91Ó°ÊÓ

Mass of the charges $=m=3.0g=3.0×{10}^{-3}kg$

Length of the thread $=l=1.0m$

Angle$=\mathrm{θ}=20°$

Let us first draw the free body diagram of one of the charged spheres to understand the forces acting on it.

Shown below is the free body diagram of the charged sphere on the right-hand side.

Where,

$FT=$Force due to tension

$F12=$Electrostatic force due to the first sphere on the second

$Fg=$Force due to gravity or weight

By the geometry of the given figure,

$$\begin{gathered} \sin \mathrm{θ}=\frac{FTx}{FT} \\ FTx=FT×\sin \mathrm{θ} \end{gathered}$$

And,

$$\begin{gathered} \cos \mathrm{θ}=\frac{FTy}{FT} \\ FTy=FT×\cos \mathrm{θ} \end{gathered}$$

The distance between the two charged spheres can also be calculated by the geometry of the figure as:

$$\begin{gathered} r=2×\sin \mathrm{θ}×1.0 \\ r=2×\sin 20 \\ r=0.68m \end{gathered}$$

Since the system is at rest, the net force acting on the body is zero.

Let us resolve the forces in X and Y directions, and solve them separately.

Considering the vertical forces, we get,

$FTy=Fg$

By replacing with the formulas, we get,

$$\begin{gathered} FT×\cos \mathrm{θ}=mg \\ FT=\frac{mg}{\cos \mathrm{θ}} \end{gathered}$$

Now, by substituting the values, we get,

$$\begin{gathered} FT=\frac{3.0×{10}^{-3}×9.81}{\cos 20} \\ FT=0.031N \end{gathered}$$

Considering the horizontal forces, we get,

$F12=FTx$

By replacing with the formulas, we get,

$\frac{K{q}^2}{r^2}=FT×\sin \mathrm{θ}$

By rearranging the terms to get q, we get,

$$\begin{gathered} q^2=\frac{FT×\sin \mathrm{θ}×r^2}{K} \\ q=\sqrt{\frac{FT×\sin \mathrm{θ}×r^2}{K}} \end{gathered}$$

Now, by substituting the values, we get,

$$\begin{gathered} q=\sqrt{\frac{0.031×\sin 20×0.{68}^2}{9.0×{10}^9}} \\ q=7.5×{10}^{-7}C=0.75\mathrm{μ}C \end{gathered}$$

Hence, the magnitude of the charge is calculated to be$0.75\mathrm{μ}C$.