Answer to Question #211503 in Electricity and Magnetism for Rocky Valmores

Question #211503

A thin glass rod is bent into a semi circle of radius R a charge Q is uniformly distributed along the upper-half and a charge −Q is uniformly distributed along the lower-half. Find the electric field E~ at the center of the semicircle. 


1
Expert's answer
2021-06-30T10:59:59-0400

The semi circle should look like the one shown below



considering the center P of the semicircle, and taking two elements of width dθd\theta the magnitude of the field at P is;

dE=14πϵo×rdθ×Q(πr/2)r2=Q2π2ϵor2dθdE=\frac{1}{4\pi \epsilon_{o}} \times \frac{rd\theta \times Q(\pi r/2)}{r^2}=\frac{Q}{2\pi ^2\epsilon_{o}r^2} d\theta

field at x due to pair of electrons=2dEsinθ=2dEsin\theta

E=0π/22dEsinθ=2×0π/2Q2πϵor2sinθdθ=Qπ2ϵor2E=_{0}\int^{\pi /2} 2dEsin\theta=2\times _{0}\int^{\pi /2}\frac{Q}{2\pi \epsilon_{o} r^2}sin\theta d\theta=\frac{Q}{\pi ^2 \epsilon_{o}r^2}




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