Chapter 24, Problem 24.62CP

(a)

To determine

The electron would be in equilibrium at centre and if displaced would experience a restoring force.

Answer to Problem 24.62CP

The electron would be in equilibrium at centre and if displaced would experience a restoring force of magnitude $-kr$.

Explanation of Solution

Given info:

Consider the field distance $r<R$ from the centre of a uniform sphere positive charge $Q=+e$ with radius $R$.

Write the expression of Guass’s law.

$\int \overrightarrow{E}.d\overrightarrow{A}=\frac{q_{\text{in}}}{{\epsilon }_{\text{o}}}$

Here,

$E$ is the Electric Field.

$A$ is the Area.

$q_{\text{in}}$ is the electric charge.

${\epsilon }_{\text{o}}$ is the electric constant.

Write the expression for area of sphere.

$A=4\pi r^2$

Here,

$r$ is the radius of sphere.

Substitute $4\pi r^2$ for $A$.

$\left(4\pi r^2\right)E=\frac{q_{\text{in}}}{{\epsilon }_{\text{0}}}$ (1)

Write the expression for electric charge.

$q_{\text{in}}=\rho V$

Here,

$\rho$ is the density of sphere.

$V$ is the volume of sphere.

Substitute $\rho V$ for $q_{\text{in}}$ in the expression (1).

$\left(4\pi r^2\right)E=\frac{\rho V}{{\epsilon }_{\text{0}}}$ (2)

Write the expression for volume of sphere.

$V=\frac{4}{3}\pi r^3$

Write the expression for density of sphere.

$\rho =\frac{m}{V}$

Here,

$m$ is the mass of a charge.

$V$ is the volume of sphere.

Substitute $+e$ for $m$ and $\frac{4}{3}\pi r^3$ for $V$.

$\rho =\frac{+e}{\frac{4}{3}\pi r^3}$

Substitute $\frac{+e}{\frac{4}{3}\pi r^3}$ for $\rho$ and $\frac{4}{3}\pi r^3$ fro $V$ in expression (2).

$\left(4\pi r^2\right)E=\left(\frac{+e}{\frac{4}{3}\pi R^3}\right)\frac{\frac{4}{3}\pi r^3}{{\epsilon }_{\text{0}}}$

So,

$E=\left(\frac{+e}{4\pi {\epsilon }_{\text{0}}{R}^3}\right)r$

The above electric field is directed outward.

The expression for the force is,

$F=qE$

Substitute $-e$ for $q$ and $\left(\frac{+e}{4\pi {\epsilon }_{\text{0}}{R}^3}\right)r$ for $E$ in above expression.

$\begin{array}{c}F=-e\left(\left(\frac{+e}{4\pi {\epsilon }_{\text{0}}{R}^3}\right)r\right)\\ =-\left(\frac{e^2}{4\pi {\epsilon }_{\text{0}}{R}^3}\right)r\\ =-kr\end{array}$

Conclusion:

Therefore, the electron would be in equilibrium at centre and if displaced would experience a restoring force of magnitude $-kr$.

(b)

To determine

To show: The constant $k$ is equal to $k_{\text{e}}{e}^2/R^3$.

Answer to Problem 24.62CP

The constant $k$ is equal to $\frac{k_{\text{e}}{e}^2}{R^3}$.

Explanation of Solution

Given info:

Write the expression for spring force.

$F=kr$

Here,

$k$ is the spring constant.

Also, the field force is,

$F=\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{R^3}r$

Equate the above two force equations.

$\begin{array}{c}kr=\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{R^3}r\\ k=\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{R^3}\end{array}$

Substitute $k_{\text{e}}$ for $\frac{1}{4\pi {\epsilon }_{\text{0}}}$.

$k=\frac{k_{\text{e}}{e}^2}{R^3}$

Conclusion:

Therefore, the constant $k$ is equal to $\frac{k_{\text{e}}{e}^2}{R^3}$.

(c)

To determine

The expression for the frequency $f$ of simple harmonic oscillations.

Answer to Problem 24.62CP

The expression for the frequency $f$ of simple harmonic oscillations  is $f=\frac{1}{2\pi }\sqrt{\frac{k_{\text{e}}{e}^2}{m_{\text{e}}{R}^3}}$.

Explanation of Solution

Given info:

According to Newton’s second law of motion,

$F=m_{\text{e}}{a}_{\text{c}}$

Here,

$a_{\text{c}}$ is the centripetal acceleration of electron.

Also, the field force is,

$F=\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{R^3}r$

Equate the above two force equations.

$m_{\text{e}}{a}_{\text{c}}=\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{R^3}r$

Rearrange the above equation to get $a_{\text{c}}$.

$a_{\text{c}}=-\left(\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{m_{\text{e}}{R}^3}\right)r$

Compare the above equation with the simple harmonic wave equation which is,

$a_{\text{c}}=-{\omega }^{2}r$

Equate the above two acceleration equations.

${\omega }^{2}r=\left(\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{m_e{R}^3}\right)r$

From the above equation $\omega$ is,

$\omega =\sqrt{\left(\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{m_{\text{e}}{R}^3}\right)}$

Write the general expression for frequency of simple harmonic motion.

$f=\frac{\omega }{2\pi }$

Substitute $\sqrt{\left(\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{m_{\text{e}}{R}^3}\right)}$ for $\omega$.

$f=\frac{1}{2\pi }\sqrt{\left(\frac{1}{4\pi {\epsilon }_{\text{0}}}\frac{e^2}{m_{\text{e}}{R}^3}\right)}$

Substitute $\frac{1}{4\pi {\epsilon }_{\text{0}}}$ for $k_{\text{e}}$.

$f=\frac{1}{2\pi }\sqrt{\frac{k_{\text{e}}{e}^2}{m_{\text{e}}{R}^3}}$

Conclusion:

Therefore, the expression for the frequency $f$ of simple harmonic oscillations  is $f=\frac{1}{2\pi }\sqrt{\frac{k_{\text{e}}{e}^2}{m_{\text{e}}{R}^3}}$.

(d)

To determine

The value of radius of orbit.

Answer to Problem 24.62CP

The value for $R$ is $1.02\text{A}$.

Explanation of Solution

Given info:

Write the expression for frequency $f$ of simple harmonic oscillations.

$f=\frac{1}{2\pi }\sqrt{\frac{k_{\text{e}}{e}^2}{m_{\text{e}}{R}^3}}$

Rearrange the above equation to get $R$.

$R^3=\frac{k_{\text{e}}{e}^2}{4{\pi }^2{f}^2{m}_{\text{e}}}$

Substitute $2.47\times {10}^{15}\text{Hz}$ for $f$, $8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2$ for $k_{\text{e}}$ $9.11\times {10}^{-31}\text{kg}$ for $m_{\text{e}}$ and $1.6\times {10}^{-19}\text{C}$ for $e$.

$\begin{array}{c}R={\left(\frac{\left(8.987\times {10}^9\text{N}\cdot {\text{m}}^2/{\text{C}}^2\right){\left(1.6\times {10}^{-19}\text{C}\right)}^2}{4{\pi }^2{\left(2.47\times {10}^{15}\text{Hz}\right)}^2\times 9.11\times {10}^{-31}\text{kg}}\right)}^{\frac{1}{3}}\\ =1.02\times {10}^{-10}\text{m}\\ \text{=1}\text{.02}\text{A}\end{array}$

Conclusion:

Therefore, the value for $R$ is $1.02\text{A}$.