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Please give solution of the given figure.  
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Asked by anshuman.anshuman090 | 12 May, 2019, 07:17: PM
answered-by-expert Expert Answer
 
(1) Electric field E(r) = 0 ,    r < a ;   Electric Field is zero inside spherical shell
 
(2) Electric field E(r) = 0 ,    a < r < b ;   electric Field is zero inside Conductor
 
(3) Electric field     E(r) =  begin mathsize 12px style fraction numerator plus 2 q over denominator 4 πε subscript 0 space straight r squared end fraction end style ,    b < r < c ;  Electric field is due to net charge on inner sphere.
                                                                           Outer spherical shell will not give electric field in this radial position.
 
(4) Electric field     E(r) = 0 ,    c < r< d ;   electric Field is zero inside Conductor
 
(5) Electric field     E(r) =  begin mathsize 12px style fraction numerator plus 6 q over denominator 4 πε subscript 0 space straight r squared end fraction end style ,    r >d  ;  Electric field is due to net charge on both sphere.
 
Graph of electrci field as a function of radial distance is shown in figure.
 
(a) charge on inner surface of small shell is zero
(b) charge on outer surface of small shell is +2q
(c) charge on inner surface of large shell is -2q  (induced charges due to charge on small shell )
(d) charge on outer surface of large shell is +6q ( net charge on large shell is +4q)
Answered by Thiyagarajan K | 12 May, 2019, 09:39: PM

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