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| ### Statement |
| ### Statement |
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| $6.1.9.$ [Insert the problem statement] |
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| ### Solution |
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| ### Statement |
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| $6.1.9.$ 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. |
| $6.1.9.$ 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. |
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| <img src="https://savchenkosolutions.com/6/6.1.9/statement.png" alt="For problem 6.1.9"> |
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| ### Solution |
| ### Solution |
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| 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 $\vec{F}_k$. | | 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 $\vec{F}_k$. |
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| Let us write Newton's second law in projections onto the horizontal and vertical axes: | | Let us write Newton's second law in projections onto the horizontal and vertical axes: |
| $$Ox: F_k - T \cdot \cos 45^\circ = 0 \quad (1)$$ | | $$Ox: F_k - T \cdot \cos 45^\circ = 0 \quad (1)$$ |
| $$Oy: T \cdot \sin 45^\circ - mg = 0 \quad (2)$$ | | $$Oy: T \cdot \sin 45^\circ - mg = 0 \quad (2)$$ |
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| From this, the equalities for the gravity force $m\vec{g}$ and the Coulomb force $\vec{F}_k$ follow: | | From this, the equalities for the gravity force $m\vec{g}$ and the Coulomb force $\vec{F}_k$ follow: |
| $$F_k = T \cdot \cos 45^\circ \quad (3)$$ | | $$F_k = T \cdot \cos 45^\circ \quad (3)$$ |
| $$mg = T \cdot \sin 45^\circ \quad (4)$$ | | $$mg = T \cdot \sin 45^\circ \quad (4)$$ |
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| Dividing one equation by the other, we obtain: | | Dividing one equation by the other, we obtain: |
| $$\frac{F_k}{mg} = \cot 45^\circ = 1 \quad (5)$$ | | $$\frac{F_k}{mg} = \cot 45^\circ = 1 \quad (5)$$ |
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| This equality can be rewritten in the form: | | This equality can be rewritten in the form: |
| $$\frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\frac{\ell}{\cos 45^\circ})^2} = mg \quad (6)$$ | | $$\frac{1}{4\pi\varepsilon_0} \frac{q^2}{(\frac{\ell}{\cos 45^\circ})^2} = mg \quad (6)$$ |
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| From where we express the charge of the balls: | | From where we express the charge of the balls: |
| $$q = \sqrt{4\pi\varepsilon_0 mg (\sqrt{2}\ell)^2} \implies q = \ell\sqrt{8\pi\varepsilon_0 mg} \quad (7)$$ | | $$q = \sqrt{4\pi\varepsilon_0 mg (\sqrt{2}\ell)^2} \implies q = \ell\sqrt{8\pi\varepsilon_0 mg} \quad (7)$$ |
