The distance covered by the proton is 48.4 m
Explanation:
The electric field produced by an electrically charged infinite plane is given by
[tex]E=\frac{\sigma}{2\epsilon_0}[/tex]
where in this case,
[tex]\sigma = -1.10\cdot 10^{-6} C/m^2[/tex] is the surface charge density
[tex]\epsilon_0 = 8.85\cdot 10^{-12} F/m[/tex] is the vacuum permittivity
Substituting,
[tex]E=\frac{-1.10\cdot 10^{-6}}{2(8.85\cdot 10^{-12})}=-5.65\cdot 10^4N/C[/tex]
And the direction is towards the plane (because the charge is negative).
The electric force on the proton due to this field is
[tex]F=qE[/tex]
where
[tex]q=1.6\cdot 10^{-19}C[/tex] is the proton charge
Substituting,
[tex]F=(1.6\cdot 10^{-19})(-5.65\cdot 10^4)=-9.0\cdot 10^{-15} N[/tex]
where the direction is toward the plane.
Now we can calculate the proton's acceleration using Newton's second law:
[tex]a=\frac{F}{m}[/tex]
where
[tex]m=1.67\cdot 10^{-27}kg[/tex] is the proton mass
Substituting,
[tex]a=\frac{-9.0\cdot 10^{-15}}{1.67\cdot 10^{-27}}=-5.4\cdot 10^{-12} m/s^2[/tex]
Now we can finally apply the following suvat equation for accelerated motion to find the distance travelled by the proton:
[tex]v^2-u^2=2as[/tex]
where
v = 0 is the final velocity
[tex]u=2.40\cdot 10^7 m/s[/tex] is the initial velocity
a is the acceleration
s is the distance covered
And solving for s,
[tex]s=\frac{v^2-u^2}{2a}=\frac{0-(2.4\cdot 10^7)^2}{2(-5.4\cdot 10^{12})}=53.3 m[/tex]
Therefore, the closest answer is 48.4 m.
Learn more about accelerated motion:
brainly.com/question/9527152
brainly.com/question/11181826
brainly.com/question/2506873
brainly.com/question/2562700
#LearnwithBrainly
