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You have a summer intern position with a company that designs and builds nanomachines. An engineer with the company is designing a microscopic oscillator to help keep time, and you’ve been assigned to help him analyze the design. He wants to place a negative charge at the center of a very small, positively charged metal ring. His claim is that the negative charge will undergo simple harmonic motion at a frequency determined by the amount of charge on the ring.

a. Consider a negative charge near the center of a positively charged ring centered on the z-axis. Show that there is a restoring force on the charge if it moves along the z-axisbut stays close to the center of the ring. That is, show there’s a force that tries to keep the charge at z=0. b. Show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3,RandQare the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a 2.0μmdiameter ring charged to 1.0×10-13C.

Short Answer

Expert verified

a. We show that there’s a force that tries to keep the charge at z=0: role="math" localid="1649239800722" F=-qQ4πε0R3z

b. We show that for small oscillations, with amplitude <<R, a particle of mass mwith charge-qundergoes simple harmonic motion with frequency f=12πqQ4πε0mR3.

c. The required oscillation frequency for the electron is2.0×1012Hz

Step by step solution

01

Given information and formula used

Given :

a. Charge near the center of a positively charged ring : Negative

Centered on the : z-axis

b. Amplitude : <<R

Mass and Charge of particle : m,-qrespectively

c. Diameter of the ring :2.0μm

Charged to :1.0×10-13C

Theory used :

The electric field is that of a positively charged ring. We consider a thin ring of radius Ris uniformly charged with total charge Qhas an electric field along the z-axisis given by the expression

(Ering)z=14πε0zQ(z2+R2)3/2

Here ε0=8.85×10-12C²/Nm²is the permittivity constant.

Hook's Law has the following equation to describe the motion of a mass m attached to a spring: F=-kx

02

Showing that there’s a force that tries to keep the charge at z=0

(a) The electric field produced by the ring at a distance zfrom its center along the central axis perpendicular to the ring's plane can be written as

(Ering)z=14πε0zQ(z2+R2)3/2

R is the ring radius with a total charge Q.

The force on a negative charge due to a positive charge on the ring is F=(-q) when a negative charge is positioned towards the center of a positively charged ring :

F=(-q)(Ering)z=-14πε0zqQ(z2+R2)3/2

We can rewrite the preceding equation as

F=-qQ4πε0R3z, when z<<R

This appears to be a restorative force with k. Thatis :

k=qQ4πε0R3

So long as... zis small enough, the electron will do simple harmonic motion.

03

Showing for small oscillations a particle undergoes simple harmonic motion with frequency f=12πqQ4πε0mR3

(b)If the charge is located closer to the ring's center, so that z<<R. In this example, z2+R2R2, and the restoring force is expressed as

F=-14πε0zqQ(z2+R2)3/2=-14πε0zqQR3=-(qQ4πε0R3)z

We can deduce from the previous calculation that the restoring force is proportional to the distance . This is analogous to Hook's law. As a result, the system behaves similarly to a mass on a spring. As a result, the previous equation can be written as F=-kz.

k=qQ4πε0R3in this case.

The frequency fof a spring with a spring constant kand a mass mwould be :

f=12πkm=12πqQ4πε0R3m=12πqQ4πε0mR3

Hence proved.

04

Evaluating the required oscillation frequency 

(c) The ring's diameter is

D=2.0μm=(2.0μm)(10-61μm)=2.0×10-6m.

R=D2is the ring's radius.

Substituting 2.0×10-6mfor Din the above equation :

R=2.0×10-6m2=1.0×10-6m

In the equation for frequency, substitute 9.11×10-31kgform, 1.602×10-19Cforq, 1.0×10-13CforQ, and 8.85×10-12C2/Nm2forε0:

f=12πqQ4πε0mR3=12π(1.602×10-19C)(1.0×10-13C)4π(8.85×10-12C2/N·m2)(9.11×10-31kg)(1.0x10-6m)3=2.0×1012Hz

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