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Question Number 212514 by ajfour last updated on 16/Oct/24

Answered by mr W last updated on 18/Oct/24

Commented by mr W last updated on 17/Oct/24

$$c^2=a^2+b^2-2ab\cos \theta$$$$F=\frac{{kq}^2}{c^2}=\frac{{kq}^2}{a^2+b^2-2ab\cos \theta }$$$$OC=h=\frac{\sqrt{2a^2+2b^2-c^2}}{2}=\frac{\sqrt{a^2+b^2+2ab\cos \theta }}{2}$$$$\frac{F}{mg}=\frac{c}{2h}=\sqrt{\frac{a^2+b^2-2ab\cos \theta }{a^2+b^2+2ab\cos \theta }}$$$$\frac{{kq}^2}{mg(a^2+b^2-2ab\cos \theta )}=\sqrt{\frac{a^2+b^2-2ab\cos \theta }{a^2+b^2+2ab\cos \theta }}$$$$\frac{{kq}^2}{mg(a^2+b^2)(1-\frac{2ab\cos \theta }{a^2+b^2})}=\sqrt{\frac{1-\frac{2ab\cos \theta }{a^2+b^2}}{1+\frac{2ab\cos \theta }{a^2+b^2}}}$$$$with\lambda =\frac{2ab\cos \theta }{a^2+b^2},\mu =\frac{{kq}^2}{mg(a^2+b^2)}$$$$\frac{\mu }{1-\lambda }=\sqrt{\frac{1-\lambda }{1+\lambda }}$$$$\Rightarrow {\lambda }^3-3{\lambda }^2+(3+{\mu }^2)\lambda +{\mu }^2-1=0$$$$let\lambda =s+1$$$$\Rightarrow s^3+{\mu }^{2}s+2{\mu }^2=0$$$$\Rightarrow s=\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}-1)}-\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}+1)}$$$$\Rightarrow \lambda =1+\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}-1)}-\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}+1)}$$$$\Rightarrow \theta ={\cos }^{-1}\left\{\frac{a^2+b^2}{2ab}[1+\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}-1)}-\sqrt[3]{{\mu }^2(\sqrt{1+\frac{{\mu }^2}{27}}+1)}]\right\}$$

Commented by ajfour last updated on 18/Oct/24

$$marvellous!illthinkofanotherway.$$

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