91Ó°ÊÓ

Mass of charges $=m=5g=5×{10}^{-3}kg$

Length of the thread $=L=1m$

Amount of charge $=q=100nC=100×{10}^{-9}C$

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.

In the free body diagram,

$T=$Tension in the string

$F_e=$Electrostatic force

$F_G=$Gravitational force or weight

Let us now resolve the forces in X and Y directions and solve for those separately.

localid="1648534062675" $F_{net}=T+F_G+F_e$

Where,

$F_{net}=$net force action of the charged sphere

By solving for the forces in the X-direction, we get,

localid="1648534623325" ${\left(F_{net}\right)}_x=T_x+{\left(F_G\right)}_x+{\left(F_e\right)}_x=0$

Since the given apparatus is in equilibrium, we have equated it with 0.

Solving the above equation further, we get,

localid="1648534429906" $$\begin{gathered} T\sin \mathrm{θ}+0+K{q}^2/r^2=0 \\ T\sin \mathrm{θ}=K{q}^2/r^2 \\ T\sin \mathrm{θ}={\mathrm{Kq}}^2/(2L\sin \mathrm{θ}{)}^2.....\left(1\right) \end{gathered}$$

Where,

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

By solving for the forces in the Y-direction, we get,

${\left(F_{net}\right)}_y=T_y+{\left(F_G\right)}_y+{\left(F_e\right)}_y=0$

Since the given apparatus is in equilibrium, we have equated it with 0.

Solving the above equation further, we get,

$$\begin{gathered} T\cos \mathrm{θ}−mg+0=0 \\ T\cos \mathrm{θ}=mg.....\left(2\right) \end{gathered}$$

By dividing equation (1) with equation (2), we get,

$$\begin{gathered} \frac{T\sin \mathrm{θ}}{T\cos \mathrm{θ}}=\frac{K{q}^2/(2L\sin \mathrm{θ}{)}^2}{mg} \\ \tan \mathrm{θ}=\frac{K{q}^2}{4L{\sin }^2\mathrm{θ}×mg} \\ \tan \mathrm{θ}{\sin }^2\mathrm{θ}=\frac{K{q}^2}{4L×mg} \end{gathered}$$

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

role="math" localid="1648535105583" $$\begin{gathered} \tan \mathrm{θ}{\sin }^2\mathrm{θ}=\frac{8.9×{10}^9×{(100×{10}^{−9})}^2}{4×1^2×5×{10}^{−3}×9.81} \\ \tan \mathrm{θ}{\sin }^2\mathrm{θ}=4.587×{10}^{-4} \end{gathered}$$

It is given that the angle is very small. For smaller angles, $\sin \mathrm{θ}=\tan \mathrm{θ}$

Hence, the above equation becomes,

$$\begin{gathered} \sin \mathrm{θ}{\sin }^2\mathrm{θ}=4.587×{10}^{−4} \\ {\sin }^3\mathrm{θ}=4.587×{10}^{−4} \\ \mathrm{θ}=4.423° \end{gathered}$$

Hence, the required angle is $\mathrm{θ}=4.423°$.