Veja a figura a seguir:
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Da figura, temos [tex3]2\alpha + 2\theta = 2\pi \therefore \alpha + \theta = \pi \Longrightarrow \alpha = \pi - \theta [/tex3]
Distância entre Q e Q:
[tex3]x^2 = L^2 + L^2- 2L^2 \cos \alpha = 2L^2 - 2L^2 \cos(\pi -\theta ) = 2L^2 ( 1+ \cos \theta ) = 2L^2 ( 1 +2 \cos^2 \frac \theta 2 -1) = 4 L^2 \cos^2 \frac \theta 2 [/tex3]
Distância entre q e q:
[tex3]y^2 = L^2 + L^2 - 2L^2 \cos \theta = 2L^2 (1 - \cos \theta ) = 2L^2 ( 1- 1 + 2 \sen^2 \frac \theta 2 ) = 4L^2 \sen^2 \frac \theta 2 [/tex3]
Equilíbrio de Q:
[tex3]2 \frac{kqQ}{L^2} \cos \frac \theta 2 + \frac{kQ^2}{4L^2 \cos^2 (\theta /2)} = 2T \cos \frac \theta 2 \therefore 2T = \frac{2kqQ}{L^2} + \frac{kQ^2}{4L^2 \cos^3 (\theta /2) }[/tex3]
(i)
Equilíbrio de q:
[tex3]2 \frac{kqQ}{L^2} \cos \frac{\alpha}{2} + \frac{kq^2}{4L^2 \sen^2 (\theta /2) } =2T \cos \frac{\alpha }{2} \\ 2T = \frac{2kQq}{L^2} + \frac{kq^2}{4L^2 \sen^2 ( \theta /2) \cos( \alpha /2) }[/tex3]
Como [tex3]\alpha = \pi - \theta \therefore \frac \alpha 2 = \frac \pi 2 - \frac \theta 2 \Longrightarrow \sen \frac \alpha 2 = \sen \left( \frac \pi 2 - \frac \theta 2 \right) = \cos \frac \theta 2 [/tex3]
Portanto, a última expressão fica [tex3]2T= \frac{2kqQ} {L^2}+ \frac{kq^2}{4L^2 \sen^3 ( \theta/2)}[/tex3]
(ii)
Igualando (i) e (ii), e simplificando
[tex3]\frac{Q^2}{\cos^3 (\theta /2)} = \frac{q^2}{\sen^3 ( \theta /2)} \Longrightarrow \tg^3 (\theta /2) = \left( \frac q Q \right)^2 = \frac 1 {64} \Longrightarrow \tg (\theta/2) = \frac 1 4 \Longrightarrow \theta = 2 \arctg\left( \frac 1 4 \right)[/tex3]
Portanto,
[tex3]\boxed{\boxed{\theta = 2 \arctg\left( \frac 1 4 \right)}}[/tex3]