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Using Gauss's law in electrostatics,obtainthe expressionfor electricfielddue to auniformlycharged thin sphericalshell at a point . (i) Outside the shell. (b) Inside the shell. |
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Answer» <P> Solution :Gauss.s theorem : The totaloutwardelectricflux passing througha closedsurface in air is`(1/epsilon_0)` times the totalchargeenclosedby it . . LET .+Q. be the chargeenclosed by a hollowconductorof radius .R.. Let .p. be a POINT at a distance .r. from thecentreof the conductorand .ds. be a smallelement of area surroundingthe point . A normal drawn from the pointcoincides withthe directionof `vecE`. Hence cos `0^@=1`. FromGauss.s theorem , `TOEF=phi=(1/epsilon_0)Q`...(1) by definitionf=E cos q Sds. where,`ads=4pr^2 , cos q=cos 0^@ =1` hence, f=E . `4pr^2`...(2) Comparing(1) & (2) we write`E=(1/(4piepsilon_0))Q/r^2` Fora point on the surface, r =R `E=(1/(4piepsilon_0))(Q/R^2)` This electric fieldintensityis maximum. Sinceelectric fluxdepends on the chargeenclosedand electric fieldintensity depends on the electric flux , electric fieldremainszero at allpoints insidethe sphericalhollow conductor. |
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