=v_{0}^{2}+2ad [/tex]
with v the final speed, [tex] v_i[/tex] the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:
[tex]v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)} [/tex]
[tex]v=138852.4 \frac{m}{s} [/tex]
