An electron is constrained to the central perpendicular axis of a ring of charge of radius 2.2 m and charge 0.021 mC. Suppose the electron is released from rest a distance 0.050 m from the ring center. It then oscillates through the ring center. Calculate its period. (The electron is always much closer to the ring center than a radius.)

T = 1.12 10⁻⁷ s

Explanation:

This exercise must be solved in parts. Let's start looking for the electric field in the axis of the ring.

All the charge dq is at a distance r

           dE = k dq / r²

Due to the symmetry of the ring, the field perpendicular to the axis is canceled, leaving only the field in the direction of the axis, if we use trigonometry

            cos θ =

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ =

                DEₓ = k dq x / r³

let's use the Pythagorean theorem to find the distance r

             r² = x² + a²

where a is the radius of the ring

we substitute

              dEₓ =

we integrate

               ∫ dEₓ =k \frac{x}{(x^2 + a^2 ) ^{3/2} }  ∫ dq

               Eₓ =

In the exercise indicate that the electron is very central to the center of the ring

                x << a

                Eₓ =

if we expand in a series

we keep the first term if x<<a

                 Eₓ =

the force is

                 F = q E

                 F =

this is a restoring force proportional to the displacement so the movement is simple harmonic,

                 F = m a

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² =

angular velocity and period are related

                 w = 2π/ T

we substitute

               4π² / T² = \frac{keQ}{m a^3}

                T = 2π  

let's calculate

                T = 2π

                 T = 2π pi

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s

statement #3 on andre"s list is false.

its a circle not a square that's simple silly.

explanation: