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+### Statement
+
+$6.1.9.$ [Insert the problem statement]
+
+### Solution
+
+### Statement
+
+$6.1.9.$ 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.
+
+<div align="center">
+ <img src="https://savchenkosolutions.com/6/6.1.9/statement.png" alt="For problem 6.1.9">
+</div>
+
+### 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:
+$$q = \sqrt{4\pi\varepsilon_0 mg (\sqrt{2}\ell)^2} \implies q = \ell\sqrt{8\pi\varepsilon_0 mg} \quad (7)$$
+
+#### Answer
+$$q = \ell\sqrt{8\pi\varepsilon_0 mg}$$
+
+#### Answer
+
+[Insert a concise answer or boxed result]