$6.1.9.$ Two identically charged balls of mass $m$, suspended from the same point on threads of length $\ell$, moved apart so that the angle between the threads became a right angle. Determine the charge of the balls.
Solution
In this case, 3 forces act on each ball: the tension force $\vec{T}$, directed along the thread (at an angle of $45^\circ$ to the horizontal); the gravity force $m\vec{g}$; and the Coulomb force $\vec{F}_k$.
Let us write Newton's second law in projections onto the horizontal and vertical axes: $$Ox: F_k - T \cdot \cos 45^\circ = 0 \quad (1)$$ $$Oy: T \cdot \sin 45^\circ - mg = 0 \quad (2)$$
From this, the equalities for the gravity force $m\vec{g}$ and the Coulomb force $\vec{F}_k$ follow: $$F_k = T \cdot \cos 45^\circ \quad (3)$$ $$mg = T \cdot \sin 45^\circ \quad (4)$$
Dividing one equation by the other, we obtain: $$\frac{F_k}{mg} = \cot 45^\circ = 1 \quad (5)$$
This equality can be rewritten in the form: $$\frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\frac{\ell}{\cos 45^\circ})^2} = mg \quad (6)$$
From where we express the charge of the balls: $$q = \sqrt{4\pi\varepsilon_0 mg (\sqrt{2}\ell)^2} \implies q = \ell\sqrt{8\pi\varepsilon_0 mg} \quad (7)$$
