Respuesta :
Answer:
a) [tex]v = 6.25\times 10^{5} m/s[/tex]
b) [tex]v = 1.73\times 10^{4} m/s[/tex]
Explanation:
Given data:
Electric field = 1.47 N/C
velocity of electron is [tex]4.55\times 10^5 m/s[/tex]
distance of point b from point A is 0.55 m
we know that acceleration of particle is given as
a) for electron
[tex]a =\frac{q E}{m}[/tex]
[tex]a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}[/tex]
[tex]a = 2.58\times 10^{11} m/s^2[/tex]
from equation of motion we have
[tex] v^2 = u^2 + 2aS[/tex]
[tex] = 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355[/tex]
[tex]v = 6.25\times 10^{5} m/s[/tex]
b) for proton
[tex]a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}[/tex]
[tex]a = -1.41\times 10^{8} m/s^2[/tex]
from equation of motion we have
[tex] v^2 = u^2 + 2aS[/tex]
[tex]= 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355[/tex]
[tex]v = 1.73\times 10^{4} m/s[/tex]
a. The speed of the electron when it reaches point B is [tex]v = 6.25 \times 10^{5} m/s[/tex]
b. The speed of the proton at point B is [tex]v = 1.73 \times 10^{4} m/s[/tex]
Calculation of the speed of electron & proton:
(a) The speed of the electron should be
[tex]a = qE \div m\\\\a = \frac{1.6\times 10^{-19}\times 1.47}{9.1 \times 10^{-31}} \\\\a = 2.58 \times 10^{11} m/s^2[/tex]
Now here we applied the equation of motion that should be
[tex]v^2 = u^2 + 2as\\\\= 20.7025 \times 10^{10} + 2\times 2.58 \times 10^{11}\times 0.355\\\\v = 6.25 \times 10^{5} m/s[/tex]
b. The speed of the proton should be
[tex]a = \frac{1.6\times 10^{-19}\times -1.47}{1.6 \times 10^{-27}} \\\\a = -1.41 \times 10^{8} m/s^2[/tex]
Now here we applied the equation of motion that should be
[tex]v^2 = u^2 + 2as\\\\= 3.8025 \times 10^{8} + 2\times 1.41 \times 10^{8}\times 0.355\\\\v = 1.73 \times 10^{4} m/s[/tex]
Learn more about electron here: https://brainly.com/question/24507731
