Question

Two identical spheres with mass m are hung from silk threads of length L $\textbf{(Fig. P21.62).}$ The spheres have the same charge, so $q_1 = q_2 = q$. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. Show that if the angle $\theta$ is small, the equilibrium separation $d$ between the spheres is $d = (q^2L/2\pi\epsilon_0 mg)^{1/3}$. ($Hint$: If $\theta$ is small, then tan $\theta \cong sin \theta$.)

Name: Problem 62, Chapter 21: University Physics with Modern Physics
Uploaded: 2019-08-13T05:24:35-08:00
Duration: 4 min 18 s
Channel: Ze-Han Lee

   Two identical spheres with mass m are hung from silk threads of length L $\textbf{(Fig. P21.62).}$ The spheres have the same charge, so $q_1 = q_2 = q$. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. Show that if the angle $\theta$ is small, the equilibrium separation $d$ between the spheres is $d = (q^2L/2\pi\epsilon_0 mg)^{1/3}$. ($Hint$: If $\theta$ is small, then tan $\theta \cong sin \theta$.)

University Physics with Modern Physics

Hugh D. Young 14th Edition

Chapter 21, Problem 62

Instant Answer

Solved by Expert Ze-Han Lee

08/13/2019

Step 1

There are three forces acting on each sphere: the gravitational force (mg), the electrostatic force (F_c), and the tension in the string (F_t). Since the spheres are in equilibrium, the net force acting on them is zero. Show more…

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Two identical spheres with mass m are hung from silk threads of length L $\textbf{(Fig. P21.62).}$ The spheres have the same charge, so $q_1 = q_2 = q$. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. Show that if the angle $\theta$ is small, the equilibrium separation $d$ between the spheres is $d = (q^2L/2\pi\epsilon_0 mg)^{1/3}$. ($Hint$: If $\theta$ is small, then tan $\theta \cong sin \theta$.)