Two identical balls with charges \pm q, rigidly connected by a weightless rod of length l, are in an electric field, which acts on them with force \pm q E. Determine the mass of the ball if the amplitude of small transverse oscillations of the balls is x_0 and the maximum velocity of the balls is v_0.

Savchenko Solutions

Solutions of Savchenko Problems in Physics
knowledge must be free

    <h3 id="back-link"><a href="/#3.1">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $3.1.8.$ Two identical balls with charges $\pm q$, rigidly connected by a weightless rod of length $l$, are in an electric field, which acts on them with force $\pm q E$. Determine the mass of the ball if the amplitude of small transverse oscillations of the balls is $x_0$ and the maximum velocity of the balls is $v_0$.

    <h3>Solution</h3>
    <p>
        Newton's second law for the normal force of the Coulomb, which creates a restoring moment of force, is found in projection onto the vertical axis

$$ma_\tau=-qE\sin\varphi$$
Since the angle $\varphi$ is negligibly small, we will use $\sin\varphi\approx\varphi$
$$-qE\varphi=-qE\frac{2x}{l}$$
From where we obtain the equation of harmonic oscillations
$$m\ddot{x}(t)+\frac{2qE}{l}x(t)=0$$

$$\ddot{x}(t)+\frac{2qE}{ml}x(t)=0$$
The solution to which we know:
$$\omega=\sqrt{\frac{2qE}{ml}}$$
From where, similarly to 3.1.6, we find the maximum speed
$$\upsilon_0=x_0\omega=x_0\sqrt{\frac{2qE}{ml}}$$
After mathematical transformations we obtain the mass of the ball
$$\boxed{m=\frac{2qEx_0^2}{l\upsilon_0^2}}$$

    </p>

    <h4>Answer</h4>
    <p>
        $$m=\frac{2qEx_0^2}{l\upsilon_0^2}$$
    </p>
	<p style="text-align: right; font-style: italic; font-size: 14;">
      Dzikan Mikita<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>