Two similar tiny balls of mass m are hung from silk threads of length L and carr

Question

Two similar tiny balls of mass m are hung from silk threads of length L and carry equal charges q as in Fig. 14. Assume that theta is so small that tan theta can be replaced by its approximate equal, sin theta. (a) To this approximation show that, for equilibrium, where x is the separation between the balls. (b) If L = 122 cm, m = 11.2 g, and x = 4.70 cm, what is the value of q? Figure 14 Problems 16, 17, and 18. If the balls of Fig. 14 are conducting, (a) what happens to them after one is discharged? Explain your answer. (b) Find the new equilibrium separation.

Explanation / Answer

Each ball is in equilibrium under the action of 3 forces: (i) its weight mg, (ii) the tension T in the thread, and (iii) the electrostatic repulsion between the balls.

Resolving forces vertically, TcosA=mg where A is the angle between each thread and the vertical.
Resolving forces horizontally, TsinA=q^2/4??x^2, the electrostatic repulsion force.

From geometry, LsinA=x/2, where L is length of each thread.

2A=?, the angle between the threads. If ? and therefore A is small, then cosA?1 and sinA?A. So T?mg, TA?q^2/4??x^2 and x?2LA.

Eliminate T and A and rearrange to find x:
(mg)(x/2L)? (q^2/4??x^2)
x^3?Lq^2/2??mg
x?[Lq^2/2??mg]^(1/3).....Answer
On putting the values : We get : q = 4.558 * 10^-8 C = 45.83 nC

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.