91Ó°ÊÓ

Mass attached to the spring$=m=1.0\mathrm{g}=1.0×{10}^{-3}\mathrm{kg}$

Unstretched length of the spring $=x=4.0\mathrm{cm}$

Stretched length due to mass localid="1648464216541" $=x\text{'}=5.0\mathrm{cm}$

Stretched length due to charges on the beads $=x\text{'}\text{'}=4.5cm$

Separation between the charged beads$=r=4.5cm=0.045m$

Force due to stretching of spring is given as:

$F_k=k\mathrm{δ}x$

Where,

$\mathrm{δ}x=$change in length of the spring

$k=$spring constant

The change in the length of the spring after the addition of mass is:

$$\begin{gathered} \mathrm{δ}x=x\text{'}-x \\ \mathrm{δ}x=5.0-4.0 \\ \mathrm{δ}x=1.0cm=0.01m \end{gathered}$$

This stretching is due to the weight of the additional mass attached to the spring, hence,

$$\begin{gathered} F_k=mg \\ k\mathrm{δ}x=mg \end{gathered}$$

By rearranging the terms, we get the spring constant as:

localid="1648467381337" $k=\frac{mg}{\mathrm{δ}x}.....\left(1\right)$

Now, the electrostatic force between the beads with the same charge can be given as:

$F_e=\frac{K{q}^2}{r^2}$

Where,

$K=\frac{1}{4{\mathrm{Ï€Î\mu }}_0}=8.9×{10}^{9}N{m}^2/C^2$

For this case, the force due to the stretching of spring is given as:

$F_k=k\mathrm{δ}x\text{'}$

The change in the length of the spring due to electrostatic force between the charged beads is:

$$\begin{gathered} \mathrm{δ}x\text{'}=x\text{'}\text{'}-x \\ \mathrm{δ}x\text{'}=4.5-4.0 \\ \mathrm{δ}x\text{'}=0.5cm=0.005m \end{gathered}$$

Since this stretching is due to the electrostatic force between the charged beads, hence,

$$\begin{gathered} k\mathrm{δ}x\text{'}=F_e \\ k\mathrm{δ}x\text{'}=\frac{K{q}^2}{r^2} \end{gathered}$$

By rearranging the terms, we get the spring constant as:

$k=\frac{K{q}^2}{\mathrm{δ}x\text{'}{r}^2}.....\left(2\right)$

By comparing the equations (1) and (2), we get,

$$\begin{gathered} \frac{K{q}^2}{\mathrm{δ}x\text{'}{r}^2}=\frac{mg}{\mathrm{δ}x} \\ {q}^2=\frac{mg\mathrm{δ}x\text{'}{r}^2}{\mathrm{δ}xK} \end{gathered}$$

By substituting the values in the above equation, we get,

$$\begin{gathered} q^2=\frac{\left(1.0×{10}^{-3}\right)\left(9.8\right)(0.005)(0.045{)}^2}{(0.01)\left(8.9×{10}^9\right)} \\ {q}^2=1.11×{10}^{-15} \\ q=3.3×{10}^{-8}C=33nC \end{gathered}$$

Hence, the magnitude of the charge on each bead is 33 nC.