A-positive-charge-of-magnigude-2-5-C-is-at-the-centre-of-an-uncharged-spherical-conducting-shell-of-inner-radius-60cm-and-an-outer-radius-of-90cm-i-find-the-charge-densities-on-thd-inner-and-outer-s

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Question Number 39792 by NECx last updated on 11/Jul/18

$$Apositivechargeofmagnigude$$$$2.5\mu Cisatthecentreofanuncharged$$$$sphericalconductingshellofinner$$$$radius60cmandanouterradius$$$$of90cm.$$$$\left(i\right)findthechargedensitiesonthd$$$$innerandoutersurfacesofthe$$$$shellandthetotalchargeoneach$$$$surface.$$$$ii)findtheelectricityfield$$$$everywhere.$$

Commented by NECx last updated on 11/Jul/18

$$pleasehelp$$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Jul/18

$$i)surfacechargedensity\left(\sigma \right)ontheinner$$$$surfaceshell{\sigma }_i=\frac{-2.5\times {10}^{-6}}{4\mathrm{\Pi }\times {\left(60\times {10}^{-2}\right)}^2}$$$$surfacechargedensity\left(\sigma \right)ontheouter$$$$surfaceshell{\sigma }_o=\frac{2.5\times {10}^{-6}}{4\mathrm{\Pi }{\left(90\times {10}^{-2}\right)}^2}$$$$electricfieldinsidetheshelliszero\dots because$$$$chargeatcentreandinducedchargedonthe$$$$innerwallofshellbalanceeachother$$$$E\times 4\mathrm{\Pi }{r}^2=\frac{2.5\mu C+(-2.5\mu C)}{{ϵ}_0}$$$$soE=0$$$$foroutsidetheshellwall..$$$$E\times 4\mathrm{\Pi }{R}^2=\frac{2.5\mu C+(-2.5\mu C)+(2.5\mu C)}{{ϵ}_0}$$$$E=\frac{2.5\times {10}^{-6}}{4\mathrm{\Pi }{ϵ}_0{\left(90\times {10}^{-2}\right)}^2}$$

Commented by NECx last updated on 11/Jul/18

$$Thisiscool.Thankyousomuch$$

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