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.)

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

Answers ( )

    0
    2021-08-05T22:08:52+00:00

    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

Leave an answer

Browse

Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )