Two identical spheres are each attached to silk threads of length L = 0.500 m and hung from a common point (Fig. P21.62). Each sphere has mass m = 8.00 g. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge q₁, and the other a different positive charge q₂; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle𝛉= 20.0° with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the given information, what can you say about the magnitudes of q₁ and q₂? Explain. (d) 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° with the vertical. Determine the original charges. (Hint: The total charge on the pair of spheres is conserved.)

The diagram shows forces on each ball

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

From the above equations get T and F:

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

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

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:

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

Apply equilibrium conditions:

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

After substitution, we get

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

Use this to find q:

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

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

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

Assume the charge lost by q’ so the gained is also q’:

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

So q₁ is

And q₂ is

$q_2=q-q\text{'}=1.0\times {10}^{-7}C$

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° with the vertical. The original charges are $q_1=4.5\times {10}^{-6}C$and $q_2=1.0\times {10}^{-7}C$.