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Question Number 34196 by ajfour last updated on 02/May/18

Commented by ajfour last updated on 02/May/18

(i)Find u if charge q (mass m)  is just able to cross the other ring.  (ii) 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   (i).

$$\left({i}\right){Find}\:{u}\:{if}\:{charge}\:{q}\:\left({mass}\:{m}\right) \\ $$$${is}\:{just}\:{able}\:{to}\:{cross}\:{the}\:{other}\:{ring}. \\ $$$$\left({ii}\right)\:{Can}\:{we}\:{find}\:{the}\:{time}\:{that}\:{it}\:{takes} \\ $$$${to}\:{reach}\:{from}\:{centre}\:{of}\:{lower} \\ $$$${ring}\:{to}\:{centre}\:{of}\:{upper}\:{ring},\: \\ $$$${with}\:{initial}\:{speed}\:{u}\:{as}\:{found}\:{in}\: \\ $$$$\left({i}\right). \\ $$

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 (−Q)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...

$${i}\:{am}\:{trying}\:{to}\:{solve}\:{it}...{let}\:{when}\:{t}={o}\:{the}\:{charge} \\ $$$${q}\:{is}\:{at}\:{the}\:{centre}\:{of}\:{lower}\:{ring}\:{of}\:{charge}\:−{Q} \\ $$$${force}\:{by}\:{every}\:{point}\:{of}\:\left(−{Q}\right){on}\:+{q}\:{cancel}\:{each} \\ $$$${other}\:{so}\:{force}\:{by}\:{upper}\:{ring}\:{comes}\:{to}\:{play}.. \\ $$$${i}\:{have}\:{solved}\:{it}\:{pls}\:{check}\:{with}\:{answer}...{i}\:{am}\: \\ $$$${attaching}\:{in}\:{image}... \\ $$

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

$${let}\:{when}\:{t}=\mathrm{0}\:{thecharge}\:+{q}\:{at}\:{the}\:{centre}\:{of}\: \\ $$$${lower}\:{ring},{the}\:{attractive}\:{force}\:{by}\:{every}\:{point} \\ $$$${of}\:{lower}\:{ring}\:{on}\:+{q}\:{balance}\:{each}\:{other} \\ $$$${for}\:{upper}\:{ring}\: \\ $$$${the}\:{electric}\:{field}\:{at}\:{the}\:{centre}\:{of}\:{lower}\:{ring} \\ $$$${isE}=\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Q}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)} \\ $$$${its}\:{vertical}\:{components}\:{E}_{{v}} ={Ecos}\theta \\ $$$${E}_{{v}} =\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Q}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}.\frac{{d}}{\sqrt{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}\:\:}\:\:\left[{cos}\theta=\frac{{d}}{\sqrt{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)}}\right] \\ $$$${net}\:{retardation}\:\left({g}+{E}_{{v}} \right) \\ $$$${let}\:{the}\:{final}\:{velocity}\:{of}\:{charge}\:{q}=\mathrm{0} \\ $$$${as}\:{it}\:{just}\:{able}\:{to}\:{cross}\:{the}\:{centre}\:{of}\:{upper}\:{ring} \\ $$$${so} \\ $$$$\mathrm{0}^{\mathrm{2}} ={u}^{\mathrm{2}} −\mathrm{2}{ad} \\ $$$${u}=\sqrt{\mathrm{2}{ad}}\:\:\:{where}\:{a}=\left({g}+{E}_{{v}} \right)\downarrow \\ $$$$\mathrm{0}={u}−{at} \\ $$$${t}={u}/{a} \\ $$$${t}=\frac{{u}}{{g}+{E}_{{v}} } \\ $$$${u}=\sqrt{\mathrm{2}\left\{{g}+\frac{\mathrm{1}}{\mathrm{4}\Pi\varepsilon_{\mathrm{0}} }\frac{{Qd}}{\left({R}^{\mathrm{2}} +{d}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} }\right\}.{d}} \\ $$$${pls}\:{check}\:{with}\:{answer} \\ $$$$ \\ $$

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 !

$${as}\:{the}\:{charge}\:{proeeeds}\:{above} \\ $$$${forces}\:{from}\:{both}\:{rings}\:{change}; \\ $$$${acceleration}\:{is}\:{not}\:{constant} \\ $$$${as}\:{you}\:{have}\:{assumed}\:! \\ $$

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