91Ó°ÊÓ

We have two charged particles one of them has charge $q_1$and another has a charge$q_2$with a distance$r$, the two particles exert forces on each other where the force decrease with the increase in the distance and increase with the increase in magnitude.

Coulomb's law describes:

$F_C=\frac{K{q}_1{q}_2}{r^2}$where K is the electrostatic constant and is given by $9.0×{10}^{9}N{m}^2/C^2$. The charge of the bead is given by $+20nC.$Convert unit into Coulomb by

$$\begin{gathered} q_1=(+20nC)\left(\frac{1×{10}^{-9}C}{1nC}\right) \\ {q}_1=+20×{10}^{-9}C \end{gathered}$$

now for

$$\begin{gathered} q_2=\frac{r^2{F}_C}{q_{1}K} \\ {q}_2=\frac{{(1.0×{10}^{-2}m)}^2(0.018N)}{(20×{10}^{-9}C)(9.0×{10}^{9}N{m}^2/C^2)} \\ {q}_2=10×{10}^{-9}C \end{gathered}$$

In nC, the charge of the ball $q_2$is

$$\begin{gathered} q_2=10×{10}^{-9}C\left(\frac{nC}{1×{10}^{-9}C}\right) \\ {q}_2=10nC \end{gathered}$$

The ball is above the bead and the force is downward, which means the glass bead attracts the ball downward, towards the bead, therefore, the two charges are not same. As the bead is positively charge, the charge of ball is negative.