Respuesta :
Answer:
a) [tex]E=364N/C[/tex]
b) No
Explanation:
A) Because the electron is affected by an acceleration force in this case by the electric field, we can use the formulas of 2-dimension movement.
We will assume the electron missed the upper plate, so we need to calculate the time to travel all the way through the plate, that is:
[tex]x=v_x*t[/tex]
[tex]where:\\x=distance\\v=speed\\t=time[/tex]
so:
[tex]t=\frac{x}{v_x}=\frac{0.02m}{1.6\cdot 10^6m/s}\\\\t=1.25\cdot10^{-8}[/tex]
the electron experiences an accelerated motion in the vertical direction, so we can obtain the acceleration of the electron:
[tex]y=\frac{1}{2}.a.t^2\\\\where:\\y=vertical\_distance\\a=acceleration\\t=time[/tex]
so:
[tex]a=\frac{2.y}{t^2}\\\\a=\frac{2*(\frac{0.01}{2}m)}{(1.25\cdot10^{-8}s)^2}\\\\a=6.4\cdot10^{13} m/s^2[/tex]
now we can use the relation:
[tex]F=m.a=E.q\\so\\E=\frac{m.a}{q}[/tex]
[tex]where:\\\\E=electric\_field\\m=electron\_mass=9.1\cdot10^{-31}kg\\q=Charge=1.6\cdot10^{-19}\\a=acceleration[/tex]
Now we can calculate the electric field:
[tex]E=\frac{9.1\cdot10^{-31}kg\cdot6.4\cdot10^{13}m/s^2}{1.6\cdot10^{-19}C}\\\\E=364N/C[/tex]
B) Because the proton has the same charge but positive it will go in the opposite direction, so because we assume the electron didn't touch the plate, the proton won't.
The magnitude of the electric field is 364 N/C if the electron misses the upper plate.
The time taken by the electron to travel through plates,
[tex]t = \dfrac {v_x}x[/tex]
Where,
[tex]v_x[/tex]- velocity = [tex]1.6 \times 10^8\; \rm m/s[/tex]
[tex]x[/tex]- distance = 0.2 m
Put the values in the formula,
[tex]t = \dfrac {0.02 }{1.6 \times 10^8}\\\\t = 1.25 \times 10^{-8}[/tex]
Now, acceleration of the electron in verticle direction,
[tex]a = \dfrac {2 \times y }{t^2}[/tex]
Where,
y - verticle distance
So,
[tex]a = \dfrac {2 \times (\frac {0.01 }2 )}{(1.25 \times 10^-8)^2 }\\\\a = 6.4 \times 10^1^3 \rm \; m/s^2[/tex]
The electrical field,
[tex]E = \dfrac {m\times a}{q}[/tex]
Where,
[tex]m[/tex] - mass of electron = [tex]9.1\times 10^{-31} \rm \; kg[/tex]
[tex]q[/tex] - charge = [tex]1.6 \times 10^{-19}\rm \; C[/tex]
So,
[tex]E = \dfrac {9.1\times 10^{-31} \rm \; kg \times 6.4 \times 10^1^3 \rm \; m/s^2}{1.6 \times 10^{-19}\rm \; C}\\\\E = 364 \rm \ N/C \\[/tex]
Therefore, the magnitude of the electric field is 364 N/C.
To know more about the electric field,
https://brainly.com/question/12757739
