Question

An electron is moving through an (almost) empty universe at a speed of 628 km,/s toward the only other object in the universe — an insulating sphere with a diameter of 4 m and charge density 3nC/m2 on its outside surface. The sphere "captures" the electron, which falls into a circular orbit.
Required:
Find the radius and period of the orbit.

Answer 1

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 = $k \frac{q_1q_2}{r^2}$

Acceleration is centripetal

        a = v² / r

we substitute

        $k \frac{q_1q_2}{r^2} = m \frac{v^2}{r}$

        r = $k \frac{q_1q_2}{m \ v^2}$          (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 $\frac{4}{3}$ π r³

        q₂ = 3 10⁻⁹ $\frac{4}{3}$ 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