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Question Number 48458 by ajfour last updated on 24/Nov/18

Commented by ajfour last updated on 24/Nov/18

$$Q.48405\left(attemptingtosolve\right)$$

Answered by ajfour last updated on 24/Nov/18

$$let\theta infigurebed\theta .$$$$Then2Td\theta =E_{0}dQ$$$$\Rightarrow 2Td\theta =E_0\left(\frac{Q}{2\pi R}\right)\left(2d\theta \right)$$$$orT=\frac{E_{0}Q}{2\pi R}......\left(i\right)$$$$\underset{-}{Andfor{E}_0}$$$${dE}_0=\frac{dq\cos \frac{\varphi }{2}}{4\pi {ϵ}_0{x}^2}$$$$xisdistancefromdqtocenter$$$$ofdQ;$$$$x=2R\cos \frac{\varphi }{2}$$$$E_0(duetochargefrom\theta to2\pi -\theta )$$$$=2{\int }_0^{\pi -\theta }\frac{\left(\lambda Rd\varphi \right)\left(\cos \frac{\varphi }{2}\right)}{4\pi {ϵ}_0\left(4R^2\cos {}^2\frac{\varphi }{2}\right)}$$$$=\frac{\lambda }{4\pi {ϵ}_{0}R}{\int }_0^{\frac{\pi }{2}-\frac{\theta }{2}}\left(\sec \frac{\varphi }{2}\right)d\left(\frac{\varphi }{2}\right)$$$$=\frac{\lambda }{4\pi {ϵ}_{0}R}\ln \mid \sec (\frac{\pi }{2}-\frac{\theta }{2})+\tan (\frac{\pi }{2}-\frac{\theta }{2})\mid$$$$=\frac{\lambda }{4\pi {ϵ}_{0}R}\ln \mid \frac{1}{\sin \frac{\theta }{2}}+\frac{\cos \frac{\theta }{2}}{\sin \frac{\theta }{2}}\mid$$$$=\frac{\lambda }{4\pi {ϵ}_{0}R}\ln \left(\cot \frac{\theta }{4}\right)$$$$Stillthiswillnotbehelpful;$$$$verydifficultquestion..$$

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