Question

Problem 3: In Figure, two tiny conducting balls of identical mass m and identical charge q hang from nonconducting threads of length L. Assume that is so small that tanê can be replaced by its approximate equal, sino 1/3 (a) Show that x= q2L 27€, mg gives the equilibrium separation x of the balls. (b) If L=120 cm, m=10 g, and x=5.0 cm, what is /q/? lee L L 9 9

Answer 1

@ Fe = electrostatic fore of repulsion = kqx? 22 IL Tuse 9 ТК fe At equillibrium state Fnet zo In vertical direction ET= tension in rope Fe Tsino Tcoso = mg (1) 2 ng along horizontal Tsino - fe= Khi (11) dividing equation (1) by cis T{sino) [kq4/72] tano = kq2 → Thoso) mg ngx2 In AOAB Sing = 쯘 for angle o which is small. very sinov tano = o A 2L B 2/ 20 Kq2 mge?

х 2L k42 х 3 11 mgu? k-L ng 4T6 — 23 ахи X 406 «EL Тема х = q-L V 2m+n) (6) {vел : е а L= 10 см с 1- 20 м аг 5см = 0. OSM ме 17 10х1.3 3 = 5-8 m/st з's Ka* "а maxs KL

2 - «921 -- mgus مه KL و فيها حقها = (ع) C KL و X6 4 : 0-15 عر 6.15 =۹|