Find the period of small vibrations of a body of mass m, whose charge is q, inside a smooth sphere of radius R, if the charge Q is fixed at the top point of the sphere.

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <h3 id="back-link"><a href="/#7.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $7.1.23^*.$ Find the period of small vibrations of a body of mass $m$, whose charge is $q$, inside a smooth sphere of radius $R$, if the charge $Q$ is fixed at the top point of the sphere.

    <h3>Solution</h3>
    <p>

Considering the fact that oscillations are small, $\varphi \ll 1$, we could use the approximations for small angle $\varphi$
$$\sin\varphi \approx \varphi; \quad\cos\varphi \approx 1\quad(1)$$
From where, we could approximate that the distance between charges remains approximately the same
$$r\approx2R$$
Thus, the сoulomb's law for two charges
$$F_c = \frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}$$
Let's write Newton's second law on a tangential trajectory, in the direction of motion of a charge of mass $m$
$$ma = mg \sin\varphi + F_c \sin\frac{\varphi}{2}$$
Considering the approximation $(1)$ and the expression
$$ma = -mg\frac{x}{R} - \frac{x}{2R} \frac{1}{4\pi\varepsilon_0}\frac{qQ}{4R^2}$$
$$a = -x\left(\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}\right)$$
Solving the equation of harmonic oscillations $(\ddot{x}+\omega^2x(t)=0)$, we could obtain the value for angular velocity
$$\omega =\sqrt{\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}}$$
From here the period of oscillation
$$\boxed{T=\frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}}}}$$
Since the expression under the square root cannot be less than zero, the ratio at which there will be no oscillations
$$\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3} > 0 \Leftrightarrow \boxed{\frac{qQ}{32\pi\varepsilon_0R^2}>-mg}$$

    <h4>Answer</h4>
  <div class="scroll-wrapper">
    <p>
        $$T=2\pi\left(\frac{g}{R}+\frac{qQ}{32\pi\varepsilon_0mR^3}\right)^{-1/2}\text{ with }\frac{qQ}{32\pi\varepsilon_0R^2}>-mg$$
    </p>
  </div>
      <p style="text-align: right; font-style: italic; font-size: 14;">
      Lutfulloyev Shukurullo<br>
      Physics Department<br>
      National University of Uzbekistan<br>
    </p>



<footer class="row container">
  <br>
    <p>
        <small> © <strong>Savchenko Solutions</strong>, 2023-2024 <br></small>
    </p>
    <p>
        <small>All rights belong to the authors. <br> Commercial use of materials - with the written permission of the authors. <br> alex@savchenkosolutions.com <br></small>
    </p>
</footer>