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CBSE Class 12-science Answered

Inner and outer space 
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Asked by alanpeter9611 | 22 Feb, 2019, 08:41: PM
answered-by-expert Expert Answer

When x is between radius of sphere r and radius of shell, i.e. x <R
we make a concentric spherical shell of radius x to apply gauss law as shown in left side figure.
 
We get from Gauss's Law, begin mathsize 12px style fraction numerator plus q over denominator epsilon subscript 0 end fraction space equals space surface integral space E. d a space equals space E open parentheses 4 πx squared close parentheses space space space end style  or  E = +q/(4πεo x2 )
when  x is outsided spherical shell,   i.e., x > R
we make a concentric spherical shell of radius x to apply gauss law as shown in right side figure.
 
We get from Gauss's Law, begin mathsize 12px style fraction numerator plus q plus Q over denominator epsilon subscript 0 end fraction space equals space surface integral space E. d a space equals space E open parentheses 4 πx squared close parentheses space space space end style  or E = (+q+Q)/(4πεo x2 )
When the conducting sphere is connected to conducting shell by wire, entire system becomes single conductor.
Inside conductor electric field E = 0 and free charges exist only at surface. Also inside spherical shell electric field is zero.
Hence when conducting sphere and shell are connected by wire, all the charges will flow towards outer surface of shell.
 
 
Answered by Thiyagarajan K | 23 Feb, 2019, 11:46: AM

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