Respuesta :
Answer:
a) 17.33 V/m
b) 6308 m/s
Explanation:
We start by using equation of motion
s = ut + 1/2at², where
s = 1.2 cm = 0.012 m
u = 0 m/s
t = 3.8*10^-6 s, so that
0.012 = 0 * 3.8*10^-6 + 0.5 * a * (3.8*10^-6)²
0.012 = 0.5 * a * 1.444*10^-11
a = 0.012 / 7.22*10^-12
a = 1.66*10^9 m/s²
If we assume the electric field to be E, and we know that F =qE. Also, from Newton's law, we have F = ma. So that, ma = qE, and E = ma/q, where
E = electric field
m = mass of proton
a = acceleration
q = charge of proton
E = (1.67*10^-27 * 1.66*10^9) / 1.6*10^-19
E = 2.77*10^-18 / 1.6*10^-19
E = 17.33 V/m
Final speed of the proton can be gotten by using
v = u + at
v = 0 + 1.66*10^9 * 3.8*10^-6
v = 6308 m/s
(A) The magnitude of electric field is 17.33 V/m
(B) The speed of proton is 6308 m/s
From the second equation of motion:
s = ut + 1/2at², where s is the distance= 1.2 cm = 0.012 m
u is the initial speed = 0 m/s
[tex]t = 3.8*10^{-6}s[/tex]
[tex]0.012 = 0 * 3.8*10^{-6} + 0.5 * a * (3.8*10^{-6})²0.012 = 0.5 * a * 1.444*10^{-11a = 0.012 / 7.22*10^{-12a = 1.66*10^9 m/s^2[/tex]
Let the electric field to be E,
the force acting on charge q in an electric field E
F =qE.
Also F = ma
ma = qE
where m = mass of proton , a = acceleration, q = charge of proton
[tex]E = ma/qE = (1.67*10^{-27} * 1.66*10^9) / 1.6*10^{-19}E = 2.77*10^{-18} / 1.6*10^{-19}E = 17.33 V/m[/tex]
The electric field intesity is 17.33 V/m
Final speed of the proton when it strikes the negatively charged plate.
[tex]v = u + atv = 0 + 1.66*10^9 * 3.8*10^{-6}v = 6308 m/s[/tex]
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