A thin glass rod is bent into a semicircle of radius r. A charge $$+Q$$ is uniformly distributed along the upper half, and a charge $$-Q$$ is uniformly distributed along the lower hall as shown in figure. The electric field E at P, the center of the semicircle, is

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Infinity Learn · Accepted Answer

Take PO as the $$x-axis$$ and PA as the $$y-axis$$. Consider two elements EF and E'F' of width dθ at angular distance θ above and below PO, respectively. The magnitude of the field at P due to either element $$isdE=14$$πε$$0rd$$θ×$$Q/($$π$$r/2)r2=Q2$$$$π$$$$2$$ε$$0r2d$$θResolving the fields, we find that the components along PO sum up to zero, and hence the resultant field is along PB. Therefore, field at P due to pair of elements is $$2d$$ E $$sin$$θ$$E=$$∫$$0$$π$$/2$$ $$2dEsin$$⁡θ$$=2$$∫$$0$$π$$/2$$ $$Q2$$πε$$0r2sin$$⁡θdθ$$=Q$$$$π$$$$2$$ε$$0r2$$

A thin glass rod is bent into a semicircle of radius r. A charge +Q is uniformly distributed along the upper half, and a charge -Q is uniformly distributed along the lower hall as shown in figure. The electric field E at P, the center of the semicircle, is

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