Question

TH Ch 22 MP TE Harrisburg (CH 22 HW Problem 22.59 Part A You have a lightweight spring whose unstretched length is 4.0 cm, First, you attach one end of the spring to the ceiling and hang a 3.0 g mass from it. This stretches the spring to a length of 5.3 cm You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.5 cm What is the magnitude of the charge (in nC) on each bead? nC

Answer 1

When the bead is hung from one end of spring,

Forces acting on the bead are gravitational force  $mg$ vertically downwards , and spring force $k(x-x_0)$ vertically upwards.

$mg=k(x-x_0)$

Spring constant $k=\frac{mg}{x-x_0}$ ( where $x_0$ is original length of spring, $x$ is stretched length of the spring)

$k=\frac{mg}{x-x_0}=\frac{(3.0*10^{-3})*9.81}{(0.053-0.040)}=2.264\,N/m$

Now two charges each of charge $q$ are attached to ends of spring.

Force of repulsion between the two charges is $F=\frac{q^2}{4\pi\epsilon_0 r^2}$ , this force is equal to spring force.

$\frac{q^2}{4\pi\epsilon_0 r^2}=k(r-x_0)$ ( $r$ is distance between the two charges )

$q=\sqrt{{(4\pi\epsilon_0 r^2)}k(r-x_0)}$

${q=\sqrt{{(4\pi(8.85*10^{-12}) (0.045)^2)}(2.264)(0.045-0.040)}=5.05*10^{-8}\,C=50.5\,nC}$

${q=50.5\,nC}$