Answer:
r = 2,026 10⁹ m and T = 2.027 10⁴ s
Explanation:
For this exercise let's use Newton's second law
F = m a
where the force is electric
F = [tex]k \frac{q_1q_2}{r^2}[/tex]
Acceleration is centripetal
a = v² / r
we substitute
[tex]k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}[/tex]
r = [tex]k \frac{q_1q_2}{m \ v^2}[/tex] (1)
let's look for the charge in the insulating sphere
ρ = q₂ / V
q₂ = ρ V
the volume of the sphere is
v = 4/3 π r³
we substitute
q₂ = ρ [tex]\frac{4}{3}[/tex] π r³
q₂ = 3 10⁻⁹ [tex]\frac{4}{3}[/tex] π 4³
q₂ = 8.04 10⁻⁷ C
let's calculate the radius with equation 1
r = 9 10⁹ 1.6 10⁻¹⁹ 8.04 10⁻⁷ /(9.1 10⁻³¹ 628 10³)
r = 2,026 10⁹ m
this is the radius of the electron orbit around the charged sphere.
Since the orbit is circulating, the speed (speed modulus) is constant, we can use the uniform motion ratio
v = x / t
the distance traveled in a circle is
x = 2π r
In this case, time is the period
v = 2π r /T
T = 2π r /v
let's calculate
T = 2π 2,026 10⁹/628 103
T = 2.027 10⁴ s
