Three 3.0 g balls are tied to 80-cm-long threads and hung from a single fixed point. Each of the balls is given the same charge q. At equilibrium, the three balls form an equilateral triangle in a horizontal plane with 20 cm sides. What is q?

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

Length of the thread $=l=80cm=0.8m$

The three balls make the vertices of an equilateral triangle

Side of the equilateral triangle formed is also the distance between each charge.

$r=20cm=0.2m$

Since the given charged spheres make an equilateral triangle, the field from the lower-left vertex will be at $0°$, and from the top vertex, the field will be at $300°$or $-60°$because positive charge fields point radially outward in all directions.

Also, the fields will have the same magnitude which can be given by the equation:

$E=Kq/r^2$

By substituting the values, the equation becomes,

$$\begin{gathered} E=9\times {10}^{9}q/{0.2}^2 \\ E=2.25\times {10}^{11}q \end{gathered}$$

Resolving all fields, by considering horizontal components, we get,

$E\times \cos 0+E\times \cos 300=1.5\times E$

By considering the vertical components, we get,

$E\times \sin 0+E\times \sin 300=-\sqrt{3}\times E/2$

Hence, the resultant field can be calculated as:

localid="1648646112281" $\sqrt{{(1.5E)}^2+{(-\sqrt{3}E/2)}^2}=\sqrt{3}Eat-30°$

Hence, the force on $q$at the lower right corner become $q\times \sqrt{3}\times E$

The balls have two forces,

Horizontal force $=\sqrt{3}\times E\times q$,and

Vertical force due to gravity $=mg$

Now, if $\theta$is the angle the string makes with the vertical

localid="1648647498989" $$\begin{gathered} \tan \theta =q\sqrt{3}E/mg \\ \tan \theta \times mg=q\sqrt{3}E.....\left(1\right) \end{gathered}$$

Let us first find the angle,

$$\begin{gathered} \sin \theta =\frac{oppositeside}{hypotenuse} \\ \sin \theta =\frac{oppositeside}{0.8} \end{gathered}$$

The opposite side can be calculated as:

$oppositeside=0.1/\cos 30=0.1155$

By substituting this value, we get the angle as:

$$\begin{gathered} \sin \theta =\frac{0.1155}{0.8} \\ \theta =8.3° \end{gathered}$$

By substituting this value of the angle in equation (1), we get,

localid="1648647570505" role="math" $$\begin{gathered} \tan \left(8.3\right)\times 0.003\times 9.8=q\sqrt{3}E \\ 4.288\times {10}^{-3}=q\sqrt{3}\times 2.25\times {10}^{11}q \\ q=1.05\times {10}^{-7}C=0.105\mu C \end{gathered}$$

Hence, the magnitude of the charge is$0.105\mu C$.