InterviewSolution
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InterviewSolution
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Using Gauss's theorem derive an expression for the electric field at any point outside a charged spherical shell of radius R and of charge densitysigma C//m^(2) |
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Answer» SOLUTION :Consider a uniformaly charged THIN spherical shell of radius R and having a CHARGE Q. To FIND electric field at a point outside the shell situated at a distance (` r gt R ) `from the centre of shell, consider a sphere of radius r as the Gaussian surface. All points on this surface are equivalent relative to given charged sheel and,thus electric ` oversetto E ` at all points of Gaussian surface has same magnitude `oversetto Eand hat n ` are parallel to each other. ` therefore ` Total electric flux over the Gaussian surface ` phi _in = int oversetto Ehat n ds=E . 4 pi r^(2) "" ...(i) ` According to Gauss.s theorem ` phi _in =(1)/( in _0)` ( charged enclosed) ` = (Q)/( in _0) ""...(II) ` Comparing (i) and ( ii),we get ` E. 4 pi r^(2)=(Q)/( in_0)or E =(Q)/( 4 pi in _0 r^(2)) ` Thus , for anypoint outside the shell, the effect is, as if whole charge Q is concentrated at the centre of teh spherical shell. If surface charge density of shell be ` sigma ` thenQ = ` 4 pi R ^(2) sigma ` and therefore ` E= ( 4 pi R ^(2) sigma)/( 4 pi in_0 r^(2)) =( sigma R^(2))/( in _0 r^(2)) ` 
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