Charging slinky

As we know,

The electrostatic force of repulsion between the two charged balls will be

$\frac{Kq_{1}q_{2}}{r^2}=0.9N$

And, by Hooke's Law, $F=kx$

(where F is the force, k is the spring constant, x is the displacement)

Here, F=0.9N and k(spring const.)=1000N/m

Hence $x=\frac{F}{k}$

$=\boxed{9 \times 10^{-4} m}$

$$ Let $F_c$ be the electrostatic force of interaction between two charged non-conducting spheres, $F_s$ be the elastic force of spring, and $\,x$ be the extension of spring. At the equilibrium state, by applying Newton $\,1^{st}$ Law, we obtain $$ $\begin{aligned} \sum F&=0\\ F_c-F_s&=0\\ F_s&=F_c\\ k_sx&=\frac{k_e q^2}{(0.1+x)^2}\\ 1000x&=\frac{9\times 10^9\cdot (1\times10^{-6})^2}{(0.1+x)^2}\\ 1000x(0.1+x)^2-0.009&=0 \end{aligned}$ $$ Truth be told, I use numerical approach to solve the polynomial. By using Newton-Raphson iterative method at $x_0 = 0.1$ , we will obtain the 'exact' solution $x=\boxed{0.000884291424478334\;\text{m}}$ only in 8 iteration. $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$

Suppose, at equilibrium, the spring has length $d$ . Since the spheres are non-conducting and everything is external, the spheres can be seen as point charges. Therefore, Coulomb's Law says that the force between the two charges is $\frac{kQ_1Q_2}{d^2}$ , where $Q_1$ and $Q_2$ are the two charges and $k_e$ is Coulomb's constant (the electrostatic constant). Since the charges have the same sign, they repel each other, so this force pulls on both sides of the spring.

Now, from Hooke's Law, the restoring force of the spring is $k(d-x)$ , where $k$ is the spring constant. This also acts at both ends of the spring, so this restoring force must be equal in magnitude to the repelling force on each of the spheres.

Therefore,

$\frac{k_eQ_1Q_2}{d^2}=k(d-x)$

Substituting $k_e=9\times10^9, Q_1=Q_2=1.0\times10^{-6}, k=1000, x=0.1$ and solving the cubic with a calculator, we get the final result $d=0.100884 m$ , so the sprint stretched $\boxed{0.000884 m}$ .