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Question Number 50994 by peter frank last updated on 23/Dec/18

Two similar ball of mass m attached by silk thread of length a and carry similar charge q.assume θ is small enough that tanθ≈sinθ to this approximation,show that X=(((qa)/(2πε_0 mg)))^(1/3) where X is distance of separation.

$${Two}\:{similar}\:{ball}\:{of}\:{mass} \\ $$$${m}\:{attached}\:{by}\:\:{silk}\:{thread} \\ $$$${of}\:{length}\:{a}\:\:{and}\:{carry} \\ $$$${similar}\:{charge}\:\:{q}.{assume}\:\theta\:{is} \\ $$$${small}\:{enough}\:{that} \\ $$$${tan}\theta\approx{sin}\theta\:{to}\:{this} \\ $$$${approximation},{show}\: \\ $$$${that}\:\:\:\:{X}=\left(\frac{{qa}}{\mathrm{2}\pi\varepsilon_{\mathrm{0}} {mg}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \: \\ $$$${where}\:\:{X}\:{is}\:{distance}\:{of} \\ $$$${separation}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

T=tension in thread Tcosθ=mg Tsinθ=F F=repulsive force =(1/(4πε_0 ))(q^2 /x^2 ) tanθ=(F/(mg)) sinθ=(F/(mg)) (x/(2a))=(1/(4πε_0 ))×(q^2 /x^2 )×(1/(mg)) x^3 =((aq^2 )/(2πε_0 ))×(1/(mg)) x=(((aq^2 )/(2πε_0 mg)))^(1/3)

$${T}={tension}\:{in}\:{thread}\: \\ $$$${Tcos}\theta={mg} \\ $$$${Tsin}\theta={F}\: \\ $$$${F}={repulsive}\:{force}\:=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }\frac{{q}^{\mathrm{2}} }{{x}^{\mathrm{2}} } \\ $$$${tan}\theta=\frac{{F}}{{mg}} \\ $$$${sin}\theta=\frac{{F}}{{mg}} \\ $$$$\frac{{x}}{\mathrm{2}{a}}=\frac{\mathrm{1}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} }×\frac{{q}^{\mathrm{2}} }{{x}^{\mathrm{2}} }×\frac{\mathrm{1}}{{mg}} \\ $$$${x}^{\mathrm{3}} =\frac{{aq}^{\mathrm{2}} }{\mathrm{2}\pi\epsilon_{\mathrm{0}} }×\frac{\mathrm{1}}{{mg}} \\ $$$${x}=\left(\frac{{aq}^{\mathrm{2}} }{\mathrm{2}\pi\epsilon_{\mathrm{0}} {mg}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$

Answered by peter frank last updated on 23/Dec/18

F=(q^2 /(4πε_0 r^2 )) [q_1 =q_(2 ) ] sin θ=(x/(2a)) tan θ=(F/(mg)) F=mgtan θ tan θ≈sin θ=θ F=mg(x/(2a))......(i) F=(q^2 /(4πε_0 r^2 )) [ r=x].....(ii) mg(x/(2a))=(q^2 /(4πε_0 x^2 )) x=[((q^2 a)/(4πε_0 mg))]^(1/3)

$${F}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}\pi\varepsilon_{\mathrm{0}} {r}^{\mathrm{2}} }\:\:\:\left[{q}_{\mathrm{1}} ={q}_{\mathrm{2}\:\:} \right] \\ $$$$\mathrm{sin}\:\theta=\frac{{x}}{\mathrm{2}{a}} \\ $$$$\mathrm{tan}\:\theta=\frac{{F}}{{mg}} \\ $$$${F}={mg}\mathrm{tan}\:\theta \\ $$$$\mathrm{tan}\:\theta\approx\mathrm{sin}\:\theta=\theta \\ $$$${F}={mg}\frac{{x}}{\mathrm{2}{a}}......\left({i}\right) \\ $$$${F}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}\pi\varepsilon_{\mathrm{0}} {r}^{\mathrm{2}} }\:\:\left[\:\:\:{r}={x}\right].....\left({ii}\right) \\ $$$${mg}\frac{{x}}{\mathrm{2}{a}}=\frac{{q}^{\mathrm{2}} }{\mathrm{4}\pi\varepsilon_{\mathrm{0}} {x}^{\mathrm{2}} } \\ $$$${x}=\left[\frac{{q}^{\mathrm{2}} {a}}{\mathrm{4}\pi\varepsilon_{\mathrm{0}} {mg}}\right]^{\frac{\mathrm{1}}{\mathrm{3}}} \\ $$

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