CBSE Class 12-science Answered
Inner and outer space
Asked by alanpeter9611 | 22 Feb, 2019, 08:41: PM
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, 
or E = +q/(4πεo x2 )
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, 
or E = (+q+Q)/(4πεo x2 )
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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