Question

Two identical spheres with mass m are hung from silk threads of length L (Fig. P21.62). The spheres have the same charge, so q₁ = q₂ = 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 u is small, the equilibrium separation d between the spheres is d = (q²L/2𝛑ε0mg)^1/3 (Hint: If𝛉is small, then tan𝛉≅sin𝛉.)

Short answer

If the angle u is small, the equilibrium separation d between the spheres is $d={\left(\frac{2L{q}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{mg}}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$.

Step-by-step solution

Component of forces

Since the charges are in equilibrium:

$$\begin{gathered} \mathbf{\sum }_{\mathbf{}}{\mathit{F}}_{\mathbf{x}}\mathbf{=}\mathit{T}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{1}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{o}}}\frac{{\mathbf{q}}^{\mathbf{2}}}{\mathbf{d}} \\ \mathbf{\sum }_{\mathbf{}}{\mathit{F}}_{\mathbf{y}}\mathbf{=}\mathit{T}\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{g} \end{gathered}$$

Step 2:

From Step 1, we get

$$\begin{gathered} \frac{T\sin \theta }{T\cos \theta }=\frac{{\displaystyle \frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{q^2}{d^2}}}{mg} \\ \tan \theta =\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\frac{q^2}{d^{2}mg} \\ {d}^2=\frac{q^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{mgtan\theta }} \end{gathered}$$

Step 3:

$\sin \theta =\frac{d}{2L}$

Sinceθ<<1, use the approximation

$$\begin{gathered} \tan \theta \simeq \sin \theta =\frac{d}{2L} \\ {d}^2=\frac{2L{q}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{mgd}} \\ {\mathrm{d}}^3=\frac{2{\mathrm{Lq}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{mg}} \\ \mathrm{d}={\left(\frac{2{\mathrm{Lq}}^2}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}\mathrm{mg}}\right)}^{1/3} \end{gathered}$$

Hence, proved.