Two small spheres, each with a mass of 20 g , are hung from a...

Let the charge on each sphere be $q$.

The forces acting on each sphere in equilibrium are the tension $T$ in the string, the weight $mg$ acting downwards, and the electrostatic force of repulsion $F_e$ acting horizontally.

Resolving the tension into horizontal and vertical components:
$T \sin \theta = F_e$
$T \cos \theta = mg$

Dividing the two equations, we get:
$\tan \theta = \dfrac{F_e}{mg}$

From the geometry of the arrangement, the distance of each sphere from the vertical axis is $\dfrac{x}{2} = \dfrac{4 \text{ cm}}{2} = 2 \text{ cm} = 0.02 \text{ m}$.
Since the angle $\theta$ is very small, $\tan \theta \approx \sin \theta = \dfrac{0.02}{0.40} = 0.05$.

The electrostatic force is given by Coulomb's law:
$F_e = \dfrac{1}{4\pi\epsilon_0} \dfrac{q^2}{x^2} = \dfrac{9 \times 10^9 \times q^2}{(0.04)^2}$

Substituting the values into the equilibrium condition:
$\dfrac{9 \times 10^9 \times q^2}{0.0016} = mg \tan \theta$

Using $m = 20 \text{ g} = 0.02 \text{ kg}$ and $g = 9.8 \text{ m/s}^2$:
$\dfrac{9 \times 10^9 \times q^2}{0.0016} = 0.02 \times 9.8 \times 0.05$

$\dfrac{9 \times 10^9 \times q^2}{0.0016} = 0.0098$

$q^2 = \dfrac{0.0098 \times 0.0016}{9 \times 10^9}$

$q^2 = \dfrac{15.68 \times 10^{-6}}{9 \times 10^9} = 1.742 \times 10^{-15} = 17.42 \times 10^{-16} \text{ C}^2$

$q = \sqrt{17.42 \times 10^{-16}} \approx 4.17 \times 10^{-8} \text{ C}$