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Chapter 4: Q62P (page 717)

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

Expert verified

If the angle u is small, the equilibrium separation d between the spheres is d=2Lq24πεomg13.

Step by step solution

01

Component of forces

Since the charges are in equilibrium:

∑Fx=Tsinθ=14ττεoq2d∑Fy=Tcosθ=mg

02

Step 2:

From Step 1, we get

TsinθTcosθ=14πεoq2d2mgtanθ=14πεoq2d2mgd2=q24πεomgtanθ

03

Step 3:

sinθ=d2L

Sinceθ<<1, use the approximation

tanθ≃sinθ=d2Ld2=2Lq24πεomgdd3=2Lq24πεomgd=2Lq24πεomg1/3

Hence, proved.

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