Question and Answers Forum

All Questions      Topic List

Electrostatics Questions

Previous in All Question      Next in All Question      

Previous in Electrostatics      Next in Electrostatics      

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

Commented by ajfour last updated on 24/Nov/18

Q.48405 (attempting to solve)

$${Q}.\mathrm{48405}\:\:\:\:\:\left({attempting}\:{to}\:{solve}\right) \\ $$

Answered by ajfour last updated on 24/Nov/18

let θ in figure be dθ. Then 2Tdθ = E_0 dQ ⇒ 2Tdθ = E_0 ((Q/(2πR)))(2dθ) or T = ((E_0 Q)/(2πR)) ......(i) And for E_0 _(−) dE_0 = ((dqcos (φ/2))/(4πε_0 x^2 )) x is distance from dq to center of dQ ; x = 2Rcos (φ/2) E_0 (due to charge from θ to 2π−θ) = 2∫_0 ^( π−θ) (((λRdφ)(cos (φ/2)))/(4πε_0 (4R^2 cos^2 (φ/2)))) = (λ/(4πε_0 R)) ∫_0 ^( (π/2)−(θ/2)) (sec (φ/2))d((φ/2)) = (λ/(4πε_0 R)) ln ∣sec ((π/2)−(θ/2))+tan ((π/2)−(θ/2))∣ = (λ/(4πε_0 R)) ln ∣(1/(sin (θ/2)))+((cos (θ/2))/(sin (θ/2)))∣ =(λ/(4πε_0 R)) ln (cot (θ/4)) Still this will not be helpful; very difficult question..

$${let}\:\theta\:{in}\:{figure}\:{be}\:{d}\theta. \\ $$$${Then}\:\:\mathrm{2}{Td}\theta\:=\:{E}_{\mathrm{0}} {dQ} \\ $$$$\Rightarrow\:\:\mathrm{2}{Td}\theta\:=\:{E}_{\mathrm{0}} \left(\frac{{Q}}{\mathrm{2}\pi{R}}\right)\left(\mathrm{2}{d}\theta\right) \\ $$$${or}\:\:\:\:{T}\:=\:\frac{{E}_{\mathrm{0}} {Q}}{\mathrm{2}\pi{R}}\:\:\:\:\:\:\:\:\:\:\:......\left({i}\right) \\ $$$$\underset{−} {{And}\:{for}\:{E}_{\mathrm{0}} } \\ $$$$\:\:\:{dE}_{\mathrm{0}} \:=\:\frac{{dq}\mathrm{cos}\:\frac{\phi}{\mathrm{2}}}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {x}^{\mathrm{2}} } \\ $$$$\:\:{x}\:{is}\:{distance}\:{from}\:{dq}\:{to}\:{center} \\ $$$${of}\:{dQ}\:; \\ $$$$\:\:\:\:\:\:\:{x}\:=\:\mathrm{2}{R}\mathrm{cos}\:\frac{\phi}{\mathrm{2}} \\ $$$$\:\:\:{E}_{\mathrm{0}} \left({due}\:{to}\:{charge}\:{from}\:\theta\:{to}\:\mathrm{2}\pi−\theta\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\int_{\mathrm{0}} ^{\:\:\pi−\theta} \:\frac{\left(\lambda{Rd}\phi\right)\left(\mathrm{cos}\:\frac{\phi}{\mathrm{2}}\right)}{\mathrm{4}\pi\epsilon_{\mathrm{0}} \left(\mathrm{4}{R}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \frac{\phi}{\mathrm{2}}\right)} \\ $$$$\:\:\:\:=\:\frac{\lambda}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {R}}\:\int_{\mathrm{0}} ^{\:\:\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}} \left(\mathrm{sec}\:\frac{\phi}{\mathrm{2}}\right){d}\left(\frac{\phi}{\mathrm{2}}\right)\: \\ $$$$\:\:\:=\:\frac{\lambda}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {R}}\:\mathrm{ln}\:\mid\mathrm{sec}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)+\mathrm{tan}\:\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)\mid \\ $$$$\:\:\:=\:\frac{\lambda}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {R}}\:\mathrm{ln}\:\mid\frac{\mathrm{1}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}+\frac{\mathrm{cos}\:\frac{\theta}{\mathrm{2}}}{\mathrm{sin}\:\frac{\theta}{\mathrm{2}}}\mid \\ $$$$\:\:=\frac{\lambda}{\mathrm{4}\pi\epsilon_{\mathrm{0}} {R}}\:\:\mathrm{ln}\:\left(\mathrm{cot}\:\frac{\theta}{\mathrm{4}}\right) \\ $$$$\:\:\:{Still}\:{this}\:{will}\:{not}\:{be}\:{helpful}; \\ $$$$\:\:\:\:\:{very}\:{difficult}\:{question}.. \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com