The spring compresses by a distance $x_{0}$. This distance $x_{0}$ can be calculated from the equality of the spring force $k x_{0}$ and the electrostatic repulsive force $K\left(q^{2} / x_{0}^{2}\right)$:

$$
k x_{0}=K\left(q^{2} / x_{0}^{2}\right)
$$

where $K=9 \cdot 10^{9} \mathrm{Vm} /$ As. From this,

$$
x_{0}=\sqrt[3]{\frac{K q^{2}}{k}}=0.1 \mathrm{~m}
$$

The electrostatic forces did as much work as the energy stored in the spring and the electrostatic energy possessed by the charge $q$ at a distance $x_{0}$ from another charge $q$. The former energy is

$$
E_{r}=(1 / 2) k x_{0}^{2}=0.045 \text { joule, }
$$

the latter is

$$
E_{p}=\int_{x_{0}}^{\infty} K\left(q^{2} / x^{2}\right) d x=K\left(q^{2} / x_{0}\right)=0.090 \text { joule. }
$$

The total work done by the forces on the second ball is

$$
W=E_{r}+E_{p}=(1 / 2) k x_{0}^{2}+K\left(q^{2} / x_{0}\right)=0.135 \text { joule }
$$

Based on the work of Gajdócsi Sándor (Bácsalmás, Hunyadi J. Gymnasium, IV. grade)
A $q=1 \mu \mathrm{C}$ charged tiny sphere, hanging from a weightless insulator spring with a spring constant of $k=9 \mathrm{~N} / m$, is weightless. A similarly charged, weightless sphere slowly approaches from infinity and stops where the first sphere originally was, after it was attached to the spring. How much work was done on the second sphere by the force acting on it?