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 θ F=((kq^2 )/c^2 )=((kq^2 )/(a^2 +b^2 −2ab cos θ)) OC=h=((√(2a^2 +2b^2 −c^2 ))/2)=((√(a^2 +b^2 +2ab cos θ))/2) (F/(mg))=(c/(2h))=(√((a^2 +b^2 −2ab cos θ)/(a^2 +b^2 +2ab cos θ))) ((kq^2 )/(mg(a^2 +b^2 −2ab cos θ)))=(√((a^2 +b^2 −2ab cos θ)/(a^2 +b^2 +2ab cos θ))) ((kq^2 )/(mg(a^2 +b^2 )(1−((2ab cos θ)/(a^2 +b^2 )))))=(√((1−((2ab cos θ)/(a^2 +b^2 )) )/(1+((2ab cos θ)/(a^2 +b^2 ))))) with λ=((2ab cos θ)/(a^2 +b^2 )), μ=((kq^2 )/(mg(a^2 +b^2 ))) (μ/(1−λ))=(√((1−λ )/(1+λ))) ⇒λ^3 −3λ^2 +(3+μ^2 )λ+μ^2 −1=0 let λ=s+1 ⇒s^3 +μ^2 s+2μ^2 =0 ⇒s=((μ^2 ((√(1+(μ^2 /(27))))−1)))^(1/3) −((μ^2 ((√(1+(μ^2 /(27))))+1)))^(1/3) ⇒λ=1+((μ^2 ((√(1+(μ^2 /(27))))−1)))^(1/3) −((μ^2 ((√(1+(μ^2 /(27))))+1)))^(1/3) ⇒θ=cos^(−1) {((a^2 +b^2 )/(2ab))[1+((μ^2 ((√(1+(μ^2 /(27))))−1)))^(1/3) −((μ^2 ((√(1+(μ^2 /(27))))+1)))^(1/3) ]}](https://www.tinkutara.com/question/Q212578.png)
$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta \\ $$$${F}=\frac{{kq}^{\mathrm{2}} }{{c}^{\mathrm{2}} }=\frac{{kq}^{\mathrm{2}} }{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta} \\ $$$${OC}={h}=\frac{\sqrt{\mathrm{2}{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }}{\mathrm{2}}=\frac{\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}}{\mathrm{2}} \\ $$$$\frac{{F}}{{mg}}=\frac{{c}}{\mathrm{2}{h}}=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}} \\ $$$$\frac{{kq}^{\mathrm{2}} }{{mg}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta\right)}=\sqrt{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}\:\mathrm{cos}\:\theta}} \\ $$$$\frac{{kq}^{\mathrm{2}} }{{mg}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\left(\mathrm{1}−\frac{\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)}=\sqrt{\frac{\mathrm{1}−\frac{\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:}{\mathrm{1}+\frac{\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}} \\ $$$${with}\:\lambda=\frac{\mathrm{2}{ab}\:\mathrm{cos}\:\theta}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} },\:\mu=\frac{{kq}^{\mathrm{2}} }{{mg}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)} \\ $$$$\frac{\mu}{\mathrm{1}−\lambda}=\sqrt{\frac{\mathrm{1}−\lambda\:}{\mathrm{1}+\lambda}} \\ $$$$\Rightarrow\lambda^{\mathrm{3}} −\mathrm{3}\lambda^{\mathrm{2}} +\left(\mathrm{3}+\mu^{\mathrm{2}} \right)\lambda+\mu^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$${let}\:\lambda={s}+\mathrm{1} \\ $$$$\Rightarrow{s}^{\mathrm{3}} +\mu^{\mathrm{2}} {s}+\mathrm{2}\mu^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow{s}=\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}−\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}+\mathrm{1}\right)} \\ $$$$\Rightarrow\lambda=\mathrm{1}+\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}−\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}+\mathrm{1}\right)} \\ $$$$\Rightarrow\theta=\mathrm{cos}^{−\mathrm{1}} \left\{\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{\mathrm{2}{ab}}\left[\mathrm{1}+\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}−\mathrm{1}\right)}−\sqrt[{\mathrm{3}}]{\mu^{\mathrm{2}} \left(\sqrt{\mathrm{1}+\frac{\mu^{\mathrm{2}} }{\mathrm{27}}}+\mathrm{1}\right)}\right]\right\} \\ $$
Commented by ajfour last updated on 18/Oct/24
$${marvellous}!\:{i}\:{ll}\:{think}\:{of}\:{another}\:{way}. \\ $$