1 \begin{problem*}{23.10}
2 Two small metallic spheres, each of mass $m=0.200\U{g}$, are suspended
3 as pendulums by light strings of length $L$ as shown if Figure~P23.10.
4 The spheres are given the same electric charge of $7.20\U{nC}$, and
5 they come to equilibrium when each string is at an angle of
6 $\theta=5.00\dg$ with the vertical. How long are the strings?
12 real theta = 20; // 5; Exaggerated for clarity
15 real x = L*Sin(theta);
16 real y = -L*Cos(theta);
18 draw((x,y)--(0,0)--(-x,y));
19 draw((0,y)--(0,0), dashed);
22 Angle t = Angle((0,y), (0,0), (x,y), "$\theta$");
25 Charge a = pCharge((-x,y));
26 Charge b = pCharge((x,y));
33 By symmetry the two spheres will be at the same height, so the
34 distance between them is $2L\sin(\theta)$ and we can draw a free body
35 diagram for the right-hand sphere:
43 Vector T = Vector((0,0), mag=E_mag/Sin(theta), dir=90+theta,
44 L=Label("$T$", position=EndPoint));
45 Vector G = Vector((0,0), mag=E_mag/Tan(theta), dir=-90,
46 L=Label("$mg$", position=EndPoint));
47 Vector E = Vector((0,0), mag=3*E_mag, dir=0,
48 L=Label("$F_E$", position=EndPoint, align=E));
57 Because the system is in equilibrium, the net force on the sphere must
60 0 &= \sum F_y = T\cos(\theta) - mg \\
61 T &= \frac{mg}{\cos(\theta)} \\
62 0 &= \sum F_x = \frac{kq^2}{[2L\sin(\theta)]^2} - T\sin(\theta) \\
63 \frac{kq^2}{[2L\sin(\theta)]^2} &= T\sin(\theta)
64 = \frac{mg}{\cos(\theta)}\sin(\theta) = mg\tan(\theta) \\
65 [2L\sin(\theta)]^2 &= \frac{kq^2}{mg\tan(\theta)} \\
66 2L\sin(\theta) &= \sqrt{\frac{kq^2}{mg\tan(\theta)}}
67 = q\sqrt{\frac{k}{mg\tan(\theta)}} \\
68 L &= \frac{q}{2\sin(\theta)}\sqrt{\frac{k}{mg\tan(\theta)}}