Question Number 34196 by ajfour last updated on 02/May/18
Commented by ajfour last updated on 02/May/18
$$\left({i}\right){Find}\:{u}\:{if}\:{charge}\:{q}\:\left({mass}\:{m}\right) \\ $$$${is}\:{just}\:{able}\:{to}\:{cross}\:{the}\:{other}\:{ring}. \\ $$$$\left({ii}\right)\:{Can}\:{we}\:{find}\:{the}\:{time}\:{that}\:{it}\:{takes} \\ $$$${to}\:{reach}\:{from}\:{centre}\:{of}\:{lower} \\ $$$${ring}\:{to}\:{centre}\:{of}\:{upper}\:{ring},\: \\ $$$${with}\:{initial}\:{speed}\:{u}\:{as}\:{found}\:{in}\: \\ $$$$\left({i}\right). \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 02/May/18
$${i}\:{am}\:{trying}\:{to}\:{solve}\:{it}…{let}\:{when}\:{t}={o}\:{the}\:{charge} \\ $$$${q}\:{is}\:{at}\:{the}\:{centre}\:{of}\:{lower}\:{ring}\:{of}\:{charge}\:−{Q} \\ $$$${force}\:{by}\:{every}\:{point}\:{of}\:\left(−{Q}\right){on}\:+{q}\:{cancel}\:{each} \\ $$$${other}\:{so}\:{force}\:{by}\:{upper}\:{ring}\:{comes}\:{to}\:{play}.. \\ $$$${i}\:{have}\:{solved}\:{it}\:{pls}\:{check}\:{with}\:{answer}…{i}\:{am}\: \\ $$$${attaching}\:{in}\:{image}… \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 02/May/18
![let when t=0 thecharge +q at the centre of lower ring,the attractive force by every point of lower ring on +q balance each other for upper ring the electric field at the centre of lower ring isE=(1/(4Πε_0 ))(Q/((R^2 +d^2 ))) its vertical components E_v =Ecosθ E_v =(1/(4Πε_0 ))(Q/((R^2 +d^2 ))).(d/( (√((R^2 +d^2 ))) )) [cosθ=(d/( (√((R^2 +d^2 )))))] net retardation (g+E_v ) let the final velocity of charge q=0 as it just able to cross the centre of upper ring so 0^2 =u^2 −2ad u=(√(2ad)) where a=(g+E_v )↓ 0=u−at t=u/a t=(u/(g+E_v )) u=(√(2{g+(1/(4Πε_0 ))((Qd)/((R^2 +d^2 )^(3/2) ))}.d)) pls check with answer](https://www.tinkutara.com/question/Q34201.png)
$${let}\:{when}\:{t}=\mathrm{0}\:{thecharge}\:+{q}\:{at}\:{the}\:{centre}\:{of}\: \\ $$$${lower}\:{ring},{the}\:{attractive}\:{force}\:{by}\:{every}\:{point} \\ $$$${of}\:{lower}\:{ring}\:{on}\:+{q}\:{balance}\:{each}\:{other} \\ $$$${for}\:{upper}\:{ring}\: \\ $$$${the}\:{electric}\:{field}\:{at}\:{the}\:{centre}\:{of}\:{lower}\:{ring} \\ $$$${isE}=\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Q}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)} \\ $$$${its}\:{vertical}\:{components}\:{E}_{{v}} ={Ecos}\theta \\ $$$${E}_{{v}} =\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Q}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}.\frac{{d}}{\:\sqrt{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}\:\:}\:\:\left[{cos}\theta=\frac{{d}}{\:\sqrt{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}\right] \\ $$$${net}\:{retardation}\:\left({g}+{E}_{{v}} \right) \\ $$$${let}\:{the}\:{final}\:{velocity}\:{of}\:{charge}\:{q}=\mathrm{0} \\ $$$${as}\:{it}\:{just}\:{able}\:{to}\:{cross}\:{the}\:{centre}\:{of}\:{upper}\:{ring} \\ $$$${so} \\ $$$$\mathrm{0}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{ad} \\ $$$${u}=\sqrt{\mathrm{2}{ad}}\:\:\:{where}\:{a}=\left({g}+{E}_{{v}} \right)\downarrow \\ $$$$\mathrm{0}={u}−{at} \\ $$$${t}={u}/{a} \\ $$$${t}=\frac{{u}}{{g}+{E}_{{v}} } \\ $$$${u}=\sqrt{\mathrm{2}\left\{{g}+\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Qd}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right\}.{d}} \\ $$$${pls}\:{check}\:{with}\:{answer} \\ $$$$ \\ $$
Commented by ajfour last updated on 02/May/18
$${as}\:{the}\:{charge}\:{proeeeds}\:{above} \\ $$$${forces}\:{from}\:{both}\:{rings}\:{change}; \\ $$$${acceleration}\:{is}\:{not}\:{constant} \\ $$$${as}\:{you}\:{have}\:{assumed}\:! \\ $$