Respuesta :
Answer:
Final velocity of electron, [tex]v=6.45\times 10^5\ m/s[/tex]
Explanation:
It is given that,
Electric field, E = 1.55 N/C
Initial velocity at point A, [tex]u=4.52\times 10^5\ m/s[/tex]
We need to find the speed of the electron when it reaches point B which is a distance of 0.395 m east of point A. It can be calculated using third equation of motion as :
[tex]v^2=u^2+2as[/tex]........(1)
a is the acceleration, [tex]a=\dfrac{F}{m}[/tex]
We know that electric force, F = qE
[tex]a=\dfrac{qE}{m}[/tex]
Use above equation in equation (1) as:
[tex]v^2=u^2+\dfrac{2qEs}{m}[/tex]
[tex]v^2=(4.52\times 10^5\ m/s)^2+2\times \dfrac{1.6\times 10^{-19}\ C\times 1.55\ N/C}{9.1\times 10^{-31}\ kg}\times 0.395\ m[/tex]
v = 647302.09 m/s
or
[tex]v=6.45\times 10^5\ m/s[/tex]
So, the final velocity of the electron when it reaches point B is [tex]6.45\times 10^5\ m/s[/tex]. Hence, this is the required solution.
