## 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 i

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.)

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6 months 2021-08-05T22:07:28+00:00 1 Answers 8 views 0

## Answers ( )

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