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 $z-axis$but 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 $m$with charge$-q$undergoes simple harmonic motion with frequency $f=\frac{1}{2\pi }\sqrt{\frac{qQ}{4\pi {\epsilon }_{0}m{R}^3}}$,$R\mathrm{and}Q$are the radius and charge of the ring.

c. Evaluate the oscillation frequency for an electron at the center of a $2.0\mu m$diameter ring charged to $1.0\times {10}^{-13}C$.

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 : $m,-q$respectively

c. Diameter of the ring :$2.0\mu m$

Charged to :$1.0\times {10}^{-13}C$

Theory used :

The electric field is that of a positively charged ring. We consider a thin ring of radius $R$is uniformly charged with total charge $Q$has an electric field along the $z-axis$is given by the expression

${\left(E_{ring}\right)}_z=\frac{1}{4\pi \epsilon }$₀zQ(z²+R²)^3/2

Here ${\epsilon }_0=8.85\times {10}^{-12}C²/Nm²$is the permittivity constant.

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

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

${\left(E_{ring}\right)}_z=\frac{1}{4\pi {\epsilon }_0}\frac{zQ}{{(z^2+R^2)}^{3/2}}$

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 :

$$\begin{gathered} F=(-q){\left(E_{ring}\right)}_z \\ =-\frac{1}{4\pi {\epsilon }_0}\frac{zqQ}{{(z^2+R^2)}^{3/2}} \end{gathered}$$

We can rewrite the preceding equation as

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

$k=\boxed{\frac{qQ}{4\pi {\epsilon }_0{R}^3}}$

So long as... $z$is small enough, the electron will do simple harmonic motion.

(b)If the charge is located closer to the ring's center, so that $z<<R$. In this example, $z^2+R^2\approx R^2$, and the restoring force is expressed as

$$\begin{gathered} F=-\frac{1}{4\pi {\epsilon }_0}\frac{zqQ}{{(z^2+R^2)}^{3/2}} \\ =-\frac{1}{4\pi {\epsilon }_0}\frac{zqQ}{R^3} \\ =-\left(\frac{qQ}{4\pi {\epsilon }_0{R}^3}\right)z \end{gathered}$$

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 $F=-kz$.

$k=\frac{qQ}{4\pi {\epsilon }_0{R}^3}$in this case.

The frequency $f$of a spring with a spring constant $k$and a mass $m$would be :

$$\begin{gathered} f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{k}{m}} \\ =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{\displaystyle \frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon }}_0{R}^3}}}{m}} \\ =\boxed{\frac{\mathbf{1}}{\mathbf{2}\mathbf{\pi }}\sqrt{\frac{\mathbf{qQ}}{\mathbf{4}{\mathbf{\pi \epsilon }}_{\mathbf{0}}{\mathbf{mR}}^{\mathbf{3}}}}} \end{gathered}$$

Hence proved.

(c) The ring's diameter is

$$\begin{gathered} D=2.0\mu m \\ =(2.0\mu m)\left(\frac{{10}^{-6}}{1\mu m}\right) \\ =2.0\times {10}^{-6}m \end{gathered}$$.

$R=\frac{D}{2}$is the ring's radius.

Substituting $2.0\times {10}^{-6}m$for $D$in the above equation :

$R=\frac{2.0\times {10}^{-6}m}{2}=1.0\times {10}^{-6}m$

In the equation for frequency, substitute $9.11\times {10}^{-31}kg\mathrm{for}m$, $1.602\times {10}^{-19}C\mathrm{for}q$, $1.0\times {10}^{-13}C\mathrm{for}Q$, and $8.85\times {10}^{-12}{C}^2/N{m}^2\mathrm{for}{\epsilon }_{\mathit{0}}$:

$$\begin{gathered} f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{qQ}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{mR}}^3}} \\ =\frac{1}{2\mathrm{\pi }}\sqrt{\frac{(1.602\times {10}^{-19}C)(1.0\times {10}^{-13}C)}{4\mathrm{\pi }(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2)(9.11\times {10}^{-31}\mathrm{kg}){(1.0\mathrm{x}{10}^{-6}\mathrm{m})}^3}} \\ =\boxed{\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{12}}\mathbf{}\mathit{H}\mathit{z}} \end{gathered}$$