1 \begin{problem*}{23.50}
2 A small sphere of charge $q_1=0.800\U{$\mu$C}$ hangs from the end of a
3 spring as in Figure P23.50a. When another small sphere of charge
4 $q_2=-0.600\U{$\mu$C}$ is held beneath the first sphere as in Figure
5 P23.50b, the spring stretches by $d=3.50\U{cm}$ from its original
6 length and reaches a new equilibrium position with a separation
7 between the charges of $r=5.00\U{cm}$. What is the force constant of
16 real frame_space = 20u;
17 real initial_stretch = 10u;
21 Surface c1 = Surface(pFrom=(ceil_width,0), pTo=(-ceil_width,0));
22 Charge a1 = pCharge((0, -initial_stretch), L="$q_1$");
23 Spring s1 = Spring(pFrom=(0,0), pTo=a1.center());
24 c1.draw(); s1.draw(); a1.draw();
26 Surface c2 = Surface(pFrom=(frame_space+ceil_width,0),
27 pTo=(frame_space-ceil_width,0));
28 Charge a2 = pCharge((frame_space, -initial_stretch-dx), L="$q_1$");
29 Charge b = nCharge(a2.center() - (0, dab), L="$q_2$");
30 Spring s2 = Spring(pFrom=(frame_space,0), pTo=a2.center());
31 c2.draw(); s2.draw(); a2.draw(); b.draw();
37 The addition downward force from electric attraction is balanced by
38 the additional upward force from spring extension, so
40 \kappa d = |F_\text{spring}| &= |F_e| = k\frac{|q_1 q_2|}{r^2} \\
41 \kappa &= k\frac{|q_1 q_2|}{dr^2}
42 = \ans{49.3\U{N/m}} \;.