Question

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.)

Answer 1

Answer:

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 =$\frac{dE_x}{dE}$

             dEₓ = dE cos θ

              cos θ = x / r

substituting

                dEₓ = $k \frac{dq}{r^2 } \ \frac{x}{r}$

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

we integrate

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

               Eₓ = $k \ Q \ \frac{x}{(x^2+a^2)^{3/2}}$

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

                x << a

                Eₓ = $k \ Q \frac{x}{a^3 \ ( 1 +(x/a)^2)^{3/2})}$

if we expand in a series

                  $(\ 1+ (x/a)^2 \ )^{-3/2} = 1 - \frac{3}{2} (\frac{x}{a} )^2$

we keep the first term if x<<a

                 Eₓ = $\frac{ k Q}{a^3} \ x$

the force is

                 F = q E

                 F = $- \frac{kQ }{a^3} \ x$

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

                 F = m a

                 $- \frac{keQ}{a^3} \x = m \frac{d^2 x}{dt^2 }$

                 $\frac{d^2 x}{dt^2} = \frac{keQ}{ma^3} \ x$

the solution is of type

                  x = A cos (wt + Ф)

with angular velocity

                w² = $\frac{keQ}{m a^3}$

angular velocity and period are related

                 w = 2π/ T

             

we substitute

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

                T = 2π  $\sqrt{\frac{m a^3 }{keQ} }$

let's calculate

                T = 2π $\sqrt{ \frac{ 9.1 \ 10^{-31} \ 2.2^3 }{9 \ 10^9 \ 1.6 \ 10^{-19} \ 0.021 \ 10^{-3} } }$

                 T = 2π pi $\sqrt{320.426 \ 10^{-18} }$

                 T = 2π  17.9 10⁻⁹ s

                 T = 1.12 10⁻⁷ s