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The diagram shows forces on each ball

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\underset{\mathbf{}}{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{x}}\mathbf{=}\mathbf{0} \\ \mathit{T}\mathit{s}\mathit{i}\mathit{n}\left(20\right)\mathbf{=}\mathit{F} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\underset{\mathbf{}}{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{y}}\mathbf{=}\mathbf{0} \\ \mathit{T}\mathit{c}\mathit{o}\mathit{s}\left(20\right)\mathbf{=}\mathit{m}\mathit{g} \end{gathered}$$

From the above equations get T and F:

$$\begin{gathered} T=\frac{8.0脳{10}^{-3}脳9.8}{\cos \left(20\right)} \\ =8.3脳{10}^{-2}N \\ F=8.3脳{10}^{-2}\sin \left(20\right) \\ =2.8脳{10}^{-2}N \end{gathered}$$

Therefore, the magnitude of the electrostatic force that acts on each sphere is $2.8脳{10}^{-2}N$and determine the tension in each thread is$9.0脳{10}^{-2}N$.

Like charges repel each other and unlike charges attract each other.

Thus, charges on both balls should be positive to produce a repulsive force between balls and be in equilibrium with tension and weight.

After connecting the wire the charges moved from the ball with much positive charge to smaller one until reaching equilibrium and both ball are equal so the electric force between them is given by:

$F=\frac{k{q}^2}{{\left(2d\right)}^2}=\frac{k{q}^2}{{\left(2L\sin \left({胃}_2\right)\right)}^2}$

Apply equilibrium conditions:

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\underset{\mathbf{}}{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{x}}\mathbf{=}\mathbf{0} \\ \mathit{T}\mathit{s}\mathit{i}\mathit{n}\left(30\right)\mathbf{=}\mathit{F} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\underset{\mathbf{}}{\mathbf{鈭�}}{\mathit{F}}_{\mathbf{y}}\mathbf{=}\mathbf{0} \\ \mathit{T}\mathit{c}\mathit{o}\mathit{s}\left(30\right)\mathbf{=}\mathit{m}\mathit{g} \end{gathered}$$

After substitution, we get

$$\begin{gathered} T=\frac{8.0脳{10}^{-3}脳9.8}{\cos \left(30\right)} \\ =9.0脳{10}^{-2}N \\ F=8.3脳{10}^{-2}\sin \left(30\right) \\ =0.18N \end{gathered}$$

Use this to find q:

$$\begin{gathered} 0.18=\frac{9.0脳{10}^9{q}^2}{{\left(2脳0.5\sin \left(30\right)\right)}^2} \\ q=2.3脳{10}^{-6}C \end{gathered}$$

From the first case the charges q1, q2 is given by:

$$\begin{gathered} 2.8脳{10}^{-2}=\frac{k{q}_1{q}_2}{r^2}=\frac{k{q}_1{q}_2}{{\left(2L\sin \left(20\right)\right)}^2} \\ {q}_1{q}_2=3.56脳{10}^{-13}{C}^2 \end{gathered}$$

Assume the charge lost by q鈥� so the gained is also q鈥�:

$$\begin{gathered} \left(q-q\text{'}\right)\left(q+q\right)=3.56脳{10}^{-13}{C}^2 \\ {\left(2.3脳{10}^{-6}\right)}^2-q{\text{'}}^2=3.56脳{10}^{-13} \\ q\text{'}=2.2脳{10}^{-6} \end{gathered}$$

So q₁ is

And q₂ is

$q_2=q-q\text{'}=1.0脳{10}^{-7}C$

A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of 30.0掳 with the vertical. The original charges are $q_1=4.5脳{10}^{-6}C$and $q_2=1.0脳{10}^{-7}C$.