Answer:
(a) x=157 m
(b) No
Explanation:
Given Data
Mass of proton m=1.67×10⁻²⁷kg
Charge of proton e=1.6×10⁻¹⁹C
Electric field E=3.00×10⁶ N/C
Speed of light c=3×10⁸ m/s
For part (a) distance would proton travel
Apply the third equation of motion
[tex](v_{f})^{2} =(v_{i})^{2}+2ax[/tex]
In this case vi=0 m/s and vf=c
so
[tex]c^{2}=(0)^{2}+2ax\\ c^{2}=2ax\\x=\frac{c^{2} }{2a}[/tex]
[tex]x=\frac{c^{2}}{2a}--------Equation (i)[/tex]
From the electric force on proton
[tex]F=qE\\where\\ F=ma\\so\\ma=qE\\a=\frac{qE}{m}\\[/tex]
put this a(acceleration) in Equation (i)
So
[tex]x=\frac{c^{2} }{2(qE/m)}\\ x=\frac{mc^{2}}{2qE} \\x=\frac{(1.67*10^{-27})*(3*10^{8})^{2} }{2*(1.6*10^{-19})*(3*10^{6})}\\ x=157m[/tex]
For part (b)
No the proton would collide with air molecule
